Calculus of Two Variables Lecture Notes 2009 Part II p63to97

8/6/2019 Calculus of Two Variables Lecture Notes 2009 Part II p63to97

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5. CRITICAL POINTS (NON-DEGENERATE)

In this brief chapter we will define, and investigate the properties of, the non-degenerate critical points of a function of two variables. Recall:

Definition 81 The critical points of a real function of a single variable, f(x), are all values of x for which

f(x) = 0.

Remarks

The phrase stationary point is synonymous with critical point. Critical points of a function of a single variable can be classified as a local maximum, a local minimum or a point of inflection. Suppose f(x) has a stationary point at x0, i.e. f(x0) = 0. Suppose further that f(x0) < 0. Then, the stationary point at x = x0 is a

local maximum.

Suppose f(x) has a stationary point at x0, i.e. f(x0) = 0. Suppose further that f(x0) > 0. Then, the stationary point at x = x0 is alocal minimum.

Suppose f(x) has a stationary point at x0, i.e. f

(x0) = 0. Suppose further that f(x0) = 0. Then, the stationary point at x = x0 could

be a local maximum, a local minimum, or a point of inflection. For example, sketch the graphs and consider the critical points of thefunctions f(x) = x4, g(x) = x4, h(x) = x3.

Our aim is to consider the generalisation of the above to functions of two variables.

Definition 82 The critical points of f(x, y), are all values of (x, y) for which

fx

= fy

= 0, i.e. f = 0.

Remarks

Once more, the phrase stationary point is synonymous with critical point. Analogously, the critical points of a function of n variables, f(x1, ..,xn), are all values of (x1,...,xn) for which f = 0.

CRITICAL POINTS (NON-DEGENERATE) 63


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Theorem 83Suppose f(x, y) has a stationary point at (x, y) = (a, b). Set fxx(a, b) = r, fxy(a, b) = s, fyy(a, b) = t. Then

(i) If r < 0, t < 0, rt s2 > 0 then f(x, y) has a local maximum at (x, y) = (a, b).(ii) Ifr > 0, t > 0, rt s2 > 0 then f(x, y) has a local minimum at (x, y) = (a, b).

(iii) Ifrt s2 < 0 then f(x, y) has a saddle point at (x, y) = (a, b).

Remarks

A saddle point entails a plot of the function f(x, y) in the region immediately around the critical point (a, b) is analogous to a horsessaddle ... the function increases in one direction and decreases in another as one moves away from the stationary point.

One often refers to rt s2 as the discriminant. Not all values of r, t, rt s2 have been considered. In particular, for the values of the variables which have not been considered above

there is degeneracy and further information is required to classify the critical point.

This is analogous to when f(x0) = 0 at a stationary point for a function of one variable.

Degenerate cases will be classified in next terms lectures.

Proof. Let n() = (cos , sin ) be the unit vector in the direction . We have the directional derivatives

df

dn

def= n f = cos fx + sin fy,

d2f

dn2def= n (n f) = n (cos fx +sin fy) = cos

x(cos fx + sin fy)+sin

y(cos fx + sin fy) = cos

2 fxx +2cos sin fxy +sin2 fyy .

64 CRITICAL POINTS (NON-DEGENERATE)


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Thus, for fixed, df /dn, d2f/dn2 give the first and second derivative for f in the direction n() = (cos , sin ). Clearly, df /dn is zero at acritical point. Additionally, at the critical point we have, by definition, fxx(a, b) = r, fxy(a, b) = s, fyy(a, b) = t and thus

d2f

dn2(a, b) = r cos2 + 2s cos sin + t sin2 = r

cos +

s

rsin

2

+1

r

rt s2 sin2 .

(i) Suppose r < 0, t < 0, rt s2 > 0. Then, we have thatd2f

dn2(a, b) < 0 .

Thus, irrespective of the direction one moves away from the critical point, the function f always decreases. Thus there is a local maximum atthe critical point.

(ii) Suppose r > 0, t > 0, rt s2 > 0. Then, we have thatd2f

dn2(a, b) > 0 .

Thus, irrespective of the direction one moves away from the critical point, the function f always increases. Thus there is a local minimum atthe critical point.

(iii) Suppose rt s2 < 0. We wish to show the critical point is a saddle point. Our proof below ASSUMESt = 0. The case where t = 0 is astraightforward alteration and left as an exercise.

