Boundary element methods in aerodynamics(lecture course)

by Dmitri V. Maklakov

Contents

1 Introduction 41.1 Historical review of development of boundary element methods . . . . . . . 41.2 Examples of computations . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Governing equations for irrotational incompressible flows 82.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Field differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Gradient of a scalar field and derivative in the given direction . . . 82.2.2 Substantial (or material) derivative . . . . . . . . . . . . . . . . . . 102.2.3 Divergence of a vector field . . . . . . . . . . . . . . . . . . . . . . . 112.2.4 Vorticity (or rotor, vector value) . . . . . . . . . . . . . . . . . . . . 112.2.5 Useful formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2.6 General property . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2.7 Integral properties of the differential operator . . . . . . . . . . . . 12

2.3 Governing equations for an inviscid incompressible fluid . . . . . . . . . . . 122.4 Kelvin’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.5 Modelling viscous flows by irrotational ones . . . . . . . . . . . . . . . . . 132.6 Velocity potential of irrotational flows, Laplace’s equation . . . . . . . . . . 142.7 Bernoulli’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.8 Logic of incompressible irrotational flows . . . . . . . . . . . . . . . . . . . 162.9 Governing equation for incompressible irrotatinal flows . . . . . . . . . . . 16

3 Basic flow potentials 163.1 Two general principals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Uniform stream . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Point source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4 Point Doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.5 Combination of a doublet and uniform stream: flow over a sphere . . . . . 20

4 Integral representations of harmonic functions 214.1 Bounded domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Unbounded domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3 Ununiqueness of integral representations . . . . . . . . . . . . . . . . . . . 25

1

2

5 Basic boundary-value problems (BVP) for harmonic functions 265.1 Formulation of basic problems . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.1.1 Dirichlet’s BVP (DBVP) . . . . . . . . . . . . . . . . . . . . . . . . 265.1.2 Neumann’s BVP (NBVP) . . . . . . . . . . . . . . . . . . . . . . . 275.1.3 Mixed BVP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5.2 Uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.3 Unsolvable BVP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.4 External and internal BVP . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.5 Mathematical formulation of the problem on movement of a body in fluid . 29

6 Surface singularity distributions 306.1 Singular integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.2 Surface distributions and their jump properties . . . . . . . . . . . . . . . 32

6.2.1 Surface source distribution . . . . . . . . . . . . . . . . . . . . . . . 326.2.2 Surface doublet distribution . . . . . . . . . . . . . . . . . . . . . . 346.2.3 Connection between the surface doublet distribution and a vortex

sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

7 Numerical computation of surface source and doublet distributions 387.1 Low-order panel methods for computing surface integrals . . . . . . . . . . 387.2 Panel method for computing the surface source distribution . . . . . . . . . 39

7.2.1 Quadrilateral source . . . . . . . . . . . . . . . . . . . . . . . . . . 397.2.2 Global and local coordinate systems . . . . . . . . . . . . . . . . . . 407.2.3 Transformation from the global to local coordinate system and back 417.2.4 Algorithm of computing the source distribution . . . . . . . . . . . 437.2.5 Test of accuracy for the source distribution . . . . . . . . . . . . . . 44

7.3 Panel method for computing the surface doublet distribution . . . . . . . . 487.3.1 Quadrilateral doublet . . . . . . . . . . . . . . . . . . . . . . . . . . 487.3.2 Test for the doublet distribution . . . . . . . . . . . . . . . . . . . . 497.3.3 General conclusions for source and doublet distributions . . . . . . 50

8 Method of integral equations for solving boundary-value problems (BVP) 508.1 Source-only and doublet-only formulations . . . . . . . . . . . . . . . . . . 50

8.1.1 Summary of possible combination of singularity distributions and BC 538.2 Morino’s formulation for solving external Neumann BVP . . . . . . . . . . 538.3 Method of reducing the external Neumann BVP to the internal Dirichlet

BVP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548.4 Discretization of integral equations for 3-D case. Influence coefficients . . . 56

9 Panel method for solving aerodynamic problems 599.1 Body in the fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599.2 Panel method for the flow over 3-D wings . . . . . . . . . . . . . . . . . . . 60

9.2.1 Simplest wake model . . . . . . . . . . . . . . . . . . . . . . . . . . 639.2.2 Integral equation for 3-D wing (Morino’s formulation) . . . . . . . . 639.2.3 Computation of the pressure and lift . . . . . . . . . . . . . . . . . 65

9.3 Six main steps in applying BEM . . . . . . . . . . . . . . . . . . . . . . . . 66

3

10 BEM for 2-D airfoils 6610.1 Problem formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6610.2 2-D versions of basic singularities . . . . . . . . . . . . . . . . . . . . . . . 67

10.2.1 Point source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6710.2.2 2-D doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6810.2.3 Point vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

10.3 Source and doublet distributions . . . . . . . . . . . . . . . . . . . . . . . . 6910.3.1 Source distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 6910.3.2 Doublet distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 6910.3.3 Constant strength source distribution along a segment (x1, x2) . . . 7010.3.4 Constant strength doublet distribution along a segment (x1, x2) . . 70

10.4 Integral representation of the 2-D potential and its discrete analog . . . . . 7110.5 Outlook of different formulations . . . . . . . . . . . . . . . . . . . . . . . 71

10.5.1 Doublet-only formulation . . . . . . . . . . . . . . . . . . . . . . . . 7110.5.2 Finding influence coefficients . . . . . . . . . . . . . . . . . . . . . . 7210.5.3 Morino’s formulation for the total potential . . . . . . . . . . . . . 7310.5.4 Super Morino’s formulation . . . . . . . . . . . . . . . . . . . . . . 7310.5.5 Computation of the pressure distribution . . . . . . . . . . . . . . . 74

11 Full potential equations for irrotational isentropic steady flows 7411.1 Isentropic process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7411.2 Deducing the full potential equation . . . . . . . . . . . . . . . . . . . . . . 7511.3 Short form of the full potential equation . . . . . . . . . . . . . . . . . . . 7611.4 Goehert compressibility correction . . . . . . . . . . . . . . . . . . . . . . . 77

12 Field panel method 8012.1 Idea of the method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8012.2 Field panel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

12.2.1 Subsonic, critical and transonic flows. Quasi-isentropic flows . . . . 8212.3 Artificial viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

13 Formulae for computing induced potentials and velocities of basic two-dimensional elements 85

14 Seminar tasks (1) 86

15 Seminar tasks (2) 87

4

1 Introduction

1.1 Historical review of development of boundary element meth-

ods

At the present time the boundary element methods (BEM) are widely used in theindustry and scientific research institutes. Usually in the aerodynamics we deal with aflow over a body and outside the body the flow is governed by a system of the nonlinearNavier-Stokes differential equations. There exists a number of numerical methods forsolving these equations. The main of them are finite difference methods (FDM) and finiteelement methods (FEM). When we are applying FDM or FEM we should generate a gridin the space around the body surface. This grid must be refined in the regions wherethe flow parameters are expected to vary rapidly, and there will generally be at leastone unknown parameter for every element of the grid. As a result a complex system ofnonlinear equations will be obtained. Solving the system numerically can be a difficult taskeven for modern computers, especially in the case of the body of complex configuration(for example, a modern aircraft).

The Navier-Stokes equations take into account viscosity and compressibility effects.BEM are based on some simplifications of these equations. To be able to apply correctlythe BEM codes for solving modern aerodynamic problems the users of such codes shouldknow the backgrounds of BEM. Just these backgrounds will be presented in this lecturecourse.

In their initial development BEM deal only with incompressible, inviscous flows. At thepresent time there exist some extensions of BEM that allow us to compute compressibleflows. These extensions will be considered at the end of these lectures.

BEM call also panel methods. What is the panel? Consider, for example, a bodyshown in fig. 1.

The body surface is represented as a collection of panels. On every panel the sin-gularities (usually, sources, σi, and doublets, µi) are distributed. The strength of thesesingularities is unknown and should be chosen so, that the panel would influence on theflow as it was a part of the body surface. So, the panels are the bricks, from whichthe body surface together with the flow around is constructed. To define σi and µi theaerodynamic problem reduces to a system of linear equations. When the mathematicalunknowns σi and µi are found, physical values of interest (velocity, pressure) can be easilydetermined at any point of the flow.

The paper by Smith and Hess (1962) laid the basis for the modern panel methods for

Figure 1:

5

calculating incompressible flows over 3D configurations. At the present time there exist anumber of panel codes. For example, VSAERO, PAN AIR, HISS are widely used in theindustry.

What makes the panel methods so popular in the world? Three reasons can be indi-cated:

• simple physical models;

• simple programming;

• the power of the modern computers.

The last point allows us to solve a very great system of linear equations, therefore, bodiesof very complex configurations can be computed.

In table 1 we show some advantages of panel methods in comparing with finite dif-ference methods. From this table one can see that the number of unknowns for panel

Table 1: Comparison between the finite difference and boundary element techniques forincompressible flows

Operation Finite difference PanelsGrid generation In the domain On the surface

Unknowns Values in nodes Panel singularitiesNumber of unknowns N3 N2

Order of the system N6 N4

Equation in domain Approx. ExactlyBoundary conditions Approx. Approx.

methods is by one order less than for FDM. This is because the initial 3D problem be-comes two-dimensional, we need to define only some values on the surface of the body,but not in the space. The second great advantage is that for panel methods the governingequation in the domain will be satisfied exactly.

6

1

8

5

3

83

33

3 8

4

98

8

2

4

8

1

63

411

7

9 5

1

44

5 94

5

7

3

3

6

3

5

10

2

7

4

6

6788

4

5

3

7

41

4

79

4 5 7

7

710

48

10

6 7

66

5

4

2

3

3

1

64

71

7

3

7

1

2

3

11

3

101191011 10104

6

101

7

6

6

35 5

115

5

265 9

6

1118

10

5

11

10

2118

111161

55

45362 23

2

10

4

5

Levelcplocal:

1-0.5

2-0.4

3-0.3

4-0.2

5-0.1

60

70.1

80.2

90.3

100.4

110.5

Mustang - ROVLM Berechnungsergebnisse bei 0,32 Grad AOA(mit Prandtl-Glauert Kompressibilitaetskorrektur, Maoo = 0.5)

Figure 2: Example. Panel representation of NNA P-51 ”Mustang” (american fighter ofthe Second World War) by David Lednicer, USA.The number of the panels N = 5000.Pressure distribution by Lutz Gebhardt. (Ma∞ = 0.5).

6

1

3

2

4

5

Diskretisierung RAE Versuchskoerper WA B2 (0) 0mit angepassten Netzen im Fluegelanschlussbereich

Figure 3: Example. Panel representation of the model RAEWAB2(0)(0), the number ofpanel N = 3384.

7

X

Y

Z

Figure 4: Example. Wake roll up on the model RAEWAB2(0)(0), computed by ROVLM.

x/L

c p

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

ROVLM, AOA 0 deg., φ= +/- 30 deg.ROVLM, AOA 2 deg., φ= +30 deg.ROVLM, AOA 2 deg., φ= -30 deg.Exp., AOA 2 deg., +30 deg.Exp., AOA 2 deg., -30 deg.Exp., AOA 0 deg., +/-30 deg.

Versuchskörper RAE WA B2 (0) 0

Druckverteilung in RumpflängsrichtungSchnitt bei φ= ± 30 Grad

xcen

c p

0 0.25 0.5 0.75 1

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Exp., η=0.400, Lower Surface, Ma=0.4, AOA=2 deg.Exp., η=0.400, Upper Surface, Ma=0.4, AOA=2 deg.Exp., η=0.400, Lower/Upper Surface, Ma=0.4, AOA=0 deg.AOA 0 deg., calc., η=0.400, RIGHT WING at it = 49AOA 2 deg., calc., η=0.400, RIGHT WING at it = 49

Versuchskörper RAE WA B2 (0) 0

Flügelschnitt bei η = 0.400

Figure 5: Example. Comparison with experiments by Treadgold, Jones and Wilson (1979);left –Cp along the fuselage, right – Cp along a wing section. The discrepancy betweenthe theoretical and experimental pressure distributions at the end of the fuselage can beexplained by not taking into account in the computation the separation at the fuselagetail.

8

1.2 Examples of computations

All computations in these examples were carried out by Lutz Gebhardt (IAG) withROVLM code developed in IAG.

2 Governing equations for irrotational incompress-

ible flows

There are three reasons why we are starting with incompressible case.

1. Historical aspects (initially BEM were developed for incompressible, irrotationalflows).

2. Main ideas of the methods can be demonstrated here.

3. For many practical applications the flow can be really considered as incompressible.

2.1 Notations

~q is a vector of the velocity field:~q = u~i + v~j + w~k;u, v, w are the components of the velocity vector;~i,~j,~k are the vectors of the unite basis;P (x, y, z) is a point in the space;x, y, z are the coordinates of the point;u = u(x, y, z, t),v = v(x, y, z, t),w = w(x, y, z, t);t is the time;abbreviation: ~q = ~q(x, y, z, t) = ~q(P, t);p is the pressure;ρ is the density;T is the temperature;δV ,∆V are small volumes of the fluid;δt,∆t are small periods of time;∇ is the symbolic ”nabla”-operator, that is very convenient to simplify mathematicalformulae:

∇ =~i∂

∂x+~j

∂

∂y+ ~k

∂

∂z.

2.2 Field differential operators

2.2.1 Gradient of a scalar field and derivative in the given direction

Scalar fields in the fluid: the density ρ(x, y, z, t), the pressure p(x, y, z, t), temperatureT (x, y, z, t), local sound velocity a(x, y, z, t) and so on.

Let us take, for example, the field of the temperature

T = T (x, y, z) = T (P ).

9

Figure 6:

Figure 7:

Definition of the gradient-vector:

grad T =∂T

∂x~i +

∂T

∂y~j +

∂T

∂z~k.

Application of ∇ to the T gives

∇T = (~i∂

∂x+~j

∂

∂y+ ~k

∂

∂z)T =~i

∂T

∂x+~j

∂T

∂y+ ~k

∂T

∂z.

So, we can writegrad T = ∇T.

Physical meaning: the gradient vector defines the direction and magnitude of the maxi-mum rate of change of the scalar property per unit length at a given point P .

Explanation. Consider a point P in the space and let S be the surface of the sphereof small radius ε with the center at the point P (see fig. 7). On the surface of this spherethere is a point P1, where the temperature T takes its maximum values. The directionof gradT coincides with the direction of the vector

−−→PP1 as ε → 0, and the module of the

gradient is

|gradT | = limε→0

Tmax − T (P )

ε.

10

Figure 8:

Figure 9:

The gradient has an important property, namely, it is perpendicular to the surfaceT (x, y, z) = const.

Let l be an axis, that goes through the point P and ~n be the unit vector of this axis(see fig. 8). Consider on l a point P1. By definition the derivative in the direction l is

∂T

∂l= lim

P1→P

T (P1) − T (P )

PP1,

where PP1 is the distance between P and P1.Connection between the derivative in the given direction and the gradient:

∂T

∂l= (∇T · ~n) (scalar product).

2.2.2 Substantial (or material) derivative

Consider a small fluid element in the space. It moves along a pathline (see fig. 9).The pathline is the trajectory of a single fluid element in the space. So, different points ofthe pathlines are plotted for different moments of time. Do not confuse the pathlines andstreamlines. The streamline consists of many fluid elements at a fixed moment of timeand for any point of the streamline the tangential direction coincides with the directionof the velocity vector ~q. Pathlines and streamlines coincide only for steady flows.

Definition of the substantial derivative: substantial derivative is the time rate ofchange of a property along the pathline. For instance,for the moment t1 the temperature of the fluid element is T1,for the moment t2 the temperature of the fluid element is T2,

11

for the moment t3 the temperature of the fluid element is T3.

So we have the function T (t) along the pathline. The substantial derivative is theordinary derivative of this function:

DT

Dt= T ′(t).

Formula for calculation of the substantial derivative:

DT

Dt=

∂T

∂t+ (~q · ∇T )

or symbolicallyD

Dt=

∂

∂t+ (~q · ∇).

The second term in these formulae calls the convective derivative.

2.2.3 Divergence of a vector field

Definition:

div ~q =∂u

∂x+

∂v

∂y+

∂w

∂z= (∇ · ~q) (scalar value).

Physical meaning. Let δV be the volume of a small fluid element. Divergence is thetime rate of change of the volume δV along the pathline per unit of the volume:

div ~q =1

δV

D(δV )

Dt.

Why we may have the change of the volume δV ? Only due to compressibility. Atdifferent moments of time the volume will be under different pressures and can be com-pressed or enlarged. If the flow is incompressible, then

D(δV )

Dt= 0 =⇒ div ~q = 0.

So, we have obtained the continuity equation for incompressible flows:

∂u

∂x+

∂v

∂y+

∂w

∂z= 0.

Another physical sense of the divergence consists in imagining any velocity field ~q asthat of incompressible fluid. In this case div ~q is the strength of a source located at thepoint P , i.e. div ~qδV is the volume quantity of fluid that appears in the small volume δVper unit of time.

2.2.4 Vorticity (or rotor, vector value)

Definition:

rot ~q =~i

(∂w

∂y− ∂v

∂z

)−~j

(∂w

∂x− ∂u

∂z

)+ ~k

(∂v

∂x− ∂u

∂y

)

12

or

rot ~q = (∇× ~q) =

∣∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

u v w

∣∣∣∣∣∣∣(vector product).

Physical meaning. Vorticity defines the angular velocity ~ω = 12rot ~q of an infinitesi-

mally small fluid element.

2.2.5 Useful formulae

div (rot ~q) = 0, rot (∇T ) = 0.

2.2.6 General property

Since all the introduced above differential operators have physical meaning, their valuesare independent of the choice of coordinate system.

2.2.7 Integral properties of the differential operator

Figure 10: Closed volume.Figure 11: Unclosed surface S boundedby a closed curve C.

Divergence theorem:∫ ∫

S

(~q · ~n)dS =∫ ∫ ∫

V

div ~qdV,

Stokes’ theorem:∫ ∫

S

(rot ~q · ~n)dS =∮

C

(~q · ~dl).

2.3 Governing equations for an inviscid incompressible fluid

Continuity equation:div ~q = 0.

Momentum equations:∂u

∂t+ (~q · ∇u) = −1

ρ

∂p

∂x,

∂v

∂t+ (~q · ∇v) = −1

ρ

∂p

∂y,

13

Figure 12:

∂w

∂t+ (~q · ∇w) = −1

ρ

∂p

∂z.

This form of the momentum equations, when the viscosity and gravity forces areneglected, calls Euler’s equations. In the vector form the Euler equations can be writtenas

∂~q

∂t+ ∇q2

2− (~q × rot ~q) = −1

ρ∇p.

Equation of state:ρ = Const.

Unknowns: u, v, w, p. So, we have four equations and four unknowns.

2.4 Kelvin’s theorem

Circulation:Γ =

∮

C

(~q · ~dl).

where ~dl is the directed line segment at a point on C (see fig. 12).From the continuity and momentum equations it can be deduced

Kelvin’s theorem:DΓ

Dt= 0.

This means that the circulation around a closed curve, consisting of the same fluid ele-ments, does not change in time.

Let us assume that the movement of the fluid starts from a static state.

Static fluid =⇒ ~q = 0, Γ = 0

Kelvin’s theorem =⇒ Γ = 0 at any moment t

Stokes’ theorem =⇒ ∫∫

S(rot ~q · ~n)dS =

∮

C(~q · ~dl) = Γ = 0. =⇒ rot ~q = 0

Conclusion: if a movement starts from a stagnant state, then the flow is irrotational(rot ~q = 0) for all points of the fluid at any moment of time.

