MA 201, Mathematics III, July-November 2018, Part II ... · Steady state heat conduction:...

MA 201, Mathematics III, July-November 2018,

Part II: Partial Differential Equations

Laplace’s equation as steady-state heat conduction

equation

Lecture 15

Lecture 15 MA 201, PDE (2018) 1 / 37

Steady state heat conduction: Laplace’s equation

Laplace’s equation in two or three dimensions

usually arises in two types of physical problems:

1. As steady state heat conduction.

2. As equation of continuity for incompressible potential flow.

However, here we will emphasize only on the first type.

Steady state solution here means

1. the solution for large time.

2. the solution does not depend anymore on time.

Lecture 15 MA 201, PDE (2018) 2 / 37

Laplace’s equation

Laplace’s equation in two dimensions and three dimensions in Cartesian

coordinates, are , respectively, given by

uxx + uyy = 0, (1)

uxx + uyy + uzz = 0. (2)

The above equations can be obtained from the two-dimensional and three-dimensional

transient heat conduction equations when u does not depend on t.

Hence Laplace’s equation models

steady heat flow in a region where the temperature is fixed on the boundary.

Lecture 15 MA 201, PDE (2018) 3 / 37

Maximum Principle

Theorem: Let u(x, y) satisfy Laplace’s equation in D, an open, bounded, connected

region in the plane; and let u be continuous on the closed domain D ∪ ∂D consisting

of D and its boundary. If u is not a constant function, then the maximum and

minimum values of u are attained on the boundary of D and nowhere inside D.

. �

This is called maximum principle theorem for Laplace’s equation.

Lecture 15 MA 201, PDE (2018) 4 / 37

Steady state heat conduction in two dimensions

We consider steady state heat conduction

in a two-dimensional rectangular region.

To be specific,

consider the equilibrium temperature inside a rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b.

Here

the temperature is a prescribed function of position on the boundary.

In general the Dirichlet BVP will be like

uxx + uyy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b

u(0, y) = g1(y), u(a, y) = g2(y), 0 ≤ y ≤ b

u(x, 0) = f1(x), u(x, b) = f2(x), 0 ≤ x ≤ a.

Lecture 15 MA 201, PDE (2018) 5 / 37


where

f1(x), f2(x), g1(y), g2(y) are given functions.

Though the equation is linear and homogenous,

the BCs are not homogenous.

Hence

the BVP is needed to be split into four BVPs with each containing one

non-homogenous BC.

Take

u = u1 + u2 + u3 + u4, 0 ≤ x ≤ a, 0 ≤ y ≤ b.

Lecture 15 MA 201, PDE (2018) 6 / 37


BVP I and BVP II:

u1,xx + u1,yy = 0; u2,xx + u2,yy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b;

u1(0, y) = 0, 0 ≤ y ≤ b; u2(0, y) = 0, 0 ≤ y ≤ b;

u1(a, y) = 0, 0 ≤ y ≤ b; u2(a, y) = 0, 0 ≤ y ≤ b;

u1(x, 0) = f1(x), 0 ≤ x ≤ a; u2(x, 0) = 0, 0 ≤ x ≤ a;

u1(x, b) = 0, 0 ≤ x ≤ a; u2(x, b) = f2(x), 0 ≤ x ≤ a;

BVP III and BVP IV:

u3,xx + u3,yy = 0; u4,xx + u4,yy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b;

u3(0, y) = g1(y), 0 ≤ y ≤ b; u4(0, y) = 0, 0 ≤ y ≤ b;

u3(a, y) = 0, 0 ≤ y ≤ b; u4(a, y) = g2(y), 0 ≤ y ≤ b;

u3(x, 0) = 0, 0 ≤ x ≤ a; u4(x, 0) = 0, 0 ≤ x ≤ a;

u3(x, b) = 0, 0 ≤ x ≤ a; u4(x, b) = 0, 0 ≤ x ≤ a.

We will consider only one of them.......take u1 = u for convenience

Lecture 15 MA 201, PDE (2018) 7 / 37


Consider the steady state heat conduction in a rectangular region

0 ≤ x ≤ a, 0 ≤ y ≤ b

where three boundaries along x = 0, x = a, y = b are kept at 00C

while

the temperature along the boundary y = 0 is f(x).