Part I. We first of all wish to show, for r, t, s fixed and satisfying the aforementioned constraint that

d2

fdn2

(a, b) = r

cos + sr

sin 2

+ 1r

rt s2 sin2 = 0 (5.1)has four real roots for [0, 2). In Part II, we show why this is sufficient to enforce a saddle point.

Equation (5.1) does not have a root for satisfying cos = 0. Suppose cos = 0. Then sin2 = 1 and hence d2f/dn2 = t = 0 (by above assumption).



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Equation (5.1) has four roots. At a root, we have cos = 0, and thus

sec2

d2f

dn2 (a, b) = r + 2s tan + t tan2

= 0

which yields

= tan1

s s2 rt

2t

.

Noting s2 rt > 0, this gives four real roots for within [0, 2).

Part II. We have that d2

f/dn2

= 0 for four values of [0, 2). Thus, on this domain, there are two non-adjacent intervals of whered2f/dn2 > 0; moving away from the critical in these directions leads to an increase in f. Similarly, there are two non-adjacent intervals of where d2f/dn2 < 0; moving away from the critical in these directions leads to a decrease in f. Thus we have a saddle point.

Remark The above is somewhat inelegant. The use of Taylors expansion for a function of two variables, and the fact a symmetric matrix canbe diagonalised via an orthogonal transformation, readily leads to a proof of the above; we do not cover these topics in this course.

Example 84 For each of the functions below find the critical points and classify them in terms of maxima, minima or saddle points:

(i) f(x, y) = x + 2x2y2 x2 + y2(ii) g(x, y) = cos(x + y)sin x, for x, y (0, ).

Solution.

(i) At a critical point we must have

fx = 1 + 4xy2 2x = 0, fy = 4x2y + 2y = y(4x2 + 2) = 0.

For fy = 0 we must have y = 0, whence x = 1/2 from the left of the above. Thus we have a single critical point at (x, y) = (1/2, 0). Notingthat

fxx = 4y2 2, fxy = fyx = 8xy, f yy = 4x2 + 2

and using the above definitions of r, s, t we have

r = 2, s = 0, t = 3, rt s2 = 6.Hence the critical point at (x, y) = (0, 1/2) is a saddle.



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(ii) At a critical point we must have



6/35



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6. PARTIAL DIFFERENTIAL EQUATIONS

6.1 Laplaces Equation

We have already mentioned Laplaces equation in earlier examples. In Cartesian co-ordinates it reads as

2

fx2

+ 2

fy2

= 0 (in the plane);

2f

x2+

2f

y2+

2f

z2= 0 (in three dimensions).

In Example 29 we showed how the equation remained the same when we changed from Cartesian to parabolic co-ordinates, and in Example 30we saw that Laplaces equation reads as

2f

r2 +

1

r

f

r +

1

r2

2f

2 = 0in planar polar co-ordinates. A function which satisfies Laplaces equation is known as harmonic.

Laplaces equation is often abbreviated to

2f = 0and the differential operator 2 is known as the Laplacian.

Laplaces equation appears in many physical situations. For example, 2

f = 0 holds true for:

the gravitational potential in a region containing no matter; the electrostatic potential in a charge-free region; the steady-state temperature in a region containing no source of heat; incompressible fluid flows whenever viscosity is negligible and there are no vortices, sources or sinks. Then the equations of motion for the

fluid can be recast into Laplaces equation for a velocity potential.

PARTIAL DIFFERENTIAL EQUATIONS 69


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Separable Solutions We will first consider separable solutions of Laplaces equation in Cartesian co-ordinates. That is we shall look forsolutions of the form

f(x, y) = X(x) Y (y) for2f

x2+

2f

y2= 0.

If we substitute f(x, y) = X(x) Y (y) into Laplaces equation we get that

X (x) Y (y) + X(x) Y (y) = 0.

If we rearrange this toX (x)

X(x)=

Y (y)

Y (y). (6.1)

The quantity, lets call it c (x, y), in (6.1) is both a function solely of x (from the LHS) and a function solely of y (from the RHS). It followsthat for any (x1, y1) , (x2, y2)

c (x1, y1) = c (x1, y2) [as c only depends on x]

= c (x2, y2) [as c only depends on y].

That is, c is constant! So we have

X (x)

X(x)= c,

Y (y)

Y (y)= c.

If c = 0 we see thatX(x) = Ax + B, Y (y) = Cy + D

for constantA,B,C,D. If c > 0 we see thatX(x) = A exp

cx

+ B expcx

Y (y) = Ccos

cy

+ D sin

cy

for constants A,B,C,D.