2.5 Modelling viscous flows by irrotational ones

In a real fluid (for example, in air and water) viscous forces are strong only in a so-called boundary layer near the body surface. The boundary layer creates the vorticities

14

in the flow that transport along the body surface and go into the wake behind the body.Beyond the boundary layer and wake Kelvin’s theorem holds and

rot ~q = 0.

To model the real situation the boundary layer and the wake are assumed to beinfinitely thin. On the wake surface there is a jump of tangential velocity components q+

s

and q−s :q+s 6= q−s .

Under these assumptions the real situation can be modelled by an irrotatinal flow.

Figure 13: Real case.Figure 14: Imaginary case.

2.6 Velocity potential of irrotational flows, Laplace’s equation

Irrotational flow:rot ~q = (∇× ~q) = 0.

Stokes’s theorem:Γ =

∮

C

(~q · ~dl) =∫ ∫

S

(rot ~q · ~n)dS = 0.

Consider two ways C1 and C2 that joint two points P0(x0, y0, z0) and P (x, y, z) in thefluid (see fig. 15). The point P0 is fixed, the point P can vary. According to the Stokes’theorem ∫

C1

(~q · ~dl) −∫

C2

(~q · ~dl) = 0.

Therefore, the line integral ∫

C

(~q · ~dl)

is independent of choice of curve C, that joins the points P0 and P , and depends only onthe location of the point P (x, y, z).

Definition of the potential:

Φ(x, y, z) =

P (x,y,z)∫

P0

(~q · ~dl) =

P (x,y,z)∫

P0

udx + vdy + wdz.

Connection between the velocity field and the potential:

u =∂Φ

∂x, v =

∂Φ

∂y, w =

∂Φ

∂z,

15

Figure 15:

or in a short form~q = ∇Φ.

Continuity equation∂u

∂x+

∂v

∂y+

∂w

∂z= 0

leads to the following equation for the potential Φ

∂2Φ

∂x2+

∂2Φ

∂y2+

∂2Φ

∂z2= 0.

This is the Laplace equation – governing equation for irrotational incompressible flows.Symbolically:

∇2Φ = 0.

Figure 16:

Remark 2.1 Let l be an axes in the fluid, ~n be the unite vector of this axes and ql be theprojection of the velocity vector ~q = ∇Φ on l (see fig. 16). Then

ql = ∇Φ · ~n =∂Φ

∂n.

2.7 Bernoulli’s equation

Vector form of the momentum equation:

∂~q

∂t+ ∇q2

2− (~q × rot ~q) = −1

ρ∇p.

16

Irrotational flow:rot ~q = 0, ~q = ∇Φ,

∂q

∂t+ ∇q2

2+ ∇p

ρ= 0 =⇒ ∇

(∂Φ

∂t+

q2

2+

p

ρ

)= 0,

∂Φ

∂t+

q2

2+

p

ρ= C(t),

where C(t) is an arbitrary function of time.This is Bernoulli’s equation, that is correct only for irrotational incompressible flows.

For the steady case ∂Φ∂t

= 0 and the Bernoulli equation takes the form

q2

2+

p

ρ= C.

2.8 Logic of incompressible irrotational flows

1. Movement from stagnant fluid + Kelvin’s theorem =⇒ rot ~q = 0.

2. rot ~q = 0 =⇒ ~q = ∇Φ (Potential).

3. ~q = ∇Φ+ Continuity equation =⇒ ∇2Φ=0 (Laplace equation).

4. ~q = ∇Φ+ Momentum equation =⇒ ∂Φ∂t

+ q2

2+ p

ρ= F (t)=0 (Bernoulli’s equation).

2.9 Governing equation for incompressible irrotatinal flows

∇2Φ = 0,

~q = ∇Φ,

∂Φ

∂t+

q2

2+

p

ρ= C(t).

3 Basic flow potentials

3.1 Two general principals

Principle 1 If we have a function Φ(x, y, z, ) that satisfies Laplace’s equation, thisfunction can be considered as the potential of a certain irrotatinal incompressible flow.

Principle 2 (of superposition). If the functions Φ1(x, y, z) and Φ2(x, y, z) satisfyLaplace’s equation then the sum and difference of them also satisfy Laplace’s equation.

17

3.2 Uniform stream

LetΦ = u∞x + v∞y + w∞z. (3.1)

We have ∂Φ∂x

= u∞, ∂2Φ∂x2 = 0. In analogous way ∂2Φ

∂y2 = 0, ∂2Φ∂z2 = 0. Therefore,

∂2Φ

∂x2+

∂2Φ

∂y2+

∂2Φ

∂z2= 0,

and Φ satisfies Laplace’s equation.Velocity field:

~q = ∇Φ = u∞~i + v∞~j + w∞

~k. (3.2)

This is the flow, whose velocities are equal in all points. Thus, the equation (3.1) definesthe potential of an uniform stream.

Let us denote~q∞ = u∞

~i + v∞~j + w∞~k, ~r = x~i + y~j + z~k,

where r is the radius-vector of the point P (x, y, z). In vector form equation (3.1) can bewritten as

Φ = (~q · ~r), (scalar product).

3.3 Point source

Consider the function

Φ(P ) = − σ

4π

1

r, where r =

√x2 + y2 + z2,

σ is a constant, r is the distance from the point P (x, y, z) to the origin of the coordinatesystem. Let us prove that Φ satisfies the Laplaces’s equation ∇2Φ = 0. Since σ = constwe should prove that ∇2(1/r) = 0. We differentiate:

∂r

∂x=

(x2 + y2 + z2)′x2√

x2 + y2 + z2=

x

r,

∂r

∂y=

(x2 + y2 + z2)′y2√

x2 + y2 + z2=

y

r,

∂r

∂z=

(x2 + y2 + z2)′z2√

x2 + y2 + z2=

z

r,

∂

∂x

(1

r

)= − 1

r2

∂r

∂x= − x

r3,

∂

∂y

(1

r

)= − 1

r2

∂r

∂y= − y

r3,

∂

∂z

(1

r

)= − 1

r2

∂r

∂z= − z

r3,

∂2

∂x2

(1

r

)=

∂

∂x

(− x

r3

)= − 1

r3− x

∂

∂x

(1

r3

)= − 1

r3+

3x

r4

∂r

∂x= − 1

r3+

3x2

r5,

18

∂2

∂y2

(1

r

)=

∂

∂y

(− y

r3

)= − 1

r3− y

∂

∂y

(1

r3

)= − 1

r3+

3y

r4

∂r

∂y= − 1

r3+

3y2

r5,

∂2

∂z2

(1

r

)=

∂

∂z

(− z

r3

)= − 1

r3− x

∂

∂z

(1

r3

)= − 1

r3+

3z

r4

∂r

∂z= − 1

r3+

3z2

r5.

Summing the last three equations we get

∇2(

1

r

)= − 3

r3+

3(x2 + y2 + z2)

r5= − 3

r3+

3r2

r5= − 3

r3+

3

r3= 0.

So, we have proved that ∇2(1/r) = 0. Therefore Φ = − σ4π

1r

is the potential of a certainflow.

The velocity field:

~q = ∇Φ = − σ

4π∇(

1

r

)= − σ

4π

(∂

∂x

(1

r

)~i +

∂

∂y

(1

r

)~j +

∂

∂y

(1

r

)~k

),

~q =σ

4π

x~i

r3+

y~j

r3+

z~k

r3

,

~q =σ

4π

~r

r3. (3.3)

It follows from equation (3.3) that ~q is directed like the radius vector r. Therefore allstreamlines are direct lines that start from the origin (0, 0, 0) of the coordinate system.

Consider in the space a sphere SR of radius R with the center at the origin (0, 0, 0).Let us compute the flux trough the surface SR. The flux through any surface is the volumequantity of fluid that goes through the surface per unit of time.

flux =∫ ∫

SR

(~q · ~n)dS,

~q =σ

4π

~r

r3, ~n =

~r

r,

f lux =∫ ∫

SR

σ

4π

(~r

r3· ~r

r

)dS =

σ

4π

∫ ∫

SR

(~r · ~r)r4

dS =σ

4π

∫ ∫

SR

r2

r4dS =

σ

4π

∫ ∫

SR

dS

r2.

On the surface of the sphere we have r = R = const. Consequently,

flux =σ

4πR2

∫ ∫

SR

dS =σ

4πR2Ssphere =

σ

4πR2(4πR2) = σ.

So, flux = σ. Since R is an arbitrary radius, this means that at the origin there is asource of fluid. The strength of the source is σ.

If σ < 0 the flux will be negative, directions of velocity vectors take the opposite signand we will have a sink instead of source.

If a source is located at the point Q(ξ, η, ζ) then

~r(P, Q) = (x − ξ)~i + (y − η)~j + (z − ζ)~k, r(P, Q) =√

(x − ξ)2 + (y − η)2 + (z − ζ)2,

19

Φ = − σ

4π

1

r(P, Q), ~q =

σ

4π

~r(P, Q)

r3(P, Q).

Calculate the magnitude of the velocity

|~q| =

∣∣∣∣∣σ

4π

~r(P, Q)

r3(P, Q)

∣∣∣∣∣ =σ

4π

r(P, Q)

r3(P, Q)=

σ

4π

1

r2(P, Q).

As the point P (x, y, z) → Q(ξ, η, ζ), we have r(P, Q) → 0, and Φ → ∞, |~q| → ∞.Such points, where Φ → ∞ or |~q| → ∞, are called singular points or singularities.

Remark 3.1 The function r(P, Q) is a function of six variables: x, y, z, ξ, η, ζ If we haveno ambiguity we shall denote ~r(P, Q) as ~r and r(P, Q) as r. The gradient with respectto the variables x, y, z we shall denote by ∇P . The gradient with respect to the variablesξ, η, ζ we shall denote by ∇Q. Thus

∇P

(1

r

)= − ~r

r3, ∇Q

(1

r

)=

~r

r3.

3.4 Point Doublet

Consider a point Q(ξ, η, ζ), where the sink with the potential

Φ1 =σ

4π

1

r(P, Q)

is located. Let l be an axes that goes through the point Q and Q1 be a point, differentfrom Q, that lies on l. At the point Q1 let the source with the potential

Φ2 = − σ

4π

1

r(P, Q1)

be located. The distance between the points Q and Q1 let us denote by ε, so ε = QQ1.Let the strength σ of the source and sink depend on the parameter µ so, that σε = µ,i.e., σ = µ/ε. Consider Φ = Φ1 + Φ2 as ε → 0:

Φ = limε→0

(− σ

4πr(P, Q1)+

σ

4πr(P, Q)

)=

− µ

4πlimε→0

1/r(P, Q1) − 1/r(P, Q)

ε= − µ

4π

∂Q

∂l

(1

r

), r(P, Q) = r.

The symbol∂Q

∂lmeans that the derivative in the direction of the l-axes is taken with

respect to variables ξ, η, ζ :∂Q

∂l= (~n · ∇Q), where ~n is a unit vector along l. So

Φ = − µ

4π

∂Q

∂l

(1

r

)= − µ

4π(~n · ∇Q

1

r) = − µ

4π

(~n · ~r

r3

).

Thus

Φ = − µ

4π

(~n · ~r

r3

)

is the potential of the point doublet.The line l is the axes of the doublet. This line is always oriented from the sink to the

source.

20

3.5 Combination of a doublet and uniform stream: flow over a

sphere

LetΦu = (~q∞ · ~r)

be the potential of the uniform stream with the velocity ~q = ~q∞,

Φd = − µ

4π

(~n · ~r

r3

)

be the potential of the doublet located at the origin of the coordinate system. Assumethat the doublet axes is directed opposite to ~q∞. Then

~n = −~q∞q∞

and Φd =µ

4πq∞

(~q∞ · ~r)r3

.

Consider the sum

Φ = Φu + Φd = (~q∞ · ~r)(

1 +µ

4πq∞r3

).

We denoteµ

4πq∞= k

and obtain that

Φ = (~q∞ · ~r)(

1 +k

r3

). (3.4)

Our goal is to define k in such a way that the potential (3.4) would be the potential ofthe flow over a sphere of radius R with the center at the origin of the coordinate system.To do so we should calculate the velocity field

~q = ∇Φ = ∇[(~q∞ · ~r)(1 + k/r3)]. (3.5)

Check it yourself that

∇(AB) = B∇A + A∇B, ∇(~q∞ · ~r) = ~q∞, ∇(1/r3) = −3~r/r5. (3.6)

Inserting formulae (3.6) into (3.5) we get

~q = ~q∞(1 + k/r3) − 3k~r

r5(~q∞ · r).

The velocity field on the surface of the sphere should be tangential to the surface. There-fore,

~q⊥~r =⇒ (~q · ~r) = 0 at r = R,

(~r · ~q) = (~q∞ · ~r)(1 + k/r3) − 3k(~r · ~r)r5

(~q∞ · ~r),

(~r · ~q) = (~q∞ · ~r)(1 − 2k/r3) = 0 at r = R,

1 − 2k/R3 = 0, k = R3/2.

21

So we obtain the final formulae. Potential for the flow over the sphere:

Φ(P ) = (~q∞ · ~r)(

1 +R3

2r3

). (3.7)

Velocity field for the flow over the sphere:

~q(P ) = ~q∞

(1 +

r3

2R3

)− 3R3

2

~r

r5(~q∞ · ~r). (3.8)

On the surface of the sphere at r = R we have

Φ = Φu

(1 +

R3

2R3)

)=

3

2Φu (3.9)

~q =3

2~q∞ − 3

2

~q∞ ·~R

R

~r

R

But since ~n = ~R/R we come to the formula

~q =3

2[~q∞ − (~q∞ · ~n)~n] (3.10)

This formula means that to compute the velocity on the surface of the sphere we shouldtake the vector 3

2~q∞ and project it onto the plane that is tangent to the sphere surface.

If the center of the sphere is located at the point Q(ξ, η, ζ) then the formulae (3.7),(3.8) are correct, but

~r = ~r(P, Q), r = r(P, Q).

The flow over the sphere ia a good test for checking the accuracy of numerical methods,because we have here the exact analytical solution (3.7), (3.8).

4 Integral representations of harmonic functions

4.1 Bounded domains

Consider two harmonic functions Φ1 and Φ2:

∇2Φ1 = 0, ∇2Φ2 = 0.

Let us calculate

div (Φ1∇Φ2) = div

(Φ1

∂Φ2

∂x~i + Φ1

∂Φ2

∂y~j + Φ1

∂Φ2

∂z~k

)= Φ1∇2Φ2 + (∇Φ1 · ∇Φ2).

Since ∇2Φ2 = 0 we obtain that

div (Φ1∇Φ2) = (∇Φ1 · ∇Φ2). (4.1)

Changing in (4.1) the numeration of indexes we get that

div (Φ2∇Φ1) = (∇Φ2 · ∇Φ1). (4.2)

22

Figure 17:

RHS’s in (4.1) and (4.2) are equal. Subtracting the formulae from each other we have

div (Φ1∇Φ2 − Φ2∇Φ1) = 0. (4.3)

Equation (4.3) is an identity which is correct for any two harmonic functions Φ1 and Φ2.Consider a vector field

~q = Φ1∇Φ2 − Φ2∇Φ1

and let this field be defined in a domain V , bounded by a closed surface S. Let ~n be aninner unit normal vector of S. Applying the divergence theorem to this field we get

∫ ∫

S

(~q · ~nout)dS =∫ ∫ ∫

V

div ~qdV,

∫ ∫

S

[(Φ1∇Φ2 − Φ2∇Φ1) · ~nout]dS = 0,

where ~nout is an outer normal unit vector of S. Since RHS is zero, it is not importantwhich of the normal vectors we use, outer or inner, because ~nout = −~n. So we can write

∫ ∫

S

[(Φ1∇Φ2 − Φ2∇Φ1) · ~n]dS = 0. (4.4)

This equation is one of Green’s identities. We should note that the Green’s identity iscertainly correct if the functions Φ1 and Φ2 depend not on x, y, z but on ξ, η, ζ . Now weset in (4.4) that

Φ1 = Φ(ξ, η, ζ) = Φ(Q), Φ2(Q) = 1/r(P, Q).

the point P (x, y, z) is a fixed point.Let us assume that the point P lies inside the domain V . The variables ξ, η, ζ are now

the variables of integration in (4.4). Since Φ2 → ∞ as Q → P , the function Φ2(Q) in(4.4) has a singularity at the point P . To exclude the singularity we surround the pointP by a sphere of a small radius ε with a surface Sε (see fig. 17). So, the integration in(4.4) is over S + Sε. The equation (4.4) takes the form

∫ ∫

S+Sε

[(Φ(Q)∇Q

1

r(P, Q)− 1

r(P, Q)∇QΦ(Q)

)· ~n]dS = 0.

23

For sake of brevity we shall write ∇ instead of ∇Q, ~r instead of ~r(P, Q), r instead ofr(P, Q) and Φ instead of Φ(Q). Thus

∫ ∫

S+Sε

Φ

(~r

r3· ~n)

dS −∫ ∫

S+Sε

1

r(∇Φ · ~n)dS = 0.

Taking into account that (∇Φ · ~n) = ∂Φ∂n

we get

∫ ∫

S

Φ

(~r

r3· ~n)

dS −∫ ∫

S

1

r

∂Φ

∂ndS +

∫ ∫

Sε

Φ

(~r

r3· ~n)

dS

︸ ︷︷ ︸I1

−∫ ∫

Sε

1

r

∂Φ

∂ndS

︸ ︷︷ ︸I2

= 0. (4.5)

Let us denote the last two integrals by I1 and I2 correspondingly. Firstly we compute I1.Since Sε is a sphere, ~n = −~r/r and

I1 =∫ ∫

Sε

Φ

(~r

r3· ~n)

dS = −∫ ∫

Sε

Φ

(~r

r3· ~r

r

)dS = −

∫ ∫

Sε

Φ1

r2dS = − 1

ε2

∫ ∫

Sε

ΦdS.

So

I1 = − 1

ε2

∫ ∫

Sε

ΦdS.

Now let ε → 0, then Φ → Φ(P ) and

I1 = − 1

ε2

∫ ∫

Sε

Φ(P )dS = −Φ(P )1

ε2

∫ ∫

Sε

dS = −Φ(P )1

ε2Sε = −Φ(P )

1

ε24πε2 = −4πΦ(P ).

Secondly, we compute I2:

I2 =∫ ∫

Sε

1

r

∂Φ

∂ndS =

1

ε

∫ ∫

Sε

(∇Φ · ~n)dS.

Application of the divergence theorem leads to∫ ∫

Sε

(∇Φ · ~n)dS =∫ ∫ ∫

Vε

div∇ΦdV =∫ ∫ ∫

Vε

∇2ΦdV = 0,

since Φ is a harmonic function. So we have just shown that

I1 = −4πΦ(P ), I2 = 0.

We insert these values of I1 and I2 in the formula (4.5) to have

4πΦ(P ) =∫ ∫

S

Φ

(~r

r3· ~n)

dS −∫ ∫

S

∂Φ

∂n

1

rdS,

or

Φ(P ) = −∫ ∫

S

Φ

(− 1

4π

~r

r3· ~n)

dS +∫ ∫

S

∂Φ

∂n

(− 1

4πr

)dS. (4.6)

24

Let us denote

fs(P, Q) = − 1

4π

1

r(P, Q), fd(P, Q) = − 1

4π

[~r(P, Q)

r3(P, Q)· ~n(Q)

].