To find the temperature at any point (x, y).

Lecture 15 MA 201, PDE (2018) 8 / 37


BVP will consist of the following:

The governing equation is two-dimensional Laplace’s equation:

uxx + uyy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b. (3)

The boundary conditions are:

u(0, y) = 0, 0 ≤ y ≤ b, (4a)

u(a, y) = 0, 0 ≤ y ≤ b, (4b)

u(x, 0) = f(x), 0 ≤ x ≤ a, (4c)

u(x, b) = 0, 0 ≤ x ≤ a. (4d)

It being a pure BVP and the solution being a function of x and y,

obviously we will not have any initial conditions.

Hence

This problem is called a steady-state problem.

Lecture 15 MA 201, PDE (2018) 9 / 37


Assume a solution of the form:

u(x, y) = X(x)Y (y). (5)

Using (5) in (3)

X ′′

X+

Y ′′

Y= 0

On separating the variables x and y,

X ′′

X= −

Y ′′

Y= k(say).

Giving us

X ′′ − kX = 0, (6)

Y ′′ + kY = 0. (7)

The zero and positive values of k will not give rise to solutions conforming to the

boundary conditions.

Lecture 15 MA 201, PDE (2018) 10 / 37


We consider only the negative values of k, say − λ2, to write the equations (6)

and (7) as

X ′′ + λ2X = 0, (8)

Y ′′ − λ2Y = 0, (9)

so that the solution u(x, y) can be written as

u(x, y) = (A cos λx+B sinλx)(C coshλy +D sinhλy). (10)

Using boundary condition (4a)

A = 0.

Using boundary condition (4b),

λn =nπ

a, n = 1, 2, 3, . . .

⇒

un(x, y) = sinnπx

a

(

An coshnπy

a+Bn sinh

nπy

a

)

.

Lecture 15 MA 201, PDE (2018) 11 / 37


Using boundary condition (4d)

Bn = −cosh nπb

a

sinh nπba

An

so that the solution u(x, y) can be written as

u(x, y) =

∞∑

n=1

un(x, y) =

∞∑

n=1

An sinnπx

a

(

coshnπy

a−

cosh nπba

sinh nπba

sinhnπy

a

)

=∞∑

n=1

An sinnπx

a

sinh nπ(b−y)a

sinh nπba

(11)

Remaining boundary condition (4c) can be used to evaluate the coefficients An:

f(x) =

∞∑

n=1

An sinnπx

a.

An is obtained as

An =2

a

∫ a

0f(x) sin

nπx

adx. (12)

The solution to the BVP described by equations (3)-(4) is given by

(11) with An given by (12).

Lecture 15 MA 201, PDE (2018) 12 / 37


Similarly we can find the other solutions u2, u3 and u4 and

write the total solution as u = u1 + u2 + u3 + u4.

This problem with Dirichlet conditions

along all boundaries is called a Dirichlet problem for a rectangle.

The problem with Neumann conditions

along all boundaries is called a Neumann problem for a rectangle.

This new problem can be solved by writing the boundary conditions as

ux(0, y) = 0, (13a)

ux(a, y) = 0, (13b)

uy(x, 0) = f(x), (13c)

uy(x, b) = 0. (13d)

TRY to solve it yourself.

Lecture 15 MA 201, PDE (2018) 13 / 37

Solution in different types of domains

Till now the problems that we have taken up are for bounded domains in

Cartesian coordinates.

There are other types of domain which occur frequently in many physical problems.

We need to rewrite our governing equations and the related conditions.

Some domains of importance are

• circular domain

• spherical domain

• cylindrical domain.

Lecture 15 MA 201, PDE (2018) 14 / 37


For two dimensional problem with (x, y)

∇2u(x, y) =∂2u

∂x2+

∂2u

∂y2.

For three dimensional problem with (x, y, z)

∇2u(x, y, z) =∂2u

∂x2+

∂2u

∂y2+

∂2u

∂z2.

For two dimensional problem with (r, θ), that is, in polar coordinates

∇2u(r, θ) =∂2u

∂r2+

1

r

∂u

∂r+

1

r2∂2u

∂θ2.