If c < 0 we have a situation similar to the c > 0 case with x and y swapped.

70 PARTIAL DIFFERENTIAL EQUATIONS


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Example 85 Find the separable solutions R (r) () to Laplaces equation in planar polar co-ordinates.

Solution. Putting f(r, ) = R (r) () into

2

fr2

+ 1r

fr

+ 1r2

2

f2

= 0

we get

R (r) () +1

rR (r) () +

1

r2R (r) () = 0.

This rearranges tor2R

R+

rR

R=

both sides of which must be constant, as previously shown. If we call this constant c then we have ...

LAPLACES EQUATION 71


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If c = k2 > 0 then we have that

If c = 0 then we have

If c = k2 < 0 then we have

6.2 Dirichlets Problem

Solving Laplaces equation

2f = 0



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in a region R, under a boundary conditionf(x) = g (x) for x R

on the boundary R of the region R, is known as Dirichlets Problem. This is named after the nineteenth century German mathematician PeterGustav Dirichlet (1805-1859). Under fairly mild conditions on the shape of the boundary R and the function g (x) it can be shown that thereexists a unique solution to Dirichlets problem.

It is beyond the scope of the course to treat any aspect of this problem in a general setting but we will look at the case of a rectangular andcircular region.

Example 86 Part I Find the most general sum of separable solutions defined on the rectangle

R = {(x, y) : 0 x a, 0 y b}which satisfies Laplaces equation in the interior of R,

2f

x2+

2f

y2= 0, 0 < x < a, 0 < y < b

and which satisfies

f(0, y) = 0, 0 < y < b, (6.2)

f(x, 0) = 0, 0 < x < a, (6.3)

f(x, b) = 0, 0 < x < a. (6.4)

Part II Hence find the solution to2f

x2+

2f

y2= 0, 0 < x < a, 0 < y < b

given the boundary conditions

f(0, y) = 0, 0 < y < b,

f(x, 0) = 0, 0 < x < a,

f(x, b) = 0, 0 < x < a,

f(a, y) = g(y)def= 2sin

y

b + sin

3y

b

, 0 < y < b.

DIRICHLETS PROBLEM 73


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Solution. Part I We need to find the most general sum of separable solutions and begin by considering the separable solutions we met earlier.These were of the form

f(x, y) = A exp cx+ B exp

cx Ccos cy+ D sin cy , c > 0; (6.5)f(x, y) =

A cos

cx+ B sin cx Cexp cy+ D exp cy , c < 0; (6.6)f(x, y) = (Ax + B) (Cy + D) . (6.7)

If we consider which of these can satisfyf(x, 0) = f(x, b) = 0, 0 < x < a,

then we see:

in (6.5) this means that



Hence the non-zero separable solutions are of the form

for some integer n. [The constant D is now no longer necessary.]

If we also require that f(0, y) = 0 for all y (0, b) then we have

It follows that the most general sum of separable solutions satisfying Laplaces equation in the rectangle R and the conditions (6.2)-(6.4) is

(6.8)



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Part II We additionally need to impose the fourth boundary condition that

f(a, y) = g (y)def= 2sin

yb

+ sin

3y

b

, 0 < y < b,

Comparing coefficients with the expression (6.8) we see that

Hence

f(x, y) =2

sinh (a/b)sinx

a

siny

b

+

1

sinh(3a/b)sin

3x

a

sin

3y

b

.

Remark 87 We will briefly comment in the next section on how to deal with a more general function g (y) in the boundary condition. However,it would also be reasonable to remark that the given boundary condition has three of its sides set to 0. This, though, is not as limiting as arestriction as one might first think. Given a general boundary condition on the four sides of the rectangle, we can treat each side separately asabove to produce four functions that match the boundary condition on a single side. Their sum is a harmonic function which meets the requiredboundary condition on all four sides.

DIRICHLETS PROBLEM 75


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Example 88 Let f(r, ) be a function, defined on the disc R = {(r, ) : r 1} which satisfies Laplaces equation in the interior of R,2f

r2+

1

r

f

r+

1

r22f

2= 0, r < 1

and satisfies the boundary condition f(1, ) = g () , 0 < 2, for some function g.