These functions are the potentials of the source of unite strength and the doublet of unitstrength correspondingly. Then the formula (4.6) takes the form

Φ(P ) = −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS. (4.7)

Conclusion: Any harmonic function can be represented as a surface distribution ofsources and doublets.

Remark 4.1 The kernel function fs(P, Q) and fd(P, Q) depend only on the shape of theboundary surface S.

Remark 4.2 The formula (4.7) holds true if ~n is the inner normal vector to the domainV . If ~n is the outer normal to V , then we need to change the signs in the representation(4.7):

Φ(P ) =∫ ∫

S

Φ(Q)fd(P, Q)dS −∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS.

Our representation is correct, if the point P (x, y, z) lies inside the domain V . Considernow the case when P lies on the inner part of the surface S. Then to exclude the singularityat the point P we surround it with an inner semi-sphere of radius ε. In this case

I1 = − 1

ε2Φ(P )(2πε2) = −2πΦ(p),

and the formula (4.7) takes the form

Φ(P ) = −2∫ ∫

S

Φ(Q)fd(P, Q)dS + 2∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS. (4.8)

If P lies outside V , then 1/r(P, Q) has no singularity. So we have no need to introducethe sphere Sε. Then we get

0 = −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS. (4.9)

Thus, if Φ(P ) is a harmonic function defined in the bounded domain V with the closedboundary surface S and ~n(Q) is an inner to V unit normal vector, then the followingidentities holds true

Φ(P ) = −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS P ∈ V,

Φ(P ) = − 2∫ ∫

S

Φ(Q)fd(P, Q)dS + 2∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS P ∈ S,

0 = −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS P 6∈ V + S,

(4.10)

where Φ(Q) and ∂Φ∂n

(Q) are the boundary values of the function Φ(P ) and its normalderivative on the surface S correspondingly.

25

4.2 Unbounded domains

Let V be an unbounded domain, for example, V is an exterior of a certain body. IfΦ(P ) is a harmonic function defined in the domain V with a closed boundary surface Sand ~n(Q) is an inner to V unit normal vector (this vector is outer with respect to thebody), then the following identities holds true

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS P ∈ V,

Φ(P ) = 2Φ∞(P ) − 2∫ ∫

S

Φ(Q)fd(P, Q)dS + 2∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS P ∈ S,

0 = Φ∞(P )−∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS P 6∈ V + S,

(4.11)where Φ∞(P ) is the function that defines the asymptotic behavior of Φ(P ) at infinity. Inaerodynamics Φ∞(P ) is usually the potential of an undisturbed flow. For example,

• if the undisturbed flow is in rest, then

Φ∞(P ) = 0;

• if it is uniform, thenΦ∞(P ) = u∞x + v∞y + w∞z.

4.3 Ununiqueness of integral representations

Consider, for example, unbounded domain V , S is the boundary surface of V , V1 isthe inner domain bounded by S. In the domain V we have the representation

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS, P ∈ V. (4.12)

If we denote

µ(Q) = −Φ(Q), σ(Q) =∂Φ

∂n(Q),

we can write

Φ(P ) = Φ∞(P ) +∫ ∫

S

µ(Q)fd(P, Q)dS +∫ ∫

S

σ(Q)fs(P, Q)dS, P ∈ V, (4.13)

where µ(Q) and σ(Q) are the strengths of doublets and sources distributed on the surfaceS. So, the function Φ(P ) is represented as the sum of source and doublet distributionson S.

Consider any harmonic function Φ(P ) in the inner domain V1, bounded by the samesurface S. For example, we can take Φ1 = const or Φ1 = ax+ by + cz (a, b, c are constant)and so on. For the boundary values Φ1(Q) and ∂Φ1

∂n(Q) of this function with accordance

to (4.10) we can write

Φ1(P ) = −∫ ∫

S

Φ1(Q)fd(P, Q)dS +∫ ∫

S

∂Φ1

∂n(Q)fs(P, Q)dS, P ∈ V1, (4.14)

26

0 = −∫ ∫

S

Φ1(Q)fd(P, Q)dS +∫ ∫

S

∂Φ1

∂n(Q)fs(P, Q)dS, P ∈ V. (4.15)

In the formulae (4.14) and (4.15) the functions fd(P, Q) and ∂Φ1

∂n(Q) are constructed by

means of the outer normal vector with respect to V . If the normal vector is inner, as in(4.12), then these functions change their signs. But since in the equation (4.15) the LHSis zero the formulae are true also for the inner normal vector. Thus we can consider thatfd(P, Q) and n are the same in (4.12) and (4.15). Now we subtract (4.15) from (4.12) toobtain

Φ(P ) = Φ∞(P ) +∫ ∫

S

[Φ1(Q) − Φ(Q)]fd(P, Q)dS +∫ ∫

S

[∂Φ

∂n(Q) − ∂Φ1

∂n(Q)

]fs(P, Q)dS,

(4.16)where P ∈ V . If we denote

µ1(Q) = Φ1(Q) − Φ(Q), σ1(Q) =∂Φ

∂n(Q) − ∂Φ1

∂n(Q),

the representation (4.16) takes the form

Φ(P ) = Φ∞(P ) +∫ ∫

S

µ1(Q)fd(P, Q)dS +∫ ∫

S

σ1(Q)fs(P, Q)dS, P ∈ V. (4.17)

Compare (4.13) and (4.17). For the same function Φ(P ) we have obtained two differentintegral representations with different source and doublet distributions.

Conclusion. The integral representation of a harmonic function in the form of thesum of source and doublet distributions is not unique. There exists an infinite set offunctions µ(Q) and σ(Q) that define the same Φ(P ).

5 Basic boundary-value problems (BVP) for harmonic

functions

5.1 Formulation of basic problems

Consider any domain V with a closed boundary S. Let ~n be an inner unit normalvector of S with respect to V .

5.1.1 Dirichlet’s BVP (DBVP)

Find a harmonic function Φ(P ):

∇2Φ = 0 (Laplace’s eq.)

such thatΦ = FD(Q), Q ∈ S,

where FD(Q) is a given function of surface points.

27

5.1.2 Neumann’s BVP (NBVP)

Find a harmonic function Φ(P ):

∇2Φ = 0

such that∂Φ

∂n= FN(Q), Q ∈ S,

where FN (Q) is a given function of surface points.We should note that the solution to Neumann’s BVP is determined to the extend of

an additive constant. So, if Φ is a solution to Neumann’s BVP, then Φ + C is a solutiontoo.

5.1.3 Mixed BVP

The surface is divided by two parts S = S1 + S2. Find a harmonic function Φ(P ):

∇2Φ = 0

such thatΦ = FD(Q), Q ∈ S1,

∂Φ

∂n= FN (Q), Q ∈ S2,

where FN (Q) and FN(Q) are given functions of surface points.The Dirichlet and Neumann BVP are particular cases of the mixed BVP.

5.2 Uniqueness theorem

Theorem 1 Mixed BVP has at most one solution except of the case of Dirichlet’s BVP,for which the solution is defined to the extend of an additive constant.

Proof. In the previous section we have proved that

div (Φ1∇Φ2) = (∇Φ1 · ∇Φ2), (4.1)

if Φ1 and Φ2 are harmonic functions. Let us set in this identity that

Φ1 = Φ2 = Φ⋆.

We obtaindiv (Φ⋆∇Φ⋆) = (∇Φ⋆ · ∇Φ⋆) = |∇Φ⋆|2. (5.1)

Let us assume that the mixed BVP has two solutions Φ1 and Φ2 and let

Φ⋆ = Φ1 − Φ2.

For the function Φ⋆ we will have only zero boundary conditions (BC):

Φ⋆ = 0, Q ∈ S1, (5.2)

28

∂Φ⋆

∂n= 0, Q ∈ S2. (5.3)

Now we apply the divergence theorem to the vector field ~q = Φ⋆∇Φ⋆:

∫ ∫

S

[(Φ⋆∇Φ⋆) · ~n]dS =∫ ∫ ∫

V

div (Φ⋆∇Φ⋆)dV. (5.4)

Firstly, we compute the LHS of (5.3):

LHS =∫ ∫

S

Φ⋆(∇Φ⋆ · ~n)dS =∫ ∫

S

Φ⋆ ∂Φ⋆

∂ndS =

∫ ∫

S1

Φ⋆ ∂Φ⋆

∂ndS +

∫ ∫

S2

Φ⋆ ∂Φ⋆

∂ndS = 0,

since we have only zero boundary conditions (5.2) on S. Now we compute RHS. It followsfrom (5.1) that

RHS =∫ ∫ ∫

V

|∇Φ⋆|2dV.

So, we have obtained

RHS = LHS =⇒∫ ∫ ∫

V

|∇Φ⋆|2dV = 0 =⇒ ∇Φ⋆ = 0 =⇒ Φ⋆ = const.

If in the mixed BVP we have Dirichlet’s BC, then on S1 we have

Φ⋆ = 0 =⇒ const = 0 =⇒ Φ⋆ = 0.

Therefore the problem can have only one solution.If we have only Neumann’s BC, then the previous reasoning cannot be used, Φ⋆ = const

and the solution is unique to the extend of an additive constant.

Remark 5.1 To make the solution of NBVP unique we can specify additionally the valueof Φ at a certain point P0

Φ(P0) = A, P0 ∈ V.

Remark 5.2 Our proof is correct only if the integral

∫ ∫ ∫

V

|∇Φ⋆|2dV

is bounded. But for unbounded domains this integral can tend to infinity. For unboundeddomains we need to specify some additional condition at infinity. For instance, ∇Φ → ~q∞as P → ∞. In this case

∇Φ⋆ = ∇Φ1 −∇Φ2 → 0, as P → ∞,

and usually ∫ ∫ ∫

V

|∇Φ⋆|2dV < ∞.

29

Figure 18:

5.3 Unsolvable BVP

It is easy to specify BC so, that BVP has no solution. Consider, for example, NBVPin the bounded domain:

∂Φ

∂n= FN(Q), Q ∈ S.

Let us compute the flux of the fluid trough the surface S:

flux =∫ ∫

S

(~q · ~n)dS =∫ ∫

S

(∇Φ · ~n)dS =∫ ∫

S

∂Φ

∂ndS =

∫ ∫

S

FN (Q)dS.

But the flux mast be zero, because we have no sources or sinks in the fluid. Therefore∫ ∫

S

FN(Q)dS = 0.

If the last integral is not zero the NBVP has no solution.We can ”correct” the problem by setting a sink in the volume with the strength

σ =∫ ∫

S

FN (Q)dS

(see fig. 18).

5.4 External and internal BVP

Let S be a closed surface. This surface divides the space by two parts. If we need tofind Φ in the part, that contains infinity, the BVP is called external, if we need to find Φin the another part the BVP is called internal.

5.5 Mathematical formulation of the problem on movement ofa body in fluid

Let a body moves in a fluid and transforms its shape in its movement. Let ~q = ~qB(Q, t)be a function that defines the velocity of the point Q of the body surface S at the momentof time t. We assume that the function ~q = ~qB(Q, t) is known. The undisturbed flow isthe uniform one with the potential

Φu = u∞x + v∞y + w∞z.

We need to find the potential Φ = Φ(P, t) of the flow.

30

1. Governing equation: ∇2Φ = 0.

2. BC at infinity: Φ − Φu → 0 as P → ∞,or in the equivalent form: ~q → ~q∞ as P → ∞.

3. BC on the surface of the body S.

Assume that at all points of the surface S we locate observers. From the point of viewof each observer the fluid moves with the velocity

~qobs = ~q − ~qB,

with ~q being the velocity of the fluid.Physical principle: No fluid can go through any parts Sp of the surface S.It follows from this principle that

flux = 0 trough any Sp ∈ S.

Therefore,

flux =∫ ∫

Sp

(~qobs · ~n)dS =∫ ∫

Sp

(~q − ~qB) · ~ndS = 0.

Since Sp is an arbitrary part of S, we get that

(~q − ~qB) · ~n = 0 =⇒ (~q · ~n) = (~qB · ~n) =⇒ (∇Φ · ~n) = (~qB · ~n) =⇒ ∂Φ

∂n= (~qB · ~n).

Finally∂Φ

∂n= (~qB · ~n) (Neumann’s BC).

So we have obtained an external Neumann BVP with an additional BC at infinity. Thisproblem always has a unique solution.

6 Surface singularity distributions

6.1 Singular integrals

Consider the source distribution∫ ∫

S

µ(Q)fs(P, Q)dS

or the doublet distribution ∫ ∫

S

σ(Q)fd(P, Q)dS.

The problem is how to compute these integrals when P lies on the surface S? Indeed, ifP ∈ S, then the kernel function fs(P, Q) and fd(P, Q) both tends to infinity as Q → P .The points in the domain of integration at which the integrand tends to infinity are calledsingular points or singularities. To explain the idea of computing surface integrals withsingularities consider firstly a more simple case of line integrals.

31

Case (A). Compute the integral1∫

−1

dx

|x|1/2.

The integrand here tends to infinity as x → 0. To compute the integral we remove thesegment [−ε1, ε2] (ε1 > 0, ε2 > 0) from the integration domain. So, by definition

1∫

−1

dx

|x|1/2= lim

ε1 → 0ε2 → 0

−ε1∫

−1

dx

|x|1/2+

1∫

ε2

dx

|x|1/2

= lim

ε1 → 0ε2 → 0

[4 − 2√

ε1 − 2√

ε2] = 4.

This procedure works with an arbitrary proportion between the positive values ε1 and ε2.Such integrals that can be computed by removing an infinitely small arbitrary segment,containing the singular point, are called improper.

Case (B)

1∫

−1

dx

x2= lim

ε1 → 0ε2 → 0

−ε1∫

−1

dx

x2+

1∫

ε2

dx

x2

= limε1 → 0ε2 → 0

[−2 +

1

ε1

+1

ε2

]→ ∞.

The limit does not exist. The integral has no sense. It is divergent.Case (C)

1∫

−1

dx

x= lim

ε1 → 0ε2 → 0

−ε1∫

−1

dx

x+

1∫

ε2

dx

x

= lim

ε1 → 0ε2 → 0

logε1

ε2.

By choosing different proportions between ε1 and ε2 we will obtain different values of theintegral. For example, ε1 = 2ε2, then the integral is log 2. The limit is undetermined.Such integrals, for which the cut off of an infinity small segment leads to undeterminedvalues, are called singular. But by virtue of symmetry we could expect that this integralis zero. Indeed, the right hand side of the graphic of the function y = 1/x equals exactlyto the left hand side, but with opposite sign. If we set that ε1 = ε2 = ε then

1∫

−1

dx

x= log

ε

ε= log 1 = 0,

and we have obtained the expected value. Such a value of integral obtained by a sym-metrical cut-off is called Cauchy’s principal value. The Cauchy principal value can bedenoted by a slash or by the letters v.p. (valeur principal, French) before the integral.So, by definition

v.p.

1∫

−1

dx

x= lim

ε→0

−ε∫

−1

dx

x+

1∫

ε

dx

x

.

On a surface symmetrical cut-off is obtained by removing a circle Sε of a small radius εwith the center at the point P . By definition

v.p.∫ ∫

S

f(Q)dS = limε→0

∫ ∫

S−Sε

f(Q)dS.

In table 2 we give the properties of integrals with different types of singularities.

32

Table 2:

Di-men-sion

Type Improper int. Singularint.(v.p.)

1-D Line int. r = 1rλ , 0 < λ < 1 ~r

r2

2-D Surface int. r = 1rλ , 0 < λ < 2 ~r

r3

3-D Volume int. r = 1rλ , 0 < λ < 3 ~r

r4

Properties No jumps Jumps

Figure 19:

6.2 Surface distributions and their jump properties

Consider a double-sided surface S in the space (see fig. 19). Let S+ and S− be thesides of S, ~n be the unite normal vector, oriented from S− to S+. If Φ(P ) is a functionin the space, then

Φ+(P ) = Φ(P ), P ∈ S+,

Φ−(P ) = Φ(P ), P ∈ S+.

So the function Φ(P ) can have different values on the sides S+ and S−. In this case weare saying that Φ(P ) has a jump across S.

6.2.1 Surface source distribution

Φ(P ) =∫ ∫

S

σ(Q)fs(P, Q)dS, fs(P, Q) = − 1

4πr(P, Q),

r(P, Q) =√

(x − ξ)2 + (y − η)2 + (z − ζ)2.

With accordance to table 2 without strict proof we can write

Property 6.1 Surface source distribution has no jump across S:

Φ+(P ) =∫ ∫

S

σ(Q)fs(P, Q)ds, P ∈ S+, (6.1)

Φ−(P ) =∫ ∫

S

σ(Q)fs(P, Q)ds, P ∈ S−, (6.2)

33

Figure 20:

Φ+(P ) − Φ−(P ) = 0. (6.3)

Corollary. The circulation along any contour is zero:

Γ =∫

C(~q · d~l) = Φ+ − Φ− = 0.

This means that no flow producing lift can be modelled with this distribution!Velocity field:

~q = ∇P

∫ ∫

S

σ(Q)fs(P, Q)ds =∫ ∫

S

σ(Q)∇Pfs(P, Q)ds,

∇P fs(P, Q) = ∇P

[− 1

4πr(P, Q)

]=

1

4π

~r(P, Q)

r3(P, Q).

Let us denote

~K(P, Q) =1

4π

~r(P, Q)

r3(P, Q).

Then~q =

∫ ∫

S

σ(Q) ~K(P, Q)dS.

Property 6.2 Normal component of ~q has the jump σ(Q) across S:

[~q +(P ) · ~n(P )] − [~q −(P ) · ~n(P )] = σ(P ), P ∈ S. (6.4)

Proof. Consider a small element ∆S of the surface S containing the point P . Weassume the element ∆S to be so small that it can be thought as flat and the velocities onthe upper and lower surfaces are constant (see fig. 20):

~q + = ~q +(P ), ~q − = ~q −(P )

for all points of ∆S. The volume quantity of fluid injected by the element per unit oftime is

flux = σ∆S.

The upper side of the element injects the flux

flux+ = [~q +(P ) · ~n]∆S.

34

The lower side injects the flux

flux− = [~q −(P ) · (−~n)]∆S = −[~q −(P ) · ~n]∆S.

We have here the sign minus, because in computing flux− the direction of the normalvector on the lower side is reversed. But

flux+ + flux− = flux,

[~q +(P ) · ~n]∆S − [~q −(P ) · ~n]∆S = σ∆S =⇒ [~q +(P ) · ~n] − [~q −(P ) · ~n] = σ.

Property 6.2 is proved.

Property 6.3 ((without proof))

~q +(P ) =σ(P )

2~n(P ) + v.p.

∫ ∫

S

σ(Q) ~K(P, Q)dS, (6.5)

~q −(P ) = −σ(P )

2~n(P ) + v.p.

∫ ∫

S

σ(Q) ~K(P, Q)dS. (6.6)

It follows from property 6.3 that the velocity vector induced by the source distributionhas a jump only in the direction normal to S, and no jump in the tangential direction.

6.2.2 Surface doublet distribution

Φ(P ) =∫ ∫

S

µ(Q)fd(P, Q)dS, fd(P, Q) = − 1

4π

[~n(Q) · ~r(P, Q)]

r3(P, Q).