Lecture 15 MA 201, PDE (2018) 15 / 37


For three dimensional problem with (r, θ, z), that is, in cylindrical coordinates

∇2u(r, θ, z) =∂2u

∂r2+

1

r

∂u

∂r+

1

r2∂2u

∂θ2+

∂2u

∂z2.

For three dimensional problem with (r, θ, φ), that is, in spherical coordinates

∇2u(r, θ, φ) =∂2u

∂r2+

2

r

∂u

∂r+

1

r2∂2u

∂θ2+

cot θ

r2∂u

∂θ+

1

r2 sin2 θ

∂2u

∂φ2.

Lecture 15 MA 201, PDE (2018) 16 / 37

Laplace’s equation in polar coordinates

For a problem involving circular disk, polar coordinates are more appropriate than

rectangular coordinates.

Let us formulate the steady-state heat flow problem in polar coordinates r, θ,

where x = r cos θ, y = r sin θ.

A circular plate of radius a can be simply represented by

r ≤ a with 0 ≤ θ ≤ 2π.

The unknown temperature inside the plate is now u = u(r, θ),

The given temperature on the boundary of the plate is

u(a, θ) = f(θ), where f is a known function.

Lecture 15 MA 201, PDE (2018) 17 / 37


Now we have the following boundary value problem:

urr +1

rur +

1

r2uθθ = 0, 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π, (14)

u(a, θ) = f(θ), 0 ≤ θ ≤ 2π. (15)

There is a periodic boundary condition which is implicit in nature:

u(r, θ) = u(r, θ + 2π). (16)

Using the separation of variables method, assume a solution:

u(r, θ) = R(r)T (θ).

Using this in equation (14),

r2R′′

R+ r

R′

R+

T ′′

T= 0 (17)

Lecture 15 MA 201, PDE (2018) 18 / 37


Separating the variables

r2R′′

R+ r

R′

R= −

T ′′

T= k (18)

which give rise to the following ODEs:

r2R′′ + rR′ − kR = 0, (19)

T ′′ + kT = 0. (20)

We cannot consider negative values of k since if k is negative,

then the ODE in T (θ) has exponential solutions, and exponential solutions cannot

satisfy periodicity conditions.

Since we are looking for a periodic solution in θ,

we must take k = λ2.

But we should keep in mind that λ = 0, corresponding to k = 0, is also an

eigenvalue with corresponding eigenfunction u0(r, θ) = constant.

Lecture 15 MA 201, PDE (2018) 19 / 37


Hence the equations reduce to

r2R′′ + rR′ − λ2R = 0, (21)

T ′′ + λ2T = 0. (22)

(22) has the general solution

T (θ) = A cos λθ +B sinλθ. (23)

The Dirichlet periodic boundary condition (16) will give us

cos 2πλ = 1,

i.e., λn = n.

Lecture 15 MA 201, PDE (2018) 20 / 37


Tn(θ):

Tn(θ) = An cosnθ +Bn sinnθ (24)

Equation (21) takes the form

r2R′′ + rR′ − n2R = 0 (25)

which is of Cauchy-Euler form and the solution is

Rn = Cnr−n +Dnr

n (26)

We set Cn = 0 since we are seeking a bounded solution in 0 ≤ r ≤ a.

and r−n is not bounded when r → 0.

Lecture 15 MA 201, PDE (2018) 21 / 37


Solution to the BVP:

u(r, θ) =A0

2+

∞∑

n=1

rn(An cosnθ +Bn sinnθ). (27)

Using the given boundary condition (15),

f(θ) =A0

2+

∞∑

n=1

an(An cosnθ +Bn sinnθ), (28)

The coefficients are given by

An =1

πan

∫ 2π

0f(θ) cosnθ dθ, n = 0, 1, 2, 3, . . . (29a)

Bn =1

πan

∫ 2π

0f(θ) sinnθ dθ, n = 1, 2, 3, . . . (29b)

(27) is the solution of Laplace’s equation with the coefficients given by (29a) and

(29b).

Lecture 15 MA 201, PDE (2018) 22 / 37


What we have just solved is called

Interior Dirichlet problem for a circle since we have used Dirichlet condition fin the

region r ≤ a.