Solution. Well begin by considering the separable solutions in r and we met earlier. These were of the form

f(r, ) =

Ark + Brk

{Ccos k + D sin k} k > 0;

f(r, ) = {A sin(k ln r) + B cos(k ln r)}

Cek + Dek

, k > 0;

f(r, ) = {A ln r + B} {Ct + D} .Note that the only solutions which are defined at r = 0 (which lies in R) are those of the form

f(r, ) = rk (Ccos k + D sin k) ,

where k is a non-negative integer (so that it is continuous on the = 0, 2 borderline.)

Again it is the case that any sum of these

f(r, ) =

k=0

rk (Ck cos k + Dk sin k)

is also a solution of the Laplaces equation. If the function g () is of the form

g () = cos2 + sin

then we can rewrite this as

g () =1

2

+1

2

cos2 + sin ,

and comparing coefficients we see that

C0 =1

2, C2 =

1

2, D1 = 1

with all other coefficients zero. Hence

f(r, ) =1

2+

1

2r2 cos2 + r sin .



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6.3 Fourier Analysis

Recall when solving Dirichlets Problem in the rectangle we produced solutions

f(x, y) =n=1

n sinhnx

b

sinny

b

which met the boundary conditions on the three sides where the function was zero.

We could meet the additional boundary condition on x = a easily when the function there, g (y) , was a finite linear combination of the functions

sin

nyb

,

as we could compare coefficients naturally.

Likewise with Dirichlets Problem in the circle we needed to solve the boundary condition

k=0

(Ck cos k + Dk sin k) = g ()

on r = 1. Again we can find the coefficients easily if the function g () is a finite linear combination of functions

cos k and sin k.

The more general problem of writing a functionF (t) as an infinite sum of functioncos kt andsin kt is the subject of Fourier analysis. Compare

this with the situation where e1, e2, . . . , en are orthonormal vectors (mutually perpendicular and of unit length) and we have

v =ni=1

iei.

In order to work out each i we look can take the component in the ei direction. If we dot both sides of the equation by ei then we find

v ei = i.

FOURIER ANALYSIS 77


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Similarly when faced with the equation

F (t) = a02

+k=1

(ak cos kt + bk sin kt) , t [0, 2] (6.9)

we are looking for a way to extract each coefficient ak or bk, somehow taking a component in the direction cos kt or sin kt.

In Fourier analysis this is done by noting for k1, k2 > 0,

20

sin(k1t)sin(k2t) dt =

=

0 if k1 = k2 if k1 = k2

2

0

cos(k1t)cos(k2t) dt =

0 if k1 = k2 if k1 = k2

20 sin(k

1

t)cos(k2

t) dt = 0 for all k1

, k2

.

So dotting (6.9) with cos kt, by which is meant multiplying both sides by cos kt and integrating over [0, 2] we find

2

0

F (t)cos kt dt =

Thus, the dotting operation eliminates all other coefficients as required.



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So the formulae for the Fourier coefficients are.

ak =1

2

0

F (t)cos kt dt (k > 0)

bk =1

2

0

F (t)sin kt dt (k > 0)

a0 =1

2

0

F (t) dt.

provided all the above calculations and the interchange of integration and summation can be justified and issues of convergence overcome! Fourier

series and their convergence will be dealt with in detail in next terms Fourier Series course.

Example 89 Find the Fourier coefficients of

f(x) = x, x [0, 2].

Consider the Fourier series convergence at x = 0, /2, , 3/2, 2 and at x + 2 compared with x.

Solution. The Fourier coefficients are

a0 =1

2

0

x dx =

x2

2

20

= 2;

ak =1

2

0

x cos kx dx =1

x sin kx

k+

cos kx

k2

2

0

= 0;

bk =

1

20 x sin kx dx =

1

x cos kxk + sin kxk2

2

0=

2

k .

So the Fourier series of f(x) = x is

F (x) =a02

+k=1

(ak cos kx + bk sin kx) = 2k=1

sin kx

k

FOURIER ANALYSIS 79


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Consider the convergence of this series.

The first thing to note is that F (x + 2) = F (x) for all x because F is a sum of sines. So certainly F does not agree with f everywhere.

If we look at F at x = 0, , 2 we see that F equals at each of these points.

At x = /2 and x = 3/2 we have

F

2

= 2

1 1

3+

1

5 1

7+

= 2

4

=

2;

F

3

2

= 2

1 + 1

3 1

5+

1

7

= 2

4

=

3

2.

So at least the Fourier series converges correctly at x = /2, , 3/2. The infinite series

1 13

+1

5 1

7+ =

4

was first proved by Leibniz.