Property 6.4 Surface doublet distribution has a jump of −µ(P ) across S. The boundaryvalue are given by the formulae:

Φ+(P ) = −µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS, (6.7)

Φ−(P ) =µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS. (6.8)

Proof. First we will proveLemma If µ(Q) = µ = const, then

Φ(P ) = −α(P )µ

4π, if P lies above S, (6.9)

Φ(P ) =α(P )µ

4π, if P lies below S (6.10)

where α(P ) is the solid angle at the vertex P of the cone with the base S.Proof of the lemma. Assume that P ia above S and locate at the point P a sink

with the strength µ (see fig. 21):

35

Figure 21:

Φsink(Q) =µ

4πr(Q, P ),

~qsink = − µ

4π

~r(Q, P )

r3(Q, P )=

µ

4π

~r(P, Q)

r3(P, Q), (6.11)

since ~r(P, Q) = −~r(Q, P ), r(P, Q) = r(Q, P ). The flux of the fluid that goes into the sinkis

flux = −α(P )µ

4π. (6.12)

On the other handflux =

∫ ∫

S

[~qsink(Q) · ~n]dS.

Using the formula (6.11) we get

flux =∫ ∫

S

[µ

4π

~r(P, Q)

r3(P, Q)· ~n]dS

Comparing this expression for flux and (6.12) we get that

∫ ∫

S

µ[~r(P, Q) · ~n(Q)]

4πr3(P, Q)︸ ︷︷ ︸

−fd(P,Q)

dS =α(P )µ

4π.

Finally ∫ ∫

S

µfd(P, Q)dS = −α(P )µ

4π, if P lies above S,

and we have proved formula (6.9). To prove (6.10) we locate at the point P a sink andrepeat the reasoning. Lemma is proved.

Now we shall prove property 6.4. Consider a small element ∆S of the surface S(see fig. 22). We assume that the element is so small that it can be thought as flat andµ(Q) = µ(P ) = const, where P is a point inside ∆S, and Q is any point of ∆S. Accordingto the lemma at the point P1 above S this element induces the potential

Φ∆S = −α(P1)µ(P )

4π.

36

Figure 22:

If P1 → P then α → 2π and

Φ∆S = −µ(P )

2(singular component).

It is worth noting that usually a small size induces a small value, but here, due to thefact that P1 → P , even infinitely small element induces a finite value of the potential.

The remaining part of S induces the potential

∫ ∫

S−∆S

µ(Q)fd(P, Q)dS.

So,

Φ+(P ) = −µ(P )

2+∫ ∫

S−∆S

µ(Q)fd(P, Q)dS.

If ∆S collapses to the point P then the surface S − ∆S tends to S and

Φ+(P ) = −µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS.

So we have proved (6.7). If P1 lies below S, then

Φ∆S =α(P1)µ(P )

4π.

Repeating previous reasoning we get

Φ−(P ) =µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS,

and the property 6.4 is proved.

6.2.3 Connection between the surface doublet distribution and a vortex sheet

Now we compute the velocity field induced by the surface doublet distribution.

~q = ∇PΦ(P ) = ∇P

∫ ∫

S

µ(Q)fd(P, Q)dS =∫ ∫

S

µ(Q)∇P fd(P, Q)dS.

For computing ~q the following property of the doublet distribution can be used.

37

Figure 23:

Property 6.5 ((without proof))

~q(P ) =1

4π

∫ ∫

S

(∇2µ × ~n) × ~r(P, Q)

r3(P, Q)dS +

1

4π

∮

C

µ(Q)

[d~l × ~r(P, Q)

r3(P, Q)

], (6.13)

where ∇2µ is the 2-D gradient (surface gradient).

If ~e1 and ~e2 are unit vectors of two perpendicular directions, tangential to the surface S,then

∇2µ =∂µ

∂e1~e1 +

∂µ

∂e2~e2.

The line C is a closed contour that bounds the surface S. The direction of integrationalong C is chosen so, that if one looks from the end of the vector ~n, this direction will becounter clock-wise (see fig. 23).

Let us denote ∇2µ × ~n = γ(Q). Then the surface integral

1

4π

∫ ∫

S

γ(Q) × ~r(P, Q)

r3(P, Q)dS

is a velocity field induced by the vortex sheet with accordance to Biot-Savar’s law: theelement of vorticity γ(Q)dS induces the velocity

γ(Q)dS × 1

4π

~r(P, Q)

r3(P, Q).

The line integral1

4π

∮

C

µ(Q)

[d~l × ~r(P, Q)

r3(P, Q)

]

represents the velocity field induced by the vortex ring C, that is the boundary of thesurface S.

So, doublet distribution is equivalent (with respect to the induced velocity) to a vortexsheet plus a vortex ring of nonconstant strength. If µ = const, then ∇2µ = 0. In thiscase doublet distribution is equivalent to a vortex ring of constant strength µ.

Property 6.6 ((without proof))

~q ±(P ) = ∓∇2µ

2+v.p.

1

4π

∫ ∫

S

(∇2µ×~n)× ~r(P, Q)

r3(P, Q)dS+

1

4π

∮

C

µ(Q)

[d~l × ~r(P, Q)

r3(P, Q)

]. (6.14)

38

Figure 24:

7 Numerical computation of surface source and dou-

blet distributions

7.1 Low-order panel methods for computing surface integrals

Let S be a surface, f(Q) be a function of surface points. We need to compute

I =∫ ∫

S

f(Q)dS.

Definition of the surface integral. Let us divide S in an arbitrary manner by Nparts (see fig. 24). Let us enumerate these parts from 1 to N . Let ∆Si be the area ofevery part and di be the diameter of every part. (Diameter of the domain is the leadingsize of the domain.) Inside every part let us choose a point Qi (also arbitrarily), computef(Qi) and construct the sum

sum =N∑

i=1

f(Qi)∆Si.

By definition

I =∫ ∫

S

f(Q)dS = limN → ∞

max di → 0

N∑

i=1

f(Qi)∆Si.

This definitions gives a direct way of computing surface integrals.

1. We need to define a number N of divisions.

2. We need to define a method of dividing the surface.

3. We need to define a method of choosing points Qi.

4. We need to define a method of computing ∆Si.

Then

I ≈N∑

i=1

f(Qi)∆Si. (7.1)

Computing the areas ∆Si can be done by the method of quadrilateral panels. A quadri-lateral is a flat quadrangle. Every part is approximated by a quadrilateral. The area of

39

quadrilateral can be computed analytically by representing it, for example, as two trian-gles. The whole approximated surface can have gapes or overlapping parts as shown onthe figure.

If we need to commute the integral, in which the integrand is a product of two func-tions, the following approximate formula is often used

∫ ∫

S

f(Q)f1(Q)dS ≈N∑

i=1

f(Qi)∫ ∫

∆Si

f1(Q)dS, (7.2)

where the integrals in the RHS are computed analytically.The formulae (7.1) and (7.2) are so-called low-order approximate formulae for com-

puting surface integrals. The main advantage of them in comparing with more truncatedformulae of higher order is their simplicity. To carry out the computation we need toknow only the values of the integrated function f(Q) at the collocation points Qi. Higherorder approximations require much more detailed information about f(Q).

7.2 Panel method for computing the surface source distribution

Φ(P ) =∫ ∫

S

σ(Q)fs(P, Q)dS, fs(P, Q) = − 1

4πr(P, Q).

By making use of (7.2) we obtain

Φ(P ) ≈N∑

i=1

σ(Qi)∫ ∫

∆Si

fs(P, Q)dS

︸ ︷︷ ︸analytically

.

If we denoteSi(P ) =

∫ ∫

∆Si

fs(P, Q)dS, (7.3)

then

Φ(P ) ≈N∑

i=1

σ(Qi)Si(P ), (7.4)

where Si(P ) is the potential, induced at the point P by the panel ∆Si with uniform sourcedistribution of unit strength.

7.2.1 Quadrilateral source

Assume that the quadrilateral S is located at the plain XOY (see fig. 25). Thiselement has a uniform source distribution of unit strength. The potential induced by theelement is

Φ(P ) = − 1

4π

∫ ∫

S

dS

r(P, Q)= − 1

4π

∫ ∫

S

dξdη√

(x − ξ)2 + (y − η)2 + z2.

This integral can be computed analytically. As a result we obtain the function of eleven

40

Figure 25:

Figure 26:

variables:Φ(P ) = Fs(x1, y1, x2, y2, x3, y3, x4, y4, x, y, z). (7.5)

There exist effective and fast programming code to compute the function Fs as well asthe partial derivatives for components of induced velocity:

u =∂Fs

∂x, v =

∂Fs

∂y, w =

∂Fs

∂z.

Properties of quadrilateral source

1. Fs is continuous with respect to x, y, z;

2. The w-component of velocity has a jump across S

w(x, y, +0) =1

2, w(x, y,−0) = −1

2, if (x, y) ∈ S.

Beyond S we have w(x, y, 0) = 0.

3. The components u and v have no jump across S, but if P tends to the edges of S,then u, v → ∞:

u ∼ log a, v ∼ log a, (7.6)

where a is the distance from the point P to the nearest edge.

7.2.2 Global and local coordinate systems

Let us consider a part ∆Si of the surface S that has not been flattened yet. Let usintroduce a local coordinate system (LCS) with basis vectors ~e1, ~e2, ~e3, so that the vector~e3 coincides with the normal: ~e3 = ~n.

41

Figure 27:

Figure 28:

Let ~r1, ~r2, ~r3, ~r4 be the angular point of ∆Si. There exist a number of ways of intro-ducing LCS and flattening the element. We describe here one of the most popular. Letus compute the diagonal vectors

~d1 = ~r2 − ~r4, ~d2 = ~r3 − ~r1.

If the panel is close approximately to a rectangle, then the vector ~d1 + ~d2 will be approx-imately parallel to the edges ~r2 − ~r1 and ~r3 − ~r4. We define

~e3 =(~d2 × ~d1)

|~d2 × ~d1|, ~e2 =

~d1 + ~d2

|~d1 + ~d2|, ~e1 = (~e2 × ~e3).

Thus, the vectors ~e1, ~e2, ~e3 are found. The origin of the local coordinate system we locateat the middle point of ∆Si:

~rc =1

4(~r1 + ~r2 + ~r3 + ~r4).

Now the local coordinate system is fully defined.

7.2.3 Transformation from the global to local coordinate system and back

If we know the components of the vectors ~e1, ~e2, ~e3, then we know the matrix

E =

e11 e12 e13

e21 e22 e23

e31 e32 e33

42

that determines these components:

~e1 = e11~i + e12

~j + e13~k,

~e1 = e11~i + e12

~j + e13~k,

~e1 = e11~i + e12

~j + e13~k.

Assume that in the global coordinate system (GCS) we have a given vector

~a = a1~i + a2

~j + a3~k,

a1, a2, a3 are the components of the vector in GCS. In the local coordinate system (LCS)this vector can be represented as

~a = a′1~e1 + a′

2~e2 + a′3~e3,

where a′1, a

′2, a

′3 are the components of ~a in LCS. The transformations from a1, a2, a3 to

a′1, a

′2, a

′3 and back are specified by the formulae

a′i =

3∑

j=1

eijaj , i = 1, 2, 3; (GCS to LCS), (7.7)

ai =3∑

j=1

ejia′j , i = 1, 2, 3; (LCS to GCS). (7.8)

Equation (7.7) means that

a′1 = e11a1 + e12a2 + e13a3,

a′2 = e21a1 + e22a2 + e23a3,

a′3 = e31a1 + e32a2 + e33a3.

(∗)

For example, we will prove the first of the relations (*). The component a′1 is the projection

of the vector ~a on the axis e1. The projection of any vector on an axis equals the scalarproduct of the vector and the unit vector of the axis:

a′1 = (a · ~e1) = (a1, a2, a3) · (e11, e12, e13) = e11a1 + e12a2 + e13a3,

and the first relation in (*) is proved. The remaining ones can be proved analogically.Equation (7.8) means that

a1 = e11a′1 + e21a

′2 + e31a

′3,

a2 = e12a′1 + e22a

′2 + e32a

′3,

a3 = e13a′1 + e23a

′2 + e33a

′3.

(∗∗)

For example, we will prove the first of the relations (**). the component a1 is the projec-tion of the vector ~a on the axis x with the unit vector ~i:

a1 = (~a ·~i) = (a′1~e1 + a′

2~e2 + a′3~e3) ·~i = a′

1 (~e1 ·~i)︸ ︷︷ ︸e11

+a′2 (~e2 ·~i)︸ ︷︷ ︸

e21

+a′3 (~e3 ·~i)︸ ︷︷ ︸

e31

,

and the first relation in (**) is proved too. The remaining ones can be proved analogically.

43

It is convenient to write the relations (7.7) and (7.8) in matrix form. Let us denote

a =

a1

a2

a3

a′ =

a′1

a′2

a′3

,

where a and a′ are so-called matrices-columns. Then

a′ = Ea,︸︷︷︸Matrixproduct

a = ET a′,︸ ︷︷ ︸Matrixproduct

where ET is the matrix transposed to E:

ET =

e11 e21 e31

e12 e22 e32

e13 e23 e33

.

So, it is easy to remember that the transformation fromGCS to LCS is done with E,LCS to GCS is done with ET .

Remark 7.1 Every element (panel) has its own matrix E.

7.2.4 Algorithm of computing the source distribution

Let P be a given point in the space. We need to compute

Φ(P ) =N∑

i=1

σ(Qi)Si(P ),

~q = ∇P Φ(P ) =N∑

i=1

σ(Qi)∇PSi(P ).

1. Every quadrilateral panel is defined by its angular points: ~r1, ~r2, ~r3, ~r4. So, for everypanel with number i we determine its own local coordinate system, i.e. the unitvectors ~e1, ~e2, ~e3:

~e3 =(~d1 × ~d2)

|~d1 × ~d2|, ~e2 =

~d1 + ~d2

|~d1 + ~d2|, ~e1 = (~e2 × ~e3),

~d1 = ~r3 − ~r1, ~d2 = ~r2 − ~r4,

and the origin of LCS:

~rc =1

4

4∑

i=1

~ri.

44

2. By means of the matrix E we transform the coordinates of the point P (x, y, z) andangular points of the panel in the LCS:

E

(x, y, z) − ~rc,~r1 − ~rc,~r2 − ~rc,~r3 − ~rc,~r4 − ~rc.

=⇒ LCS.

3. Because it is possible that the element is not flat, all z components of the cornerpoints in LCS are set to zero.

4. Using a standard code for Fs(x, y, z) we compute the induced potential Si(P ) andthe induced velocity ~q = ∇Si(P ).

5. We transform the induced velocity vector ~q = uloc~e1 + vloc~e2 + wloc~e3 to the GCS:

ET

uloc

vloc

wloc

=⇒ GCS.

6. We find the sums of all induced potentials and velocities.

7.2.5 Test of accuracy for the source distribution

Consider a sink of strength 4π at the origin. The potential of the sink is

Φ =1√

x2 + y2 + z2=

1

r.

The potential is defined to the extend of an additive constant. So we can set

Φ =1√

x2 + y2 + z2− 1 =

1

r− 1.

Consider a sphere S of unit radius. Then Φ = 0 on S and Φ is a harmonic functionoutside S. From the integral representation (4.11) we obtain

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)︸ ︷︷ ︸

0

fd(P, Q)dS +∫ ∫

S

∂Φ

∂n︸︷︷︸−1

fs(P, Q)dS,

Φ =1

r− 1 =⇒ Φ∞ = −1,

∂Φ

∂n=

∂Φ

∂r= − 1

r2= −1 at r = 1.

As a result we get

1

r− 1 = −1 −

∫ ∫

S

fs(P, Q)dS, P is outside S,

45

(3D) 26 Oct 2003 shere shape

-1

-0.5

0

0.5

1

Z

-1

-0.5

0

0.5

1

X

-1

-0.5

0

0.5

1

Y

X Y

Z

(3D) 26 Oct 2003 shere shape

Figure 29: Panel representation of the sphere. The number of panels N = 900.

0 = −1 −∫ ∫

S

fs(P, Q)dS, P is inside S.

We find from the last formulae the test for the potential:

∫ ∫

S

fs(P, Q)dS = −1

r, P is outside S,

∫ ∫

S

fs(P, Q)dS = −1, P is inside S,

and the test for the velocity field:

∇P

∫ ∫

S

fs(P, Q)dS =~r

r3, P is outside S,

∇P

∫ ∫

S

fs(P, Q)dS = 0, P is inside S.

The test computations were carried out for a sphere of unit radius with the centerlocated at the origin. The vector of velocity at infinity ~q∞ = (−1, 0, 0). The panelrepresentation of the sphere is shown in fig. 29. The representation has been obtained inthe following manner. In every section that goes through the x-axes we have 2 ·nx panels

46

(XY) 27 Oct 2003

x 30o

30o

1

z

0

Region ofcheck

(XY) 27 Oct 2003

Figure 30: Region of the accuracy check. The bottom of the region consists of 5 panels.

distributed uniformly on the circumference of unit radius. So the angular size of all panelsis the same and at nx = 30 is 6. In every section parallel to the coordinate plane Y OZwe have nr panel, that also are distributed uniformly along the circumference boundedthe section. The total number of the panel is N = nx ·nr. We set nx = 30, nr = 30. Thusthe total number of panels is N = 900. Now we pass the section through the x-axes insuch a manner that the section goes trough the collocation points. By virtue of symmetryia all such sections the velocity field is the same. To investigate the accuracy of panelmethod we choose the region shown in fig. 30. In this region we compute the potentialΦ and the module of the velocity vector by means of the panel method and analyticalformulae. The distribution of the related errors of the panel method is shown in fig. ??and fig. ??.

As one can see the results for the potential Φ, obtained by the low-order panel method,are accurate everywhere in the flow. For the velocity field ~q the low-order panel methodgives less accurate results, especially, if the point is very close to the panel edge. But yetabove and below collocation points the results are accurate enough for practical applica-tions.

To explain this less accuracy for the velocity field we should recall that the potentialgiven by the quadrilateral source is continuous, but the velocity components have slightsingularities, defined by the formulae (7.6), as the point P tends to the edge of the panel.

47

tet

r

30 36 42 48 54 60

1.01

1.02

1.03

1.04

1.05

1.06

1.07

1.08

1.09

1.1

phierr0.050.0450.040.0350.030.0250.020.0150.010.0050

Frame 001 5 Jun 2003 Sourse errorsFrame 001 5 Jun 2003 Sourse errors

Figure 31: Typical distribution of errors of potential for quadrilateral sources along 5panels.

tet

r

30 36 42 48 54 60

1.01

1.02

1.03

1.04

1.05

1.06

1.07

1.08

1.09

1.1

qerr0.050.0450.040.0350.030.0250.020.0150.010.0050

Frame 001 5 Jun 2003 Sourse errorsFrame 001 5 Jun 2003 Sourse errors

Figure 32: Typical distribution of errors of velocity for quadrilateral sources along 5panels.

48

7.3 Panel method for computing the surface doublet distribu-

tion

We need to compute

Φ =∫ ∫

S

µ(Q)fd(P, Q)dS, where fd(P, Q) = −~n(Q) · ~r(P, Q)

r3(P, Q).

By discretizing we get

Φ =N∑

i=1

µ(Qi)∫ ∫

∆Si

fd(P, Q)dS

︸ ︷︷ ︸analytically

.

If we denoteDi(P ) =

∫ ∫

∆Si

fd(P, Q)dS, (7.9)

then

Φ(P ) ≈N∑

i=1

µ(Qi)Di(P ). (7.10)

7.3.1 Quadrilateral doublet

The element is located in the plane XOY . On this element we have a uniform doubletdistribution of unit strength:

Φ(P ) =∫ ∫

S

fd(P, Q)dS =∫ ∫

S

−~k · [(x − ξ)~i + (y − η)~j + (z − 0)~k]

4π[(x − ξ)2 + (y − η)2 + (z − 0)2]3/2dξdη.