If we change only the condition to Neumann

we have what is called Interior Neumann problem for a circle.

Now if we change the region to r > a with Dirichlet condition

we have what is called Exterior Dirichlet problem for a circle.

If in above condition is replaced by Neumann

we have what is called Exterior Neumann problem for a circle.

Lecture 15 MA 201, PDE (2018) 23 / 37

Diffusion in a disk

Consider a circular, planar disk of radius a for which

• initial temperature is a function of the radial distance r alone

• boundary is held at zero degrees.

Intuition tells that

the temperature u in the disk depends only on time and the distance r from the centre.

To be precise, the initial temperature u will be the same for some r = rn irrespective

of what value of θ is assigned.

That is, if the initial temperature u is, say, u1 for some r = r1, then it is imperative

that the temperature is same for that specific r.

Consider now that the initial temperature u is, say, u2 for some r = r2, then it is

imperative that the temperature is same for that specific r.

which, in turn means that

the heat flow will take place either from r = r1 to r = r2 or from r = r2 to r = r1

depending on which has higher temperature.

Lecture 15 MA 201, PDE (2018) 24 / 37

Diffusion in a disk

It then allows us to consider u simply as

u = u(r, t) though the given equation initially seemed to contain r, θ, t as independent

variables.

This assumption looks perfectly alright

because there is nothing in the initial condition or boundary condition to cause heat to

diffuse in an angular direction

Heat will flow only along rays emanating from the origin.

Now it is obvious that

the diffusion equation looks simpler than what it was originally.

Lecture 15 MA 201, PDE (2018) 25 / 37

Diffusion in a disk

Diffusion equation is now as simple as follows:

ut = α(urr +1

rur), 0 ≤ r ≤ a, t > 0. (30)

Boundary condition

u(a, t) = 0, t > 0, (31)

Initial condition

u(r, 0) = f(r), 0 ≤ r < a, (32)

f is a given initial radial temperature distribution.

Lecture 15 MA 201, PDE (2018) 26 / 37

Diffusion in a disk

Assume a solution in the form:

u(r, t) = R(r)T (t).

From given equation

R′′

R+

1

r

R′

R=

T ′

αT= k.

k = −λ2 gives rise to the following pair of ODEs:

r2R′′ + rR′ + λ2r2R = 0, (33)

T ′ + αλ2T = 0. (34)

Equation (34) can be easily solved to write as

T (t) = Ce−αλ2t. (35)

Can you recognize equation (33)??

Lecture 15 MA 201, PDE (2018) 27 / 37

Diffusion in a disk

Equation (33) is Bessel’s equation of order 0.

Its solution can be written as

R(r) = AJ0(λr) +BY0(λr), (36)

where

J0 and Y0 are, respectively, Bessel’s function of first kind and second kind of order zero.

We are looking for a bounded solution as r → 0,

we must take B = 0 as Y0(λr) → −∞ as r → 0.

Lecture 15 MA 201, PDE (2018) 28 / 37

Diffusion in a disk

Solution u(r, t) can be written as

u(r, t) = AJ0(λr)e−αλ2t. (37)

Applying boundary condition (31),

0 = AJ0(λa) implying

J0(λa) = 0.

Hence λna = νn,

where νn are the zeros of J0.

Hence the eigenvalues are given by

λn =νn

a. (38)

Lecture 15 MA 201, PDE (2018) 29 / 37

Diffusion in a disk

⇒

un(r, t) = AnJ0

(νn

ar)

e−α

ν2n

a2t.

The solution u(r, t) is

u(r, t) =

∞∑

n=1

un(r, t)

=∞∑

n=1

AnJ0

(νn

ar)

e−α

ν2n

a2t. (39)

Now using the initial condition (32):

f(r) =∞∑

n=1

AnJ0

(νn

ar)

(40)

Note the difference between the orthogonal properties of sine/cosine functions

and Bessel functions.

Lecture 15 MA 201, PDE (2018) 30 / 37

Diffusion in a disk

Bessel functions {Jµ(λr)} form an orthogonal set with respect to the weight function

r.