In fact, the series F (x) converges to x on the interval (0, 2). At multiples of 2 the series converges to more generally a Fourierseries converges at multiples of 2 to the average of f(0) and f(2) , and will do similarly at any discontinuity in the interval (0, 2).These observations are general if the function is piecewise smooth.

The graph of F (x) is below:



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Example 90 By considering the Fourier series of x at /4 find the sum of the infinite series

S = 1 +1

3

1

5

1

7+

1

9+

1

11

1

13

1

15+

Solution. If we put x = /4 into F (x) we get

4= F

4

=

Thus S = /

2

2

.

6.4 Poissons Equation

Poissons equation is the inhomogeneous Laplace equation

2f = g.It commonly appears in gravitational theory: Laplaces equation 2 = 0 dictates how the gravitational potential behaves in the absence ofany matter, but, more generally, when there is matter present, distributed with density function (x,y,z) , then Poissons equation

2 = 4G

describes the potentials behaviour. Poissons equation occurs in numerous other physical scenarios, such as electrostatics in the presence ofcharge distributions.

Whilst we will not be specifically interested in gravity, electrostatics or other physical applications in this course, solving Poissons equationraises the same issues that will again appear in later courses.

POISSONS EQUATION 81


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Example 91 Find the circularly symmetric solutions of Poissons equation in the plane 2f = g where

g (r) = r for r < a0 for r > a

Solution. Recall that the Laplacian is given by

2f = 2f

r2+

1

r

f

r+

1

r22f

2

in planar polar co-ordinates. We are only interested in circularly symmetric functions of the form f(r); so for r < a we need to solve

f (r) + r1f (r) = r.

If we solve the equation similarly in the region r > a then we find

There are several things to note though. Firstly if this function is to be defined at r = 0 it must be the case that A = 0. But also there are issuesof continuity and smoothness at the r = a border. For our solution f to be continuous at r = a we need

a3/9 + B = Cln a + D

and for it to be smooth at r = a (that is df /dr agrees on both sides of r = a) we need a2/3 = C/a. Hence

A = 0, C = a3/3, D =

a3/9

(1 3 ln a) + B.So the most general circularly symmetric solution of this Poisson equation is

f(r) =

r3

9+ B for r a,

a3

9ln r + a

3

9(1 3 ln a) + B for r a.

To specify the solution uniquely then we might, for example, require a further condition such as f(0) = 0 which would give B = 0.



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6.5 The Wave Equation

Derivation We will consider the vibrations of an elastic uniform string under certain simplifying assumptions:- the string undergoes small vibrations: by this we mean that second order terms, such as y2, y2x, 21 and 22 in the analysis below will be

considered to be negligible;

the vibrations are entirely transverse, so that a point at distance x0 along the string remains on the line x = x0 throughout the motion; the string is at constant tension T and the density of the string is uniformly ; the effects of gravity and air resistance are negligible compared with the tension in the string.

Consider the vibrations of a small section of the string from x to x + x. We shall denote the angles the string makes with horizontal at x andx + x by 1 and 2 respectively, as shown in the diagram below.

I. Taking components of the forces in the y-direction and invoking Newtons second law (Force = Mass Acceleration) we get

where x0 is the x-co-ordinate of the centre of mass of the segment, x is the segments mass and ytt its vertical acceleration. Note that x0 willlie between x and x + x as the entirety of the segment of string lies in this range.

There is no gravity term as the weight of the string is considered negligible compared with the tension involved.

THE WAVE EQUATION 83


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II At the level of approximation we are considering (i.e. neglecting second order terms), we have

sin 1 = 1

31

6

+ . . .

1

tan 1 = 1 +

31

3

+ . . .

and similarly for 2. Hence at the level of approximation we are considering

tan 2 tan 1x

=

T

2y

t2(x0, t) .

III Now

tan 1 =y

x(x, t) ,

tan 2 =y

x(x + x, t) =

where the final expression can be derived by a Taylor expansion (assuming yxx is continuous and differentiable; see other courses this year fordetails of Taylor expansions). Thus

2y

x2(x, t)

T

2y

t2(x0, t) = O(x).

Letting x tend to 0, whereupon x0 x, yields2y

x2(x, t) =

T

2y

t2(x, t) .

Note that /T has units (kg/m)/(kgms2) = (s/m)2 so that c =

T / has the units of velocity. The wave-equation then reads

c22y

x2=

2y

t2.