Finally, the potential of the quadrilateral doublet is

Φ(P ) = − 1

4π

∫ ∫

S

zdξdη

[(x − ξ)2 + (y − η)2 + z2]3/2.

The integral in the RHS can be computed analytically. As a result we get the function of11 variables.

Φ(P ) = Fd(x1, y1, x2, y2, x3, y3, x4, y4, x, y, z). (7.11)

For computation of this function also exists a fast and effective codes.Properties of quadrilateral doublet.

1. Fd(x, y, z) has a jump across S:

Fd(x, y,±0) = ∓1

2on S.

2. The velocity field ~q = ∇P [Fd(x, y, z)] is equivalent to the velocity field induced bya vortex ring:

~q =1

4π

∮

Cd~l × ~r(P, Q)

r3(P, Q),

49

where C is the contour that bounds the quadrilateral. Because of that the veloc-ity field is continuous across S, but near the edges of the element ~q has a strongsingularity:

|~q(P )| ∼ 1

a,

where a is the distance from the point P to the nearest edge of the S.

The algorithm of computing the surface doublet distribution is the same as for thesurface source distribution. The only difference is that instead of the standard code forFs(x, y, z) we are using the code for Fd(x, y, z).

7.3.2 Test for the doublet distribution

Consider a flow over a sphere of unit radius. S is the surface of the sphere,

~q∞ = u∞~i + v∞~j + w∞

~k

is the velocity vector at infinity. For this flow we obtained earlier the exact solution:

Φ(P ) = Φu(1 +1

2r3),

where Φu is the potential of the uniform stream with the velocity ~q∞:

Φu = u∞x + v∞y + w∞z = (~q∞ · ~r), ~r = x~i + y~j + z~k.

On the surface S we have ∂Φ∂n

= 0, Φ = 32Φu for the exact solution. By the integral

representation (4.11) for harmonic functions in an infinite domain we get

Φ(P ) = Φ∞(P )︸ ︷︷ ︸

Φu

−∫ ∫

S

Φ(Q)︸ ︷︷ ︸

3

2φu

fd(P, Q)dS +∫ ∫

S

∂Φ

∂n︸︷︷︸0

fs(P, Q)dS,

Φ(P ) = Φu(P ) −∫ ∫

S

3

2Φufd(P, Q)dS

︸ ︷︷ ︸doublet distribution

.

Test for doublet distribution:

Φu(1 +1

2r3) = Φu(P ) −

∫ ∫

S

3

2Φu(Q)fd(P, Q)dS, P is outside S,

0 = Φu(P ) −∫ ∫

S

3

2Φu(Q)fd(P, Q)dS, P is inside S.

Check it yourself that

~q = ∇[Φu(1 +

1

2r3

]= ~q∞(1 +

1

2r3) − 3(~q∞ · ~r)~r

2r5.

Test for the velocity field induced by doublet distribution:

~q = ~q∞ −∫ ∫

S

3

2Φu(Q)∇P [fd(P, Q)]dS, P is outside S,

50

0 = ~q∞(P ) −∫ ∫

S

3

2Φu(Q)∇P [fd(P, Q)]dS, P is inside S.

The test computations has been carried out in the region shown in fig. 30. Theyhave demonstrated that the results for the potential Φ, obtained by the low-order panelmethod, are accurate if the point P is not too close to the edges of the panels. Forexample, at the collocation points we obtained accurate values of Φ.

But for velocity field ~q the accuracy of computation decreases catastrophically, if thepoint P is close to the surface of the body. The explanation can be found in the formula(6.14):

~q ±(P ) = ∓∇2µ

2+v.p.

1

4π

∫ ∫

S

(∇2µ×~n)× ~r(P, Q)

r3(P, Q)dS+

1

4π

∮

C

µ(Q)

[d~l × ~r(P, Q)

r3(P, Q)

]. (6.14)

On every panel the strength of doublets is constant ∇2µ = 0. Because of that the singularcomponent ∓∇2µ/2 disappears as the point P is on the surface S. This leads to the errorsthat have the order of |∇2µ|.

7.3.3 General conclusions for source and doublet distributions

Let dmax be the maximum diameter of the elements, that represent the surface S.There exists a layer of width approximately dmax above and below S where approximateformulae can give wrong results. The behavior of low-order panel approximations in thisboundary layer of possible errors is demonstrated in table 3.

Table 3: The behavior of low-order panel approximations in the boundary layer of errors

Potential VelocityAccuracy of results forsource distribution

Everywhere Above and below colloca-tion points

Accuracy of results fordoublet distribution

Above and below colloca-tion points

Nowhere in the layer(disappearance of thesingular component)

In practical aerodynamic problems we need very often to compute the velocity field ~qdirectly on the surface of the body. To make such computations correctly we should definethe singular component ∓∇2µ/2 by some other way (e.m., by finite difference method)and add it to the sum of the induced velocities obtained by the low-order panel method.

8 Method of integral equations for solving boundary-

value problems (BVP)

8.1 Source-only and doublet-only formulations

Problems of aerodynamics usually reduce to external Neumann’s BVP. So, as an ex-ample we consider the external NBVP. Let S be a closed surface, ~n be the external unitnormal vector, FN(Q) be a function of surface points.

51

tet

r

30 36 42 48 54 60

1.01

1.02

1.03

1.04

1.05

1.06

1.07

1.08

1.09

1.1

phierr0.050.0450.040.0350.030.0250.020.0150.010.0050

Frame 001 5 Jun 2003 doublet errorsFrame 001 5 Jun 2003 doublet errors

Figure 33: Typical distribution of errors of potential for quadrilateral doublets along 5panels.

tet

r

30 36 42 48 54 60

1.01

1.02

1.03

1.04

1.05

1.06

1.07

1.08

1.09

1.1

qerr0.050.0450.040.0350.030.0250.020.0150.010.0050

Frame 001 5 Jun 2003 doublet errorsFrame 001 5 Jun 2003 doublet errors

Figure 34: Typical distribution of errors of velocity for quadrilateral doublets along 5panels.

52

External Neumann’s BVP:

∂2Φ

∂2x2+

∂2Φ

∂2y2+

∂2Φ

∂2z2= 0 (Laplace’s eq.), (8.1)

∂Φ

∂n= FN(P ), P ∈ S, (8.2)

Φ − Φ∞(P ) → 0 as P → ∞. (8.3)

Integral representation (4.13):

Φ = Φ∞(P ) +∫ ∫

S

σ(Q)fs(P, Q)dS +∫ ∫

S

µ(Q)fd(P, Q)dS. (4.13)

For the representation (4.13) equation (8.1) will be satisfied automatically. The integralsin the RHS tend to zero as P → ∞. Therefore the boundary condition at infinity (8.3)for (4.13) will be satisfied automatically too. Thus we need to find σ(Q) and µ(Q) so,that the boundary condition (8.2) would be fulfilled.

Let us set in (4.13) that µ(Q) = 0. In this case we obtain the source-only formulation:

Φ = Φ∞(P ) +∫ ∫

S

σ(Q)fs(P, Q)dS, (8.4)

∂Φ

∂n= [∇PΦ · ~n(P )], P ∈ S,

∇Φ = ∇Φ∞(P ) +∫ ∫

S

σ(Q)~r(P, Q)

4πr3(P, Q)︸ ︷︷ ︸

~K(P,Q)

dS,

∇Φ = ∇Φ∞(P ) +∫ ∫

S

σ(Q) ~K(P, Q)dS.

The last formula is correct if P lies outside S. If P ∈ S+, then with accordance to thejump property of the source distribution (formula (6.5)) we obtain

∫ ∫

S

σ(Q) ~K(P, Q)dS

+

=σ(P )

2~n(P )

︸ ︷︷ ︸singular component

+v.p.∫ ∫

S

σ(Q) ~K(P, Q)dS,

∇P Φ = ∇P Φ∞(P ) +σ(P )

2~n(P ) + v.p.

∫ ∫

S

σ(Q) ~K(P, Q)dS,

[∇P Φ · ~n(P )] = FN(P ) (BC),

[∇PΦ∞(P ) · ~n(P )] +σ(P )

2[~n(P ) · ~n(P )]︸ ︷︷ ︸

=1

+v.p.∫ ∫

S

σ(Q)[ ~K(P, Q) · ~n(P )]dS = FN(P ),

σ(P )

2+ v.p.

∫ ∫

S

σ(Q)K1(P, Q)dS = FN(P ) − [∇P Φ∞(P ) · ~n(P )]︸ ︷︷ ︸

RHS is known

, (8.5)

53

whereK1(P, Q) = [~n(P ) · ~K(P, Q)]

is a kernel function.Equation (8.5) ia a Fredholm integral equation of the second kind (”good” equation,

after discretization we shall have a well conditioned linear system).In a similar manner we can obtain an integral equation for solving Dirichlet’s BVP.

Φ(P ) = FD(P ) (Dirichlet BC).

We recall that the source distribution is continuous across S. Therefore

Φ∞(P ) +∫ ∫

S

σ(Q)fs(P, Q)dS = FD(P ),

∫ ∫

S

σ(Q)fs(P, Q)dS = FD(P ) − Φ∞(P ).

This is a Fredholm’s integral equation of the first kind. (”Bad” equation, after discretiza-tion we will have possibly a full matrix that is not diagonally dominant.)

To reduce a BVP to an integral equation we can use doublet-only formulation. In thiscase σ(Q) = 0 and

Φ(P ) = Φ∞(P ) +∫ ∫

S

µ(Q)fd(P, Q)dS.

Repeating previous reasoning we will obtain again an integral equation for finding thedoublet distribution µ(Q).

8.1.1 Summary of possible combination of singularity distributions and BC

1. External Neumann’s BVP. Source-only distribution is unique, doublet-only distri-bution is not unique.

2. Internal Neumann BVP. Neither source-only nor doublet-only distributions areunique.

3. External Dirichlet BVP. Source-only distribution is unique, doublet-only distribu-tion is not unique.

4. Internal Dirichlet BVP. Both source-only and doublet-only distributions are unique.

8.2 Morino’s formulation for solving external Neumann BVP

Consider again external NBVP (8.1)-(8.3). BC:

∂Φ

∂n= (∇Φ · ~n) = FN(P ), P ∈ S, (Neumann’s BC).

The integral representation (4.12) for a harmonic function is

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂nfs(P, Q)dS

︸ ︷︷ ︸∂Φ

∂n=FN (Q), (from BC)

,

54

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)fd(P, Q)dS

︸ ︷︷ ︸doublets

+∫ ∫

S

FN (Q)fs(P, Q)dS

︸ ︷︷ ︸sources

. (8.6)

The representation (8.6) is correct if P lies outside S. Now let P → S. We recall thatthe source distribution is continuous across S. Therefore

∫ ∫

S

FN (Q)fs(P, Q)dS

+

=∫ ∫

S

FN(Q)fs(P, Q)dS

︸ ︷︷ ︸no singular component

.

But the doublet distribution has a jump across S. So according to (6.7) we have

∫ ∫

S

Φ(Q)fd(P, Q)dS

+

= −Φ(P )

2+∫ ∫

S

Φ(Q)fd(P, Q)dS.

It follows from (8.6) that

Φ(P ) = Φ∞(P ) +Φ(P )

2−∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

FN (Q)fs(P, Q)dS.

Bringing out unknowns in LHS we get

Φ(P )

2+∫ ∫

S

Φ(Q)fd(P, Q)dS = Φ∞(P ) +∫ ∫

S

FN (Q)fs(P, Q)dS

︸ ︷︷ ︸B(P )

,

Φ(P )

2+∫ ∫

S

Φ(Q)fd(P, Q)dS = B(P ). (8.7)

Equation (8.7) is called Morino’s integral equation. (Fredholm’s equation of the secondkind, good for numerical solution). Equation (8.7) is very convenient for numerical com-putations. It allows us to find the distribution of the potential Φ(Q) on the body surfaceS. In the source-only or doublet-only formulations we are satisfying directly the NeumannBC on the surface S. In Morino’s formulation instead of BC we have a mathematical equa-tion (8.7). BC are hidden in this mathematical equation. So, we are satisfying here theBC indirectly. Since we can only solve (8.7) approximately, the BC will be also fulfilledonly approximately.

8.3 Method of reducing the external Neumann BVP to the in-

ternal Dirichlet BVP

Consider inside S any harmonic function Φin(P ). Stress your attention to the fact thatwe are choosing Φin(P ) ourself. For example, we can set Φin(P ) = 0 or Φin(P ) = Φ∞(P ).So, Φin(P ) is given and the velocity field ~qin = ∇Φin(P ) is also given inside S. Nowconsider the function

G(P ) = Φ∞(P ) +∫ ∫

S

σ(Q)fs(P, Q)dS +∫ ∫

S

µ(Q)fd(P, Q)dS. (8.8)

55

Here σ(Q) and µ(Q) are unknown. We will try to find them in such a manner that

G(P ) = Φ(P ), P lies outside S, (8.9)

G(P ) = Φin(P ), P lies inside S. (8.10)

First of all we are computing the jump of ~q = ∇P (G) across S. We are using here theformulae for the jumps that were deduced earlier for source and doublet distributions insection 5. We obtain

~q +(P ) − ~q −(P ) = σ(P )~n − ∇2µ(P )︸ ︷︷ ︸

2-D gradient

. (8.11)

Now we calculate the normal component of this jump. In so doing we are taking intoaccount that 2-D gradient is tangential to the surface S. Therefore ∇2µ⊥~n, and (∇2µ·~n) =0. Thus

[~q +(P ) − ~q −(P )] · ~n(P ) = σ(P )(~n · ~n) = σ(P ).

So, for ~q = ∇G we have[~q +(P ) − ~q −(P )] · ~n = σ(P ). (8.12)

On the other hand we can compute the jump by (8.9), (8.10). Indeed,

(~q + · ~n) = [∇G(P ) · ~n]+ = (∇Φ · ~n) =∂Φ

∂n= FN(Q),

(~q − · ~n) = [∇G(P ) · ~n]− = (∇Φin · ~n) = (~qin · ~n),

[~q +(P ) − ~q −(P )] · ~n = FN(P ) − (~qin · ~n). (8.13)

Comparing (8.12) and (8.13) we can see that

σ(Q) = FN (Q) − (~qin · ~n). (8.14)

Thus in (8.8) σ(Q) is known, and we need to find only µ(Q) (doublet distribution). Tofind µ(Q) we shall use (8.10). But to provide (8.10) it is enough to fulfill

G−(P ) = Φin(P ), P ∈ S− (8.15)

Indeed, equation (8.15) is Dirichlet BC. Since DBVP has a unique solution, equation(8.10) follows from (8.15). We satisfy (8.15) by the method of integral equations. For thedoublet distribution

∫ ∫

S

µ(Q)fd(P, Q)dS

−

=µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS.

For the source distribution

∫ ∫

S

σ(Q)fs(P, Q)dS

−

=∫ ∫

S

σ(Q)fs(P, Q)dS,

G−(P ) = Φ∞(P ) +µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS +∫ ∫

S

σ(Q)fs(P, Q)dS,

56

Φin(P ) = Φ∞(P ) +µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS +∫ ∫

S

σ(Q)fs(P, Q)dS,

µ(P )

2+∫ ∫

S

µ(Q)fd(P, Q)dS = B1(P ),

whereB1(P ) = Φin(P ) − Φ∞(P ) −

∫ ∫

S

σ(Q)fs(P, Q)dS,

σ(Q) = FN (Q) − (∇Φin · ~n).

This is again Morino’s integral equation but the RHS B1(P ) is different. From this newformulation Morino’s formulation can be obtained if we set Φin(P ) = 0, µ(Q) = −Φ(Q)on S. In the general case

µ(Q) = −[Φ(Q) − Φin(Q)].

Advantage of this general formulation compared with ordinary Morino’s formulation iscan be obtained by choosing the function Φin(P ) so, that |B1(P )| < |B(P )|. The smallerRHS leads to the smaller function µ(Q), and the functions with smaller values are easierto determine by numerical methods.

8.4 Discretization of integral equations for 3-D case. Influence

coefficients

Consider Morino’s integral equation

Φ(P )

2+∫ ∫

S

Φ(Q)fd(P, Q)dS = B(P ),

B(P ) = Φ∞(P ) +∫ ∫

S

FN(Q)fs(P, Q)dS.

By making use of the formula

∫ ∫

S

Φ(Q)fd(P, Q)dS

−

=Φ(P )

2+∫ ∫

S

Φ(Q)fd(P, Q)dS, P ∈ S

we rewrite the Marino integral equation in the form

∫ ∫

S

Φ(Q)fd(P, Q)dS

−

= Φ∞(P ) +∫ ∫

S

FN(Q)fs(P, Q)dS.

So, we have included the singular component Φ(P )/2 in the integral. To compute theLHS and RHS of the Morino equation we use the approximate formulae (7.3), (7.4), (7.9),(7.10) of the panel method, that were obtained in the previous section:

∫ ∫

S

FN(Q)fs(P, Q)dS =N∑

j=1

FN (Qj)Sj(P ), Sj(P ) =∫ ∫

∆Sj

fs(P, Q)dS,

57

Figure 35:

∫ ∫

S

Φ(Q)fd(P, Q)dS

−

=N∑

j=1

Φ(Qj)D−j (P ), Dj(P ) =

∫ ∫

∆Sj

fd(P, Q)dS.

To apply the above formulae we approximate the surface S by N elements ∆Sj . On everyelement we locate the collocation point Qj, j = 1, 2, . . . , N , and assume that on everyelement the sources and doublets of unit strength are distributed uniformly.

Consider two elements with numbers i and j (see fig. 35). The influence coefficientD+

ij is the potential induced by the doublet element with the number j at the collocationpoint Qi of the element with number i on the surface S+. The influence coefficient D−

ij isthe potential induced by the doublet element with the number j at the collocation pointQi of the element with number i on the surface S−.

By the function Dj(P ), introduced above, the coefficients D±ij express as follows

D±ij = D±

j (Qi).

Connection between D+ij and D−

ij :

D+ij = D−

ij if i 6= j,D+

ij − D−ij = −1 if i = j.

This is due to the jump properties of the doublet distribution.The influence coefficients for sources are

Sij = Sj(Qi).

The sense is the same, Sij is the potential induced by the source element of unite strengthwith the number j at the collocation point Qi. Since the source distribution is continuousacross S, we have

Sij = S+ij = S−

ij .

58

Now we will satisfy the Marino integral equation only at the collocation point Qi. If wedenote Φj = Φ(Qj) we shall have

∑

j=1

D−ijΦj = Φ∞(Qi) +

N∑

j=1

SijFN (Qj), i = 1, 2, . . . , N.

This is a linear system of N equations with N unknowns Φj .To find numerically the influence coefficients D±

ij and Sij we shall use the followingalgorithm. Let Qi be a collocation point and j be the number of an element.

1. We know the coordinates of the collocation point Qi and the corner points r1j , r2j , r3j, r4j

of the element in GCS.

2. We transform Qi, r1j , r2j, r3j , r4j from the GCS to LCS.

3. Using standard codes for the potentials induced by the quadrilateral source anddoublet we compute the potentials induced at the point Qi. These potentials arejust the influence coefficients that we need: Sij and D±

ij .