For finding the coefficient An, we multiply (40) by rJ0

(

νnar)

and then integrate with respect to r from 0 to a to get

An =

∫ a

0 rf(r)J0(

νnar)

dr∫ a

0 r(

J0(

νnar))2

dr. (41)

The following orthogonality property is used:∫ a

0rJ0

(νn

ar)

J0

(νm

ar)

dr = 0, n 6= m.

Lecture 15 MA 201, PDE (2018) 31 / 37

Steady-state heat conduction in a circular cylinder

Consider a right circular cylinder of radius a and height l having

(a) its convex surface and base in the xy-plane at temperature 00C,

(b) the top end z = l is kept at temperature f(r)0C.

To find the steady-state temperature at any point of the cylinder.

The governing equation for this problem will be Laplace’s equation in r, θ, z.

∇2u(r, θ, z) =∂2u

∂r2+

1

r

∂u

∂r+

1

r2∂2u

∂θ2+

∂2u

∂z2.

For simplicity, we will consider Radially Symmetric Solution for the Laplace’s

equation.

Radially symmetric solution means that u(r, θ, z) = u(r, z) that is the solution doesn’t

depend on the polar angle θ.

In other sense, solutions are symmetric under rotation.

Lecture 15 MA 201, PDE (2018) 32 / 37


But assuming that the cylinder is symmetrical about its axis, Laplace’s equation

takes the form:

urr +1

rur + uzz = 0, 0 < r ≤ a, 0 ≤ z ≤ l. (42)

The boundary conditions are:

(on the side) u(a, z) = 0, 0 ≤ z ≤ l (43a)

(on the bottom) u(r, 0) = 0, 0 < r ≤ a (43b)

(on the top) u(r, l) = f(r), 0 < r ≤ a. (43c)

Assume a solution in the form

u(r, z) = R(r)Z(z)

Applying it to the governing equation (42):

R′′

R+

1

r

R′

R+

Z ′′

Z= 0.

Lecture 15 MA 201, PDE (2018) 33 / 37


By separating the variables:

R′′

R+

1

r

R′

R= −

Z ′′

Z= k.

Observing that only the negative value of the separation constant will give rise to

nontrivial solutions,

we get the following ODEs by considering k = −λ2:

Z ′′ − λ2Z = 0, (44)

R′′ +1

rR′ + λ2R = 0, (45)

The solutions of the above equations are, respectively, given by

Z(z) = A cosh λz +B sinhλz, (46)

R(r) = CJ0(λr) +DY0(λr), (47)

Lecture 15 MA 201, PDE (2018) 34 / 37


The solution u(r, z):

u(r, z) = (A cosh λz +B sinhλz)(CJ0(λr) +DY0(λr)) (48)

We are looking for a bounded solution in 0 ≤ r ≤ a,

we must take D = 0 since Y0 → −∞ as r → 0.

Equation (48) can be written as

u(r, z) = J0(λr)(A cosh λz +B sinhλz). (49)

Now applying the boundary condition (43a), we get 0 = AJ0(λa)

implying

J0(λa) = 0.

Lecture 15 MA 201, PDE (2018) 35 / 37


Hence

λna = νn,

where νn are the zeros of J0.

The eigenvalues are given by

λn =νn

a. (50)

un(r, z) = AnJ0

(νn

ar)

coshνn

az +BnJ0

(νn

ar)

sinhνn

az.

By superimposing all the solutions,

u(r, z) =

∞∑

n=1

(

AnJ0

(νn

ar)

A coshνn

az +BnJ0

(νn

ar)

sinhνn

az)

. (51)

Lecture 15 MA 201, PDE (2018) 36 / 37


Using the boundary condition (43b),

we get A = 0 thereby reducing the solution to

u(r, z) =∞∑

n=1

BnJ0

(νn

ar)

sinhνn

az. (52)

The coefficient Bn can be obtained by using the boundary condition (43c):

f(r) =∞∑

n=1

BnJ0

(νn

ar)

sinhνn

al (53)

giving us

Bn =

∫ a

0 rf(r)J0(

znar)

dr

sinh νnal∫ a

0 r(

J0(

znar))2

dr. (54)

Lecture 15 MA 201, PDE (2018) 37 / 37

MA 201, Mathematics III, July-November 2018, Part II ... · Steady state heat conduction:...

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