4. To define D+ii (influence of an element on itself) we proceed in the following manner.

We compute the integral Di(Qi) not at the point Qi directly but at the point Q′i =

Qi + ε~n, where ε is a small value, for example, ε = 10−9. To define D−ii we compute

the integral at the point Q′i = Qi − ε~n. Proceeding in this manner provides us the

jump D+ij − D−

ij = −1.

The system of linear equations can be solved by standard routines. After solving thesystem we obtain the values of potential at the collocation points Qj :

Φj = Φ(Qj), j = 1, 2, . . . , N.

If we use another formulation, then the influence coefficients have a different sense.For example, consider the source-only formulation:

Φ = Φ∞(P ) +∫ ∫

S

σ(Q)fs(P, Q)dS,

where σ(Q) is an unknown source distribution. Earlier, for finding σ(Q) we deduced theintegral equation

σ(P )

2+ v.p.

∫ ∫

S

σ(Q)K1(P, Q)dS = FN (P ) −∇Φ∞(P ) · ~n(P ),

K1(P, Q) =~r(P, Q) · ~n(P )

4πr3(P, Q).

The LHS of the equation is the normal component of the velocity vector induced by thesource distribution at the point P of the surface S+. We define the influence coefficient W±

ij

as the normal components of the velocity vector at the point Qi, induced by the sourcesof unite strength distributed on the element with the number j. Using this definition weget

N∑

j=1

W+ij σj = FN(Qi) − [(∇Φ∞(Qi)) · ~n(P )]︸ ︷︷ ︸

RHS is known

, i = 1, 2, . . . , N.

where σj = σ(Qj), j = 1, 2, . . . , N are unknown. To determine σj we can again usestandard routines for solving linear systems.

59

9 Panel method for solving aerodynamic problems

9.1 Body in the fluid

Consider a body that moves in the fluid with constant velocity ~qB. The body moveswithout changing its shape. The fluid at infinity is at rest. Let us introduce a coordinatesystem in the body frame of reference. Then the body is at rest and the flow at infinityhas the velocity ~q∞ = −~qB. We need to find the potential Φ = Φ(x, y, z). In section 4we reduced this problem to the external BVP:

∇2Φ = 0 (Laplace’s eq.), (9.1)

∂Φ

∂n= 0 (BC on the surface S), (9.2)

Φ − u∞x − v∞y − w∞z → 0, as P → ∞ (BC at infinity). (9.3)

To solve the external NBVP we can use the method of integral equations with discretiza-tion by the panel method. For Morino’s formulation we obtain the equation

Φ(P )

2+∫ ∫

S

Φ(Q)fd(P, Q)dS = u∞x + v∞y + w∞z, (9.4)

where Φ(P ) is the distribution of the potential on the body surface.Φ(P ) is the total potential defined by the integral representation

Φ(P ) = −∫ ∫

S

Φ(Q)fd(P, Q)dS + u∞x + v∞y + w∞z.

Here we have no source distribution∫ ∫

S

∂Φ

∂nfs(P, Q)dS,

since ∂Φ∂n

= 0.Very often for solving hydrodynamic problems instead of the total potential Φ(x, y, z)

it can be used a so-called perturbation potential

φ(x, y, z) = Φ(x, y, z) − u∞x − v∞y − w∞z.

If we rewrite the BVP (9.1)-(9.3) in terms of the perturbation potential we get

∇2φ = 0, (9.5)

∂φ

∂n= −(~q∞ · ~n) (check it yourself!), (9.6)

φ → 0 as P → ∞ (BC at infinity). (9.7)

Morino’s integral equation for external Neumann BVP (9.5)-(9.7) takes the form

φ(P )

2+∫ ∫

S

φ(Q)fd(P, Q)dS = −∫ ∫

S

(~q∞ · ~n)fs(P, Q)dS. (9.8)

60

Figure 36:

If to compare equation (9.4) and (9.8) we can see that in (9.4) we have no integral in theRHS. Numerical solution of this equation seems to be easier. But in equation (9.8) theintegrand (~q∞ ·~n) and the RHS are small if the body is slender and moves at a small angleattack. This means that the function to be found φ(Q) will be also small. This can giveus a numerical advantage and greater accuracy.

Remark 9.1 Because in panel methods integral equations are fulfilled only in the finitenumber of collocation points we will have leakage through the body surface S. The stream-lines near S can go in S and leave out it.

9.2 Panel method for the flow over 3-D wings

The difference between flows over a smooth body and a wing is that behind thetrailing edge line of the wing a wake forms (see fig. 36). The wake in the potential flowis an infinitely thin surface on which the potential Φ and the tangential component ofvelocity ~q have a jump. The BC on the surface SB of the wing are the same as for asmooth body. If φ(x, y, z) is the perturbation potential then

∂φ

∂n= −(~q∞ · ~n).

For the total potential ∂Φ∂n

= 0.The boundary condition for the steady wake:

1. The wake is impermeable, i.e.

∂Φ

∂n+= 0,

∂Φ

∂n−= 0.

2. The wake is unloaded, i.e.p+ = p−,

where p+ and p− are the pressures on the S+w and S−

w .

61

Let us write the integral representation for the perturbation potential

φ(P ) = Φ(P ) − u∞x − v∞y − w∞z.

In so doing we should take into account that the wake surface Sw has two sides S+w and

S−w . On these sides we have the different normals ~n+ = ~n and ~n− = −~n. The values of

the potential φ on these sides are also different: φ+ and φ−. This gives

φ(P ) = −∫ ∫

SB

φfdP, QdS +∫ ∫

SB

∂φ

∂n︸︷︷︸(−~q∞·~n)

fs(P, Q)dS−

∫ ∫

S+w

φfdP, QdS −∫ ∫

S−

w

φfd(P, Q)dS

︸ ︷︷ ︸(∗)

+

∫ ∫

S+w

∂φ

∂n+fs(P, Q)dS +

∫ ∫

S−

w

∂φ

∂n−fs(P, Q)dS

︸ ︷︷ ︸(∗∗)

,

fd(P, Q) = −(~n+ · ~r)4πr3

= −(~n · ~r)4πr3

on S+w

fd(P, Q) = −(~n− · ~r)4πr3

=(~n · ~r)4πr3

on S−w ,

fs(P, Q) = − 1

4πron S+

w and S−w .

Now

(∗) = −∫ ∫

Sw

(φ+ − φ−)fd(P, Q)dS, fd(P, Q) = −(~n · ~r)4πr3

.

On the wake surface from the impermeability conditions we have

∂Φ

∂~n+= 0 =⇒ ∂φ

∂~n+= −(~q∞ · ~n+) = −(~q∞ · ~n),

∂Φ

∂~n−= 0 =⇒ ∂φ

∂~n−= −(~q∞ · ~n−) = (~q∞ · ~n).

Therefore,

(∗∗) =∫ ∫

Sw

[−(~q∞ · ~n) + (~q∞ · ~n)]fs(P, Q)dS = 0,

and finally

φ(P ) = −∫ ∫

SB

φfd(P, Q)dS −∫ ∫

SB

(~q∞ · ~n)fs(P, Q)dS −∫ ∫

Sw

µw(Q)fd(P, Q)dS

︸ ︷︷ ︸wake term

,

where µw = φ+ − φ−.

62

We see that the wake can be modelled by a doublet distribution. Now we recall that

φ = Φ−u∞x − v∞y − w∞z︸ ︷︷ ︸has no jump

,

µw = φ+ − φ− = Φ+ − Φ−.

It follows from here that µw is the circulation along the contour C that intersects thewake:

µw = Φ+ − Φ− =∮

C

(~q · d~l).

We should note that the wake surface Sw is unknown in advance and is to be foundin solving. Because of that we have two conditions on the wake. So far we have usedthe impermeability conditions. Now we will use the condition that the wake is unloaded:p+ = p−. From Bernoulli’s equation it follows that

p+ + ρ(q+)2

2= p− + ρ

(q−)2

2,

where q+ and q− are the modula of the velocity vectors ~q + and ~q − on the surface Sw.Let us introduce the average velocity

~qw =~q + + ~q −

2on Sw.

This average velocity is also called the wake velocity. The vector ~qw forms a velocity fieldon Sw. This velocity field forms some streamlines on Sw. These streamlines are calledmean streamlines.

Theorem 2 The doublet distribution µw = φ+ − φ− = Φ+ − Φ− is constant along meanstreamlines.

Proof. It follows from Bernoulli’s equation that

(q+)2 = (q−)2.

Consider on Sw two perpendicular directions e1 and e2 that go trough the point P ∈ Sw.We expand ~q + and ~q − along this directions:

~q + = q+1 ~e1 + q+

2 ~e2, ~q − = q−1 ~e1 + q−2 ~e2,

q±1 and q±2 are the components of the velocity vector ~q ±.

(q+)2 = (q−)2 =⇒ (q+1 )2 + (q+

2 )2 = (q−1 )2 + (q−2 )2,

(q+1 − q−1 )(q+

1 + q−1 ) + (q+2 − q−2 )(q+

2 + q−2 ) = 0,∣∣∣∣ : 2

(q+1 − q−1 )

q+1 + q−1

2+ (q+

2 − q−2 )q+2 + q−2

2= 0. (9.9)

We note that

~qw =q+1 + q−1

2~e1 +

q+2 + q−2

2~e2. (9.10)

63

Compute 2-D gradient of µw:

∇2µw = ∇2(φ+ − φ−) = ∇2(Φ

+ − Φ−) =

∂(Φ+ − Φ−)

∂e1~e1 +

∂(Φ+ − Φ−)

∂e2~e2,

∇2µw = (q+1 − q−1 )~e1 + (q+

2 − q−2 )~e2. (9.11)

It follows from (9.9)-(9.11) that

(∇2µw · ~qw) = 0 =⇒ ∇2µw⊥~qw,

~e = ~qw/|~qw| is the direction of the mean streamline that goes through the point P . Then

(∇2µw · ~e) = 0 =⇒ ∂µw

∂e= 0.

This means that µw is constant along the mean streamline. Theorem is proved.

9.2.1 Simplest wake model

1. The shape of the wake is given.

2. The mean streamlines on the wake are given two.

In this case it is possible to continue µw from the trailing edge to the whole wake. thevalues of µw on the trailing edge can be determined by the Kutta-Joukovsky conditions:the velocity at the trailing edge of the wing should be finite.

Consequences:

1. The wake surface bisects the trailing edge angle.

2. Since the T.E. line is common to the wing and wake we have

(φ+ − φ−)∣∣∣∣T.E. of wing= (φ+ − φ−)∣∣∣∣T.E. of wake

. (9.12)

9.2.2 Integral equation for 3-D wing (Morino’s formulation)

φ(P ) = −∫ ∫

SB

φ(Q)fd(P, Q)dS −∫ ∫

SB

(~q∞ · ~n)fs(P, Q)dS −∫ ∫

Sw

µwfd(P, Q)dS.

Let P → SB, then

∫ ∫

SB

φ(Q)fd(P, Q)dS

+

= −φ(P )

2︸ ︷︷ ︸singularcomponent

+∫ ∫

SB

φ(Q)fd(P, Q)dS.

64

Figure 37:

By means of this formula we get

φ(P )

2+∫ ∫

SB

φ(Q)fd(P, Q)dS +∫ ∫

Sw

µwfd(P, Q)dS = −∫ ∫

SB

(~q∞ · ~n)fs(P, Q)dS

︸ ︷︷ ︸B(P )

or ∫ ∫

SB

φ(Q)fd(P, Q)dS

−

+∫ ∫

Sw

µwfd(P, Q)dS = B(P ). (9.13)

To discretize equation (9.13) we represent the surface of the wing as a collection ofN panels in a manner that the upper and lower panels, that adjoin to the trailing edge,have a common edge (see fig. 37). The wake surface Sw is represented as a collection ofpanels and on every wake row µw = const. Now we need to find the influence coefficientsD−

ij and Sij. The influence coefficient Sij for the source distribution are defined as for asmooth body:

Sij = Sj(Qi).

But to define D−ij we should take into account the influence of the wake. In computing

D−ij we have three options.

1. The number j is not the number of the panel that adjoins to the trailing edge. Inthis case

D−ij = D−

j (Qi).

2. The number j is the number of an upper panel that adjoins to the trailing edge. Tocompute D−

ij we should add to D−j (Qi) the potential induced at the point Qi by the

wake row, that adjoins to the j-panel. All panel in the wake row have the strength+1.

65

3. The number j is the number of an lower panel that adjoins the trailing edge. Tocompute D−

ij we should add to D−j (Qi) the potential induced at the point Qi by the

wake row, that adjoins to the j-panel. All panel in the wake row have the strength−1.

The discrete form of Morino’s equation is

N∑

j=1

D−ijφj = −

N∑

j=1

(~q∞ · ~n),

where φj = φ(Qj), j = 1, 2, . . . , N .The wake disappears in this equation. Its influence is included in the influence coeffi-

cients.

9.2.3 Computation of the pressure and lift

The discrete form of the integral representation for φ is

φ = −N∑

j=1

φj

∫ ∫

∆Sj

fd(P, Q)dS −Nw∑

j=1

µwj

∫ ∫

∆Sj

fd(P, Q)dS −N∑

j=1

(~q∞ ·~n)∫ ∫

∆Sj

fs(P, Q)dS, (9.14)

where ∆Swj are the elements of the wake, Nw is the number of elements of the wake,

Φ = φ + u∞x + v∞y + w∞z, (9.15)

~q = ∇P Φ = ∇Pφ + ~q∞.

To compute the pressure distribution we will use Bernoulli’s equation:

p +ρq2

2= p∞ +

ρq2∞

2.

Pressure coefficient

Cp =p − p∞ρq2

∞/2= 1 − q2

q2∞

.

To determine q we need to find ∇P φ on the surface of the wing. But in our integralrepresentation (9.14) we have the doublet distribution, so the method of summing inducedvelocities leads to great errors. To compute correctly q on the surface of the wing (in thelayer of error) it is possible to use the finite difference method. Indeed after solvingthe Morino equation we know the accurate values of the perturbation potential φ atthe collocation points. By making use of (9.15) we can find the values of the totalpotential Φ at the same points. Since the surface of the wing is impermeable we have∂Φ∂n

= 0 and therefore q = |∇2Φ|. Finding of this 2-D gradient can be made by FDM. Forexample, if the surface of the wing is represented by equidistant rectangles and x, y arelocal coordinates whose axes are parallel to the edges of the rectangle with the number i,then

∂Φ

∂x=

Φi3 − Φi1

2h,

∂Φ

∂y=

Φi2 − Φi4

2h, q =

√√√√(

∂Φ

∂x

)2

+

(∂Φ

∂y

)2

.

Lift:

~L = −ρq2∞

2

N∑

j=1

Cpj∆Sj~n(Qj),

where ∆Sj is the areas of elements, Cpj is the pressure coefficient, computed at the pointQj .

66

Figure 38:

9.3 Six main steps in applying BEM

1. Choice of singularity elements.

2. Geometric representation of the body.

3. Computation of influence coefficients.

4. Computation of the RHS of the linear equation system.

5. Solving the linear equation system.

6. Computation of the lift, induced drag, pressure distribution, velocity field and soon.

10 BEM for 2-D airfoils

10.1 Problem formulation

For 2-D case the potential Φ depends only on two variables x, y:

Φ = Φ(x, y), ~q = ∇Φ, ~q = u~i + v~j.

Boundary conditions (see fig. 38):On the contour C we have the impermeability condition:

∂Φ

∂n= 0.

On the wake Cw we have two conditions.

1. The wake is impermeable:∂Φ

∂n= 0 on Cw.

It follows from here that the wake is a streamline.

67

2. The wake is unloaded: p+ = p−. From Bernoulli’s equation we get

p+ +ρ(q+)2

2= p− +

ρ(q−)2

2=⇒ (q+)2 = (q−)2 =⇒ q+ = q−.

Since the wake is a streamline we obtain that ~q + = ~q −. Therefore, we have no jump ofthe velocity vector across the wake. We know that on the wake the jump of the potential

µw = Φ+ − Φ− = const (10.1)

along mean streamlines (for proof see 3-D case). For 2-D case the mean streamlinescoincide with ordinary streamlines. Consequently, equation (10.1) is fulfilled along thewake.

Since the velocity field is continuous and µw = Φ+ − Φ− = const along the wake theshape of the wake does not influence the velocity field. (This fact we give without proof).This means that as the wake we can take any line that starts from the trailing edge andgoes to infinity. For example we can take as the wake a direct semi-infinite segment. Asa result we come to the following boundary-value problem

∂2Φ

∂x2+

∂2Φ

∂y2= 0 (Laplace’s eq. for 2-D case), (10.2)

∂Φ

∂n= 0 on C, (10.3)

µw = Φ+ − Φ− = const along Cw, (10.4)

∇Φ → ~q∞ as P → ∞. (10.5)

Beside equations (10.2)-(10.5) we have the Kutta condition that the velocity is finite atthe trailing edge.

10.2 2-D versions of basic singularities

Notation:P (X, Y ) is a point in the domain;Q(ξ, η) is a point on the line of integration;

~r(P, Q) =−→QP = (x − ξ)~i + (y − η)~j;

r(P, Q) = |−→QP | =√

(x − ξ)2 + (y − η)2;

10.2.1 Point source

Assume that the source is located at the point Q(ξ, η). The potential of the source is

Φ =σ

2πln r,

σ is the strength of the source. Compare this 2-D potential with the 3-D one

Φ = − σ

4πr.

68

Figure 39:

If to integrate the uniform 3-D source distribution along the line that is parallel to thez-axes we will obtain the 2-D case from the 3-D one.

2-D velocity field is

~q = ∇P Φ =σ

2π

~r(P, Q)

r2(P, Q),

Q(ξ, η) is a singular point, where Φ → ∞, |~q| → ∞.

10.2.2 2-D doublet

The potential is

Φ = − µ

2π

[~n · ~r(P, Q)]

r2(P, Q).

~n here is the unit vector of the axis of the doublet. The axis of the doublet is alwayspointed from the sink to the source. Let the axis be parallel to the line OY . For thisdoublet ~n = ~j, (~n · ~r) = y − η

Φ = − µ

2π

y − η

(x − ξ)2 + (y − η)2.

10.2.3 Point vortex

The vortex is located at the point Q(ξ, η). The streamlines are circumferences (seefig. 39). The potential is

Φ = − γ

2πθ,

where θ is the angle between the vector ~r(P, Q) =−→QP and the x-axis.

Let L be the cut that starts from the point Q and infinitely goes to the right. Weshow now that

Φ+ − Φ− = γ = const. (10.6)

Indeedon L+ θ = 0, Φ+ = 0,

on L− θ = 2π, Φ− = − γ

2π2π = −γ.

It follows from here that (10.6) holds true.By definition

γ =∮

C

(~q · d~l) = Φ+ − Φ−

69

is the circulation of the vortex. Here C is a closed contour that surrounds the point Q.How to compute θ. To make it correctly we should take into account the quadrant in

which the point P lies. Let us denote

Ω = arctany − η

x − ξ.

Then

1. x − ξ > 0, y − η > 0, (I) =⇒ θ = Ω,

2. x − ξ > 0, y − η < 0, (IV) =⇒ θ = Ω + 2π,

3. x − ξ < 0, x − ξ 6= 0, (II,III) =⇒ θ = Ω + π,

4. x − ξ = 0, y − η > 0 =⇒ θ = π/2,

5. x − ξ = 0, y − η < 0 =⇒ θ = 3π/2.

The simpler way is to use the Fortran atan2(y, x) function. This function defines theangle α between the point P (x, y) and the x-axes with the cut that goes infinitely to theleft. Applying this function we get

θ = atan2(η − y, ξ − x) + π.

(Check it yourself!).

10.3 Source and doublet distributions

10.3.1 Source distribution

Φ =∫

C

σ(Q)fs(P, Q)dl,

where

fs(P, Q) =1

2πln r(P, Q).

The source distribution is continuous, but the normal component of the velocity vectorhas a jump: (

∂Φ

∂n

)+

−(

∂Φ

∂n

)−

= (∇Φ)+ · ~n − (∇Φ)− · ~n = σ(P ).

10.3.2 Doublet distribution

Φ =∫

C

µ(Q)fd(P, Q)dl,

where

fd(P, Q) = −~n · ~r(P, Q)

2πr2(P, Q).

The doublet distribution has a jump across C:

Φ+ − Φ− = −µ.

70

Figure 40:

10.3.3 Constant strength source distribution along a segment (x1, x2)

Let us assume that on the segment (x1, x2) the sources with the strength σ are dis-tributed:

Φ = σ

x2∫

x1

fs(P, Q)dξ =σ

2π

x2∫

x1

ln√

(x − ξ)2 + y2dξ.

The potential Φ isΦ = σFs(x, y, x1, x2),

where Fs(x, y, x1, x2) is the function of four variables:

Fs(x, y, x1, x2) =1

2π

x2∫

x1

ln√

(x − ξ)2 + y2dξ.

The function Fs can be computed analytically. The velocity vector

~q = ∇P Φ = σ∇PFs

can also be determined analytically.

10.3.4 Constant strength doublet distribution along a segment (x1, x2)

Assume that the axes of all doublet be parallel to OY . The potential of this distribu-tion is

Φ = − µ

2π

x2∫

x1

(y − η)dξ

(x − ξ)2 + y2.

The integral in the RHS can be computed analytically:

Φ = µFd(x, y, x1.x2), Fd = − 1

2π

x2∫

x1

(y − η)dξ

(x − ξ)2 + y2.

It can be shown that the potential Φ is equivalent with respect to induced velocities totwo vortexes located at the points x1 and x2 and having the opposite strengths −µ andµ correspondingly (see fig. 40):

Φ = − µ

2π(θ2 − θ1).

71

10.4 Integral representation of the 2-D potential and its discrete

analog

In the 2-D case integral representation of the total potential Φ takes the form

Φ(P ) = −∫

C

Φ(Q)fd(P, Q)dl

︸ ︷︷ ︸doublets

+∫

C

∂Φ

∂nfs(P, Q)dl

︸ ︷︷ ︸sources

−∫

Cw

µwfd(P, Q)dl

︸ ︷︷ ︸doublets

+Φ∞(P ),

where µw = Φup(B)−Φlow(B) = const (Kutta condition), Φup and Φlow are the values ofthe potential at the trailing edge B.

If we denote µ(Q) = −Φ(Q), σ(Q) = ∂Φ∂n

, then

Φ(P ) =∫

C

µ(Q)fd(P, Q)dl +∫

C

σ(Q)fs(P, Q)dl − µw

∫

Cw

fd(P, Q)dl + Φ∞(P ),

µw = −µup(B) + µlow(B) on the wake.

Let us denote as in the 3-D case

Dj(P ) =∫

∆Cj

fd(P, Q)dl, Sj(P ) =∫

∆Cj

fs(P, Q)dl,

µj = −Φ(Qj), σj =∂Φ

∂n(Qj), j = 1, 2, . . . , N.

On the wake we denote µN+1 = −µw. Taking into account that µup(B) = µN , µlow(B) =µ1 we obtain the Kutta’s condition in the form

µN+1 = µN − µ1.

In the discrete form the representation for Φ(P ) can be written as

Φ(P ) =N+1∑

j=1

µjD j(P ) +N+1∑

j=1

σjSj(P ) + Φ∞(P ). (10.7)

In this representation µj and σj are unknowns. Since

Φ∞(P ) = u∞x + v∞y

we get

~q = ∇P Φ(P ) =N+1∑

j=1

µj∇P D j(P ) +N+1∑

j=1

σj∇P Sj(P ) + ~q∞. (10.8)

10.5 Outlook of different formulations

10.5.1 Doublet-only formulation

Let us set all source strengths to zero. Then

Φ(P ) =N+1∑

j=1

µjD j(P ) + Φ∞(P ).

72

Boundary conditions:

∂Φ

∂n= 0 =⇒ ∂Φ

∂n(Qi) = 0, i = 1, 2, . . . , N,

Qi are the collocation points.

∂Φ

∂n(Qi) = ∇PΦ(Qi) · ~n(Qi) = 0, i = 1, 2, . . . , N,

N+1∑

j=1

µj [∇P D j(Qi) · ~n(Qi)]︸ ︷︷ ︸Eij

+[~q∞ · ~n(Qi)] = 0.

And we obtain the system of linear equations

N+1∑

j=1

Eijµj = −(~q∞ · ~n(Qi)). (10.9)

The unknowns are µ1, µ2, . . . µN , µN+1, Eij are influence coefficients. But we have only Nequation. An additional equation is the Kutta-Joukovsky condition:

µN+1 = µN − µ1.

10.5.2 Finding influence coefficients

Consider an element with the number j and the point P in the space. At the point Pwe have a given vector

~n = nx~i + ny

~j.

On the element we have the uniform doublet distribution of unit strength. We introducethe local CS:

~e1 =

−−→AjAj+1

|−−→AjAj+1|= e11

~i + e12~j,

~e2⊥~e1 =⇒ ~e2 = −e12~i + e11

~j,

~a = a1~i + a2

~j (global CS),

~a = a′1 ~e1 + a′

2 ~e2 (local CS),

a′1 = e11a1 + e12a2

a′2 = −e12a1 + e11a2

global to local CS,

a1 = e11a′1 − e12a

′2

a2 = e12a′1 + e11a

′2

local to global CS.

We need to find [∇P Dj(P ) · ~n]. This is the influence coefficient at the point P. Theinfluence coefficient Eij of equation (10.9) will be obtained if we set P = Qi, ~n = ~n(Qi).

1. We should find the vector−−→O′P in the global CS.

2. We transform the vector−−→O′P to the local CS. Let its coordinates in LCS be (x′, y′).

73

3. We are finding the induced velocity (u′, v′) at the point (x′, y′) in the local CS.

4. We transform the velocity (u′, v′) to the global CS to get the velocity components(u, v).

5. We find the influence coefficient at the point P :

infl.coef. = nxu + nyv.

10.5.3 Morino’s formulation for the total potential

Φ(P ) =N+1∑

j=1

µjDj(P ) +N∑

j=1

σj︸︷︷︸0

Sj(P ) + Φ∞(P ).

Let P → Qj. We have the BC

∂Φ

∂n= 0 =⇒ σj = 0.

We recall that µj = −Φ(Qj) and obtain a discrete analog of Morino’s equation

−µi =N+1∑

j=1

µjD+j (P ) + Φ∞(Qi), i = 1, 2, . . . , N,

µN+1 = µN − µ1 (Kutta condition).

10.5.4 Super Morino’s formulation

Now we assume that σj 6= 0

1. We satisfy the Morino’s integral equation.

2. We satisfy directly the BC ∂Φ∂n

= 0.

Morino’s equation:

−µi =N+1∑

j=1

µjD+j (Qi) +

N∑

j=1

σj︸︷︷︸6=0

Sj(Qi) + Φ∞(Qi).

BC: ∇Φ(Qi) · ~n(Qi) = 0 =⇒

N+1∑

j=1

µj[∇P Dj(Qi) · ~n(Qi)] +N∑

j=1

σj [∇P Sj(Qi) · ~n(Qi)]+ + ~q∞ · ~n(Qi) = 0,

µN+1 = µN − µ1 (Kutta condition)

74

10.5.5 Computation of the pressure distribution

From Bernoulli’s equation we find that

Cp = 1 − q2

q2∞

.

We need to compute the velocity q on the surface of the airfoil. If we have only thesource distribution (µ(Q) = 0) we can compute the induced velocity directly at thecollocation points. But if we use the formulation in which µ(Q) 6= 0, for example, Morino’sformulation the direct computation of the induced velocities leads to significant errors. ForMorino’s formulation Φj = −µj , so we know the values of the potential in the collocationpoints. To compute q we proceed as follows

1. We compute the lengths:

|−−→BQ1|, |−−→Q1Q2|, . . . , |

−−−→QNB|.

2. We locate these lengths on the s-axis. Now we are using the formula

q =dΦ

ds.

Approximate differentiation can be made, for example, by means of the cubic splineinterpolation.

11 Full potential equations for irrotational isentropic

steady flows

11.1 Isentropic process

1. Adiabatic process. One in which no heat is added to or removed from the system.

2. Reversible process. One in which no dissipative phenomena occur, in particularviscosity forces are absent.

3. Isentropic process. One which is both adiabatic and reversible.

For the isentropic process the entropy

S =∫

dq

T= const

and the square of the velocity of sound is defined by the formula

a2 =dp

dρ. (11.1)

Moreover, the following isentropic relation holds

a2

a2∞

=1

M2∞

+κ − 1

2

(1 − q2

q2∞

)(11.2)

and connects the local velocity of sound a and the local velocity of the fluid q. In (11.2)M∞ = a∞/q∞ is the Mach number at infinity, κ = 1.4 (for air) is the isentropic exponent.

75

11.2 Deducing the full potential equation

We assume that the flow is irrotational, i.e. rot ~q = 0, steady and isentropic. Thecontinuity equation gives

∂(ρu)

∂x+

∂(ρv)

∂y+

∂(ρw)

∂z= 0 =⇒ div (ρ~q) = 0,

(∇ρ · ~q) + ρdiv ~q = 0,

div (~q) = −(∇ρ · ~q)ρ

. (11.3)

The momentum equation in the vector form :

∇q2

2− (~q × rot ~q︸ ︷︷ ︸

0

) = −∇p

ρ

or

∇q2

2= −∇p

ρ. (11.4)

The relation

a2 =dp

dρ=⇒ dp = a2dρ.

It follows from here that∇p = a2∇ρ. (11.5)

Indeed,

dp =∂p

∂xdx +

∂p

∂ydy +

∂p

∂zdz,

a2dρ = a2 ∂ρ

∂xdx + a2 ∂ρ

∂ydy + a2∂ρ

∂zdz.

By comparing these differential forms we get

∂p

∂x= a2 ∂ρ

∂x,

∂p

∂y= a2 ∂ρ

∂y,

∂p

∂z= a2 ∂ρ

∂x,

and, therefore, equation (11.5) holds true.Let us take the momentum equation (11.4) and multiply it by ~q (scalar product):

(~q · ∇q2)

2= −(∇p · ~q)

ρ.

Instead of ∇P we insert in the RHS the expression (11.5) to obtain

(~q · ∇q2)

2= −a2(∇ρ · ~q)

ρ

or(∇ρ · ~q)

ρ= −(~q · ∇q2)

2a2.

76

Now making use of the continuity equation (11.3) gives

div ~q =(~q · ∇q2)

2a2. (11.6)

In equation (11.6) we have excluded p and ρ. The connection between a and ~q is givenby the equation (11.2). For the irrotational flow

rot ~q = 0 =⇒ ~q = ∇Φ,

where Φ is the total potential. Inserting ~q = ∇Φ in (11.6) and carrying out the differen-tiation, we obtain the following equation for Φ

∂2Φ

∂x2+

∂2Φ

∂y2+

∂2Φ

∂z2=

u2

a2

∂2Φ

∂x2+

v2

a2

∂2Φ

∂y2+

w2

a2

∂2Φ

∂z2+

+2uv

a2

∂2Φ

∂x∂y+

2uw

a2

∂2Φ

∂x∂z+

2vw

a2

∂2Φ

∂y∂z.

(11.7)

The equation (11.7) calls the full potential equation. A mnemonic rule to remember theRHS of this equation is in comparing it with the formula

(u + v + w)2 = u2 + v2 + w2 + 2uv + 2uw + 2vw.

If we denote the RHS of the equation by σv, we obtain

div ~q = σv.

If to consider ~q(P ) as the velocity field of an incompressible fluid, then the divergenceis the strength of a source, that is located at the point P . Thus (div ~q(P ) · ∆V ) is avolume quantity of fluid that appears (or disappears, depending on the sign div ~q) in thesmall volume ∆V arrounding the point P per unit of time. This gives us the possibilityto model compressible flows by incompressible ones by allowing the fluid to appear ordisappear in the flow domain.

11.3 Short form of the full potential equation

Taking into account that

∇q2 =∂q2

∂x~i +

∂q2

∂y~j +

∂q2

∂z~k = 2q

∂q

∂x~i + 2q

∂q

∂y~j + 2q

∂q

∂z~k = 2q∇q

and using (11.6) we can write

div ~q =[~q ·

scalar︷ ︸︸ ︷(2q) ∇q]

2a2=

2q(~q · ∇q)

2a2=

q2

a2

(~q

q· ∇q

).

Consider any streamline of the flow and let ~s be the unit vector of the streamline. Then

div ~q =q2

a2(~s · ∇q) =

q2

a2

∂q

∂s.

77

On the streamline we have q = ∂Φ∂s

, where ∂∂s

is the derivative in the direction of ~s. Notingthat q/a = M is the local Mach number, we get

div ~q = M2 ∂2Φ

∂s2,

∇2Φ = M2 ∂2Φ

∂s2. (11.8)

Equation (11.8) is the short form of the full potential equation.The laplacian ∇2Φ is invariant with respect to the system of coordinates. Let us

introduce the coordinate system that is connected with the streamline. Then we have

∂2Φ

∂s2+

∂2Φ

∂m2+

∂2Φ

∂n2= M2 ∂2Φ

∂s2,

(1 − M2)∂2Φ

∂s2+

∂2Φ

∂m2+

∂2Φ

∂n2= 0. (11.9)

As one can see from (11.9) for M < 1 the full potential equation is of elliptic type, butfor M > 1 the one is of hyperbolic type. Shock waves in the compressible fluid are theresult of the change in the type of the equation.

11.4 Goehert compressibility correction

Consider the case of a small perturbation flow. This means that the body in the flowis slender, the angle of attack is small. Let us choose the system of coordinates so thatthe flow at infinity is parallel to the x-axis, i.e. ~q∞ = u∞

~i. In this case it is possible toshow that

∂2Φ

∂s2≈ ∂2Φ

∂x2, M ≈ M∞.

The full potential equation takes the form

∇2Φ = M2∞

∂2Φ

∂x2=⇒ (1 − M2

∞)∂2Φ

∂x2+

∂2Φ

∂y2+

∂2Φ

∂z2= 0. (11.10)

This equation is called the linearized full potential equation. Let us introduce the pertur-bation potential

Φ(P ) = Φ∞(P ) + φ(P ), φ(P ) = Φ(P ) − u∞x,

∂φ

∂x=

∂Φ

∂x− u∞,

∂2φ

∂x2=

∂2Φ

∂x2,

∂2φ

∂y2=

∂2Φ

∂y2,

∂2φ

∂z2=

∂2Φ

∂z2.

Inserting these derivatives in (11.10) we see that the linearized full potential equation forφ is the same as for Φ:

(1 − M2∞)

∂2φ

∂x2+

∂2φ

∂y2+

∂2φ

∂z2= 0. (11.11)

Assume that the flow is subsonic: 0 < M∞ < 1 and denote β2 = 1 − M2∞. Make the

change of variablesx = x, y = βy, z = βz, φ = β2φ,

φ(x, y, z) = β2φ(x, y, z), φ =φ

β2.

78

Differentiation:∂φ

∂x=

∂φ

∂x

∂x

∂x︸︷︷︸1

=1

β2

∂φ

∂x,

∂2φ

∂x2=

∂

∂x

(1

β2

∂φ

∂x

)=

1

β2

∂

∂x

(∂φ

∂x

)∂x

∂x︸︷︷︸1

=1

β2

∂2φ

∂x2,

∂2φ

∂x2=

1

β2

∂2φ

∂x2.

∂φ

∂y=

∂φ

∂y

∂y

∂y︸︷︷︸β

=1

β2

∂φ

∂yβ =

1

β

∂φ

∂y,

∂2φ

∂y2=

∂

∂y

(1

β

∂φ

∂y

)=

∂

∂y

(1

β

∂φ

∂y

)∂y

∂y︸︷︷︸β

=∂2φ

∂y2,

∂2φ

∂y2=

∂2φ

∂y2.

∂φ

∂z=

1

β

∂φ

∂z,

∂2φ

∂z2=

∂2φ

∂z2.

Inserting these derivatives in (11.11) we obtain for φ the Laplace’s equation:

∂2φ

∂x2+

∂2φ

∂y2+

∂2φ

∂z2= 0 (Laplace’s equation).

BC at infinity is

∇φ → 0 as P (x, y, z) → ∞ =⇒ ∇φ → 0 as P (x, y, z) → ∞.

BC on the surface of the body is

∂φ

∂n= −(~q∞ · ~n) =⇒ (∇φ · ~n) = −(~q∞ · ~n),

where ~n is the outer normal to the body surface. Assume that the surface of the body isdefined by the equation

z = f(x, y).

Because the body is slender, we have

∣∣∣∣∣∂φ

∂x

∣∣∣∣∣ << 1,

∣∣∣∣∣∂f

∂x

∣∣∣∣∣ << 1,

∣∣∣∣∣∂f

∂y

∣∣∣∣∣ = O(1).

If we denotef(x, y) − z = F (x, y, z), then F (x, y, z) = 0,

∂F

∂x=

∂f

∂x=⇒

∣∣∣∣∣∂F

∂x

∣∣∣∣∣ << 1.

The normal unit vector to the surface is defined by the formula

~n =∇F

|∇F | .

79

BC on the surface of the body takes the form(∇φ · ∇F

|∇F |

)= −

(~q∞ · ∇F

|∇F |

),

or(∇φ · ∇F ) = −(~q∞ · ∇F ).

Taking into account that ~q∞ = u∞~i we get

(∇φ · ∇F ) = −u∞

∂F

∂x,

∂φ

∂x

∂F

∂x+

∂φ

∂y

∂F

∂y+

∂φ

∂z

∂F

∂z= −u∞

∂F

∂x.

Because ∣∣∣∣∣∂φ

∂x

∣∣∣∣∣ << 1,

∣∣∣∣∣∂F

∂x

∣∣∣∣∣ << 1

the first term ∂φ∂x

∂F∂x

≈ 0 and the BC can be written as

u∞

∂F

∂x+

∂φ

∂y

∂F

∂y+

∂φ

∂z

∂F

∂z= 0. (11.12)

Equation (11.12) is the linearized impermeability condition on the surface of the body.Now we rewrite this equation in terms of x, y, z. Differentiation gives

∂φ

∂y=

1

β

∂φ

∂y,

∂φ

∂z=

1

β

∂φ

∂z,

∂F

∂y=

∂F

∂y

∂y

∂y= β

∂F

∂y,

∂F

∂z=

∂F

∂z

∂z

∂z= β

∂F

∂z,

∂F

∂x=

∂F

∂x.

And finally the BC is

u∞

∂F

∂x+

∂φ

∂y

∂F

∂y+

∂φ

∂z

∂F

∂z= 0. (11.13)

Comparing (11.12) and (11.13) one can see that they are of the same form. This meansthat if we consider φ as the perturbation potential of a certain imaginary incompressibleflow, this potential satisfies the impermeability condition on the surface of a new bodydefined by the equation F (x, y, z) = 0.

In the linear theory the pressure coefficient Cp can be calculated by the formula

Cp = −2∂φ∂x

u∞

.

Since∂φ

∂x=

1

β2

∂φ

∂x

80

we have

Cp = −2∂φ

∂x

u∞β2=

Cp

β2,

where by Cp we denote the pressure coefficient of the imaginary flow.Now we can formulate the Goehert rule:

1. Obtain a new body by means of the transformations

x = x, y = βy, z = βz.

2. Consider the imaginary flow of an incompressible fluid over this new body andcompute Cp on its surface.

3. The real Cp coefficient is given by

Cp =Cp

β2.

12 Field panel method

12.1 Idea of the method

Consider a body in an incompressible fluid. Let Φ(P ) be the total potential. Then wehave the following BVP:

∇2Φ = 0, (Laplace’s equation), (12.1)

∂Φ

∂n= 0, (BC on the body surface), (12.2)

∇Φ → ~q∞ as P → ∞. (12.3)

We know that

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS

orΦ(P ) = Φ∞(P ) +

∫ ∫

S

µ(Q)fd(P, Q)dS +∫ ∫

S

σ(Q)fs(P, Q)dS, (∗)

and the representation (*) for Φ satisfies automatically (12.1) and (12.3) for any sourceand doublet distributions σ(Q) and µ(Q). These distributions should be found so, thatthe BC (12.3) would be satisfied. To find them we can use any of formulations that wehave studied (Morino’s formulation, source-only formulation and so on).

Now consider the same body but in a compressible fluid. The BC for Φ are the sameas for the incompressible flow, so the potential Φ must satisfy (12.2) and (12.3). But inthe flow domain we have the full potential equation:

∇2Φ = σv (12.4)

81

with

σv =u2

a2

∂2Φ

∂x2+

v2

a2

∂2Φ

∂y2+

w2

a2

∂2Φ

∂z2+

2uv

a2

∂2Φ

∂x∂y+

2uw

a2

∂2Φ

∂x∂z+

2vw

a2

∂2Φ

∂y∂z.

(12.5)

Using the same method as in section 4 it is possible to show that

Φ(P ) = Φ∞(P ) −∫ ∫

S

Φ(Q)fd(P, Q)dS +∫ ∫

S

∂Φ

∂n(Q)fs(P, Q)dS +

∫ ∫ ∫

V

(∇2Φ)fs(P, Q)dV.

(12.6)The formula (12.6) is correct for any function that has second continuous partial deriva-tives. Let us assume that σv(P ) is a known function. Then

Φ(P ) = Φ∞(P ) +∫ ∫

S

µ(Q)fd(P, Q)dS +∫ ∫

S

σ(Q)fs(P, Q)dS +∫ ∫ ∫

V

σv(Q)fs(P, Q)dV.

If we denoteΩ∞(P ) = Φ∞(P ) +

∫ ∫ ∫

V

σv(Q)︸ ︷︷ ︸known

fs(P, Q)dV, (12.7)

we obtain

Φ(P ) = Ω∞(P )︸ ︷︷ ︸known

+∫ ∫

S

µ(Q)fd(P, Q)dS +∫ ∫

S

σ(Q)fs(P, Q)dS. (∗∗)

Comparing (*) and (**) one can see that we have no difference. So to satisfy the BC(12.2) we can use the panel method and find the source and doublet distributions σ(Q)and τ(Q). The only equation that will not be satisfied is equations (12.5), because wehave specified the volume source distribution σv(P ) ourselves. To find σv(P ) we applythe iterative process:

1. Specify σv(Q), e.g. σv(Q) = 0.

2. Using the formula (12.7) find Ω∞(P ).

3. Using (**) and any of known formulations find µ(Q), σ(Q).

4. Using equation (12.5) find the new σv(Q).

5. Go to step 2.

To compute the volume integral Sv(P ) =∫∫∫

Vσv(Q)fs(P, Q)dV we shall use field panels.

12.2 Field panel

We represent the space around the body by a collection of parallelepipeds, that arecalled field panels. Let us denote every field panel by ∆Vi. Every field panel is specifiedby the location of one angular point Ai(xi, yi, zi) and by the lengths of edges ai, bi, ci (seefig. 41). The basic field panel potential is

82

Figure 41:

Svi(P ) =∫ ∫ ∫

∆Vi

fs(P, Q)dS.

This integral can be computed analytically. As a result of integration we obtain that thefunction Svi depends on nine variables:

Svi = Fvs(xi, yi, zi, ai, bi, ci, x, y, z).

Inside every panel we choose the collocation point

Qi(xi + ai/2, yi + bi/2, zi + ci/2).

So Qi is the centroid of the panel. Now we set σvi = σv(Qi) to get the discrete represen-tation of the volume integral

Sv(P ) =Nv∑

i=1

σviSvi.

12.2.1 Subsonic, critical and transonic flows. Quasi-isentropic flows

The full potential equation was deduced under the assumption that the flow is isen-tropic. But if we have the shock wave in the flow we will have the jump ∆S of theentropy.

It is known that the jump ∆S is a function only the Mach number M1 before theshock. In the region 1 < M1 < 1.3 the jump ∆S is less than 0.02. So, the flow can beconsidered approximately isentropic. The flows with the local Mach number less than 1.3calls quasi-isentropic. For such flows we can still use the full-potential equation.

12.3 Artificial viscosity

Numerical experiments show that the described above iterative procedure of findingthe volume distribution of sources σv converges only if the flow is entirely subsonic. If inthe flow supersonic regions appears the procedure should be corrected to take into accountthe jumps of the flow properties across the shocks. This can be made by the method ofartificial viscosity that comes from finite difference method (FDM). Consider again thefull potential equation

∇2Φ = M2 ∂2Φ

∂s2,

83

Figure 42:

where s is the direction of the streamline, that goes trough the point P . In FDM the ∂2Φ∂s2

can be computed by

• central differences(

∂2Φ∂s2

)C;

• left (retarded) differences(

∂2Φ∂s2

)L;

• right differences(

∂2Φ∂s2

)R.

Analytical computations we shall denote by(

∂2Φ∂s2

)A. The formulae for central and left

differences are (∂2Φ

∂s2

)C

i

=Φi+1 − 2Φi + Φi−1

(∆s)2,

(∂2Φ

∂s2

)L

i

=Φi − 2Φi−1 + Φi−2

(∆s)2=

(∂2Φ

∂s2

)C

i−1

.

The idea of the artificial viscosity method is to use left differences in the supersonic zone(M > 1) and more accurate central differences in the subsonic zone (M < 1). This allowsthe jump of the derivative ∂2Φ

∂s2 across the shock to be computed. In the field panel method

we shall use the analytical computation of ∂2Φ∂s2 in subsonic zone, but in the supersonic

zone we add to the analytical derivative a special term that calls artificial viscosity andequivalent to use of the left differences in the supersonic zone (see fig. 42).

Consider the difference

(∂2Φ

∂s2

)C

i

−(

∂2Φ

∂s2

)L

i

=

(∂2Φ∂s2

)C

i−(

∂2Φ∂s2

)C

i−1

∆s∆s =

∂3Φ

∂s3∆s,

(∂2Φ

∂s2

)C

i

=

(∂2Φ

∂s2

)L

i

+∂3Φ

∂s3∆s,

(∂2Φ

∂s2

)L

i

=

(∂2Φ

∂s2

)C

i

− ∂3Φ

∂s3∆s.

84

The central difference gives the most accurate result because of that we can set that(

∂2Φ

∂s2

)C

i

=

(∂2Φ

∂s2

)A

i

and therefore (∂2Φ

∂s2

)L

i

=

(∂2Φ

∂s2

)A

i

− ∂3Φ

∂s3∆s.

Now from the full potential equation we deduce

∂2Φ

∂s2+

∂2Φ

∂m2+

∂2Φ

∂n2= M2 ∂2Φ

∂s2,

∂2Φ

∂m2+

∂2Φ

∂n2= (M2 − 1)

∂2Φ

∂s2,

∂2Φ

∂m2+

∂2Φ

∂n2=

(M2 − 1)∂2Φ

∂s2, M < 1,

(M2 − 1)∂2Φ

∂s2− (M2 − 1)

∂3Φ

∂s3∆s, M ≥ 1.

In the last formula we have taken into account that in the supersonic zone we use the leftdifferences to compute ∂2Φ

∂s2 .

∂2Φ

∂s2+

∂2Φ

∂m2+

∂2Φ

∂n2=

M2 ∂2Φ

∂s2, M < 1,

M2 ∂2Φ

∂s2− (M2 − 1)

∂3Φ

∂s3∆s, M ≥ 1,

∂2Φ

∂s2+

∂2Φ

∂m2+

∂2Φ

∂n2=

M2 ∂2Φ

∂s2, M < 1,

M2[∂2Φ

∂s2−(1 − 1

M2

)∂3Φ

∂s3∆s

︸ ︷︷ ︸artificial viscosity

], M ≥ 1. (12.8)

Equation (12.8) is the full potential equation with a special term in the supersonic zonethat is the artificial viscosity. How to compute the ∂3Φ

∂s3 ? We denote

σv = M2 ∂2Φ

∂s2.

It follows from here that∂2Φ

∂s2=

σv

M2,

∂3Φ

∂s3=

∂

∂s

(∂2Φ

∂s2

)=

∂

∂s

(σv

M2

)=(∇ σv

M2

)· ~s.

Here ~s is the unite vector along the streamline. Therefore ~s = ~q/q and

∂3Φ

∂s3=

∂

∂x

(σv

M2

)u

q+

∂

∂y

(σv

M2

)v

q+

∂

∂z

(σv

M2

)w

q.

The derivatives∂

∂x

(σv

M2

),

∂

∂y

(σv

M2

),

∂

∂z

(σv

M2

)

can be computed by some simple finite difference method. The value of ∆s must beproportional to the length of the edges of the field panel, e.g. ∆s = max(ai, bi, ci).

85

13 Formulae for computing induced potentials and

velocities of basic two-dimensional elements

Figure 43: a) Nomenclature for calculating the influence of the point source, doublet andvortex; b) the same for the constant-strength doublet and source

Two-dimensional uniform flow

Φ = (~q∞ · ~r) = u∞x + v∞y, ~q = ~q∞ = u∞~i + v∞~j

Two-dimensional point source

Φ =σ

2πln r, ~q =

σ

2π

~r

r2,

where~r = (x − ξ)~i + (y − ξ)~j, r =

√x − ξ)2 + (y − η)2.

Two-dimensional point doublet

Φ = − µ

2π

(~n · ~r)r2

, ~q = − µ

2πr2

[~n − 2(~n · ~r)~r

r2

]

Two-dimensional point vortex

Φ = − γ

2πθ

θ = atan2(η − y, ξ − x) + π

u =γ

2π

y − η

(x − ξ)2 + (y − η)2

v = − γ

2π

x − ξ

(x − ξ)2 + (y − η)2

Two-dimensional constant-strength source distribution

θk = atan2(−y, xk − x) + π, k = 1, 2

rk =√

(x − xk)2 + y2

86

Φ =σ

4π

[(x − x1) log r2

1 − (x − x2) log r22 + 2y(θ2 − θ1)

]

u =σ

4πlog

r21

r22

, v =σ

2π(θ2 − θ1)

On the element

u(x,±0) =σ

2πlog

x − x1

x2 − x, v(x,±0) = ±σ

2

Two-dimensional constant-strength doublet distribution

Φ = − µ

2π(θ2 − θ1)

u =µ

2π

[y

r22

− y

r21

], v = − µ

2π

[x − x2

r22

− x − x1

r21

]

On the element

Φ(x,±0) = ∓µ

2, u(x,±0) = 0, v(x,±0) = − µ

2π

(1

x − x2

− 1

x − x1

)

Combination of the uniform stream and 2D-doublet

Φ = (~q∞ · ~r) − µ

2π

(~n · ~r)r2

For the flow over cylinder of radius R:

~n = −~q∞q∞

, µ = 2πq∞R2

14 Seminar tasks (1)

Variant 0 (example)

Investigate the streamlines and equipotential lines for the combination of an uniformstream and a doublet .

Variant 1

Investigate the streamlines and equipotential lines for the combination of an uniformstream and two sources. Stress a special attention to the case of a source and sink ofequal strengths.

Variant 2

Investigate the streamlines and equipotential lines for the combination of an uniformstream and vortexes. Stress a special attention to the case the vortexes with equal butopposite circulations.

87

Variant 3

Investigate the streamlines and equipotential lines for the combination of an uniformstream and an uniform source distribution on the segment (0.1) .

Variant 4

Investigate the streamlines and equipotential lines for the combination of the uniformstream and an uniform doublet distribution on the segment (0.1) .

Variant 5

Investigate the streamlines and equipotential lines for the combination of an uniformstream and a vortex and source . Stress a special attention to the case of the vortex andsource located at the same point.

Variant 6

Investigate the streamlines and equipotential lines for the combination of an uniformstream and a doublet and source. Stress a special attention to the case for which the axisof the doublet is opposite to the uniform flow and the source locates on this axis behindthe doublet.

Variant 7

Investigate the streamlines and equipotential lines for the combination of the uniformstream and three sources . Stress a special attention to the case of sources with zero sumof strengths.

15 Seminar tasks (2)

Variant 0 (example)1. By means of the code gukvx generate the shape of a Jukovskiy airfoil for N = 60.2. Using Morino’s formulation for the total potential Φ, create the panel code that

finds the pressure distribution at the collocation points and the lift coefficient for anarbitrary airfoil.

3. Setting N=500 in the gukvx code find an accurate pressure distribution on theairfoil surface at the angle of attack α = 5o.

4. Prepare the graphs and compare the results of the panel codes for different Nwith the accurate pressure distribution, obtained by guk.vx for N = 500. Pay a specialattention to the suction peak near the leading edge. Carry out the comparison by settingN = 30, 50, 70, 90 in the panel code.

Variant 11. By means of the code gukvx generate the shape of a Jukovskiy airfoil for N = 60.2. Using Morino’s formulation for the perturbation potential φ, create the panel code

that finds the pressure distribution at the collocation points and the lift coefficient for anarbitrary airfoil.

88

3. Setting N=500 in the gukvx code find an accurate pressure distribution on theairfoil surface at the angle of attack α = 5o.

4. Prepare the graphs and compare the results of the panel codes for different Nwith the accurate pressure distribution, obtained by gukvx for N = 500. Pay a specialattention to the suction peak near the leading edge. Carry out the comparison by settingN = 30, 50, 70, 90 in the panel code.

Variant 21. By means of the code gukvx generate the shape of a Jukovskiy airfoil for N = 60.2. Using the doublet-only formulation create the panel code that finds the pressure

distribution at the collocation points and the lift coefficient for an arbitrary airfoil.3. Setting N=500 in the gukvx code find an accurate pressure distribution on the

airfoil surface at the angle of attack α = 5o.4. Prepare the graphs and compare the results of the panel codes for different N

with the accurate pressure distribution, obtained by gukvx for N = 500. Pay a specialattention to the suction peak near the leading edge. Carry out the comparison setting byN = 30, 50, 70, 90 in the panel code.

Variant 31. By means of the code gukvx generate the shape of a Jukovskiy airfoil for N = 60.2. Using the formulation with prescribed doublet strengths create the panel code

that finds the pressure distribution at the collocation points and the lift coefficient for anarbitrary airfoil shape. The strength of doublet along the airfoil surface should be a linearfunction of the arc-length and the function should be chosen so, that Kutta’s conditionwould be satisfied. To satisfy Kutta’s condition create a wake panel at the trailing edgethat bisects the trailing edge angle. On this panel the impermeability condition must befulfilled.

3. Setting N=500 in the gukvx code find an accurate pressure distribution on theairfoil surface at the angle of attack α = 5o.

4. Prepare the graphs and compare the results of the panel codes for different Nwith the accurate pressure distribution, obtained by gukvx at N = 500. Pay a specialattention to the suction peak near the leading edge. Carry out the comparison by settingN = 30, 50, 70, 90 in the panel code.

Variant 41. By means of the code gukvx generate the shape of a symmetrical Jukovskiy airfoil

for N = 60.2. Using the source-only formulation create the panel code that finds the pressure

distribution at the collocation points for an arbitrary symmetrical airfoil.3. Setting N=500 in the gukvx code find an accurate pressure distribution on the

airfoil surface at the angle of attack α = 0o.4. Prepare the graphs and compare the results of the panel codes for different N

with the accurate pressure distribution, obtained by gukvx at N = 500.Carry out thecomparison by setting N = 30, 50, 70, 90 in the panel code.

89

Variant 51. By means of the code gukvx generate the shape of a Jukovskiy airfoil for N = 60.2. Using super Morino’s formulation for the total potential Φ, create the panel code

that finds the pressure distribution at the collocation points and the lift coefficient for anarbitrary airfoil.

3. Setting N=500 in the gukvx code find an accurate pressure distribution on theairfoil surface at the angle of attack α = 5o.

4. Prepare the graphs and compare the results of the panel codes for different N withthe accurate pressure distribution, obtained by gukvx for N = 500. Stress a specialattention to the suction peak near the leading edge. Carry out the comparison setting byN = 30, 50, 70, 90 in the panel code.

Variant 61. By means of the code gukvx generate the shape of a Jukovskiy airfoil for N = 60.2. Using super Morino’s formulation for the perturbation potential φ, create the panel

code that finds the pressure distribution at the collocation points and the lift coefficientfor an arbitrary airfoil.

3. Setting N=500 in the gukvx code find an accurate pressure distribution on theairfoil surface at the angle of attack α = 5o.

4. Prepare the graphs and compare the results of the panel codes for different N withthe accurate pressure distribution, obtained by gukvx for N = 500. Stress a specialattention to the suction peak near the leading edge. Carry out the comparison by settingN = 30, 50, 70, 90 in the panel code.

90

x

y

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Frame 001 27 May 2003 Frame 001 27 May 2003

Figure 44: Streamlines for the 2D-doublet: q∞ = 0, ~n = (−1, 0), µ = π/2.

x

y

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Frame 001 27 May 2003 Frame 001 27 May 2003

Figure 45: Streamlines for the combination of the uniform stream and 2D-doublet: q∞ =(1, 0), ~n = (−1, 0), µ = π/2.

91

x

y

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Frame 001 27 May 2003 Frame 001 27 May 2003

Figure 46: Streamlines for the combination of the uniform stream and 2D-doublet: ~q∞ =(1, 0), ~n = (−

√2/2,

√2/2), µ = π/2.

x

y

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

fi10.90.80.70.60.50.40.30.20.10

-0.1-0.2-0.3-0.4-0.5-0.6-0.7-0.8-0.9-1

Frame 001 28 May 2003 Frame 001 28 May 2003

Figure 47: Streamlines and equipotential lines for the combination of the uniform streamand 2D-doublet: ~q∞ = (1, 0), ~n = (−

√2/2,

√2/2), µ = π/2.

92

x

y

0.25 0.5 0.75 1-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Frame 001 13 Jun 2003 ComparisonFrame 001 13 Jun 2003 Comparison

Figure 48: The shape of the computed Joukovskiy airfoil: gam=15, H=0.1, D=0.03

x

y,C

p,C

pin

0.25 0.5 0.75 1

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

Frame 001 13 Jun 2003 ComparisonFrame 001 13 Jun 2003 Comparison

Figure 49: Comparisons of the pressure distributions. Adding to the sum of the inducedvelocities the lost singular component.

93

x

y,C

p,C

pin

0.25 0.5 0.75 1

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

Frame 001 13 Jun 2003 ComparisonFrame 001 13 Jun 2003 Comparison

Figure 50: Comparisons of the pressure distributions. Without adding the singular com-ponent.