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Proposition 92 (DAlembert 1746) The general solution of the wave equation

2y

t2

= c22y

x2

is

y (x, t) = f(x ct) + g (x + ct) .

Proof. We introduce two new variables

= x ct and = x + ct.

Then by Theorem 28, that is the chain rule, we have

yxx = yxx + yxx + y (x)2 + 2yxx + y (x)

2

= y 0 + y 0 + y (1)2 + 2y 1 1 + y (1)2= y + 2y + y

and

ytt = ytt + ytt + y (t)2 + 2ytt + y (t)

2

= y 0 + y 0 + y (c)2 + 2y (c) c + y (c)2= c2 (y 2y + y)

Hence c2yxx = ytt if

c2 (y + 2y + y) = c2 (y 2y + y)

so that y = 0. We know, from Example 12(i), that the general solution of this is

y = f() + g () = f(x ct) + g (x + ct) .



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Remark 93 Consider a solution of the form y (x, t) = f(x ct), with g = 0. Note that this solution at time t + T resembles the solution attime t, but translated to the right by cT. This is a wave moving to the right with speed c and likewise a wave of the form g (x + ct) is one whichis moving to the left at speed c.

Example 94 Find the solution to the wave equation when

y (x, 0) = 1 |x| 1 < x < 1

0 otherwise,

y

t(x, 0) = 0.

Sketch your solutions at ct = 0, 12

, 1, 32

.

Solution. We know that the solution has the form y (x, t) = f(x ct) + g (x + ct) and so the initial conditions give that

f(x) + g (x) =

1 |x| 1 < x < 1,

0 otherwise,

cf (x) + cg (x) = 0.

Integrating the second equation we see that f(x) = g (x) + K where K is a constant. We can, without any loss of generality take K to be 0.[Keeping K in the calculation would yield to the same solution with slightly different f and g.] So f = g and we have

y (x, t) = f(x ct) + f(x + ct)where

f(u) =

1

2(1 |u|) 1 < u < 1,

0 otherwise.



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Sketches of the solution at different t are given below.



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Example 95 Find all the separable solutionsy (x, t) = X(x) T (t) (6.10)

of the wave equation ytt = c2yxx which satisfies the boundary conditions

y (0, t) = y (L, t) = 0 for all t.

Solution. Substituting y (x, t) = X(x) T (t) into Equation (6.10) we get

X(x) T (t) = c2X (x) T (t) = X (x)

X(x)=

T (t)

c2T (t)= k (constant).

We then have

X(x) =

depending on the sign of k.

As in Example 86 the only one of these solutions which can meet the boundary conditions X(0) = X(L) = 0 without being everywhere zero is

the third of these.For

X(t) = Ecoskx+ F sinkx

to satisfy X(0) = 0 then we must have E = 0.

In order to X(L) = 0 without having F = 0 it must be the case that

k =

for some integer n. So the separable solutions are of the form

y (x, t) = X(x) T (t) = sinnx

L

n cos

nct

L

+ n sin

nct

L

.



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Remark 96 The solutions of the form

y (x, t) = A sinnx

L

sin

nct

L

are known as the natural or normal modes; when n = 1 this is known as the fundamental mode. The quantities nc/L are called thenatural frequencies of the string.

Remark 97 Using Fourier analysis as in Section 6.3 it is possible to write down any solution of the wave equation with these boundaryconditions in terms of the separable solutions. To specify uniquely the solution initial conditions describing y (x, 0) and yt (x, 0) can also begiven.



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7. EPILOGUE

7.1 The Heat Equation

We will consider here the problem of modelling heat flow in a one-dimensional uniform bar. If two nearby points on the rod, separately by asmall distance , are at temperatures T1 on the left and T2 on the right, then the heat flow from left to right between these points is proportional

to the temperature difference and inversely proportional to the distance. That is

Amount of heat per unit time = T1 T2

(7.1)

where the constant of proportionality is called the thermal conductivity and which (we assume) depends only on the material that makes upthe rod. Equation (7.1) is Fouriers law of heat conduction. Note that this heat is signed in the sense that it is negative if T2 > T1 and theheat is actually flowing from right to left.

Consider a small section of the rod between x = x0 and x = x0 + x. Well consider the flux of heat into this section. The rate of heat transfer from left-to-right past the pointx = x0 equals

lim0

T (x0 /2, t) T (x0 + /2)

= T

x(x0, t)

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and similarly the rate at which heat flows from left to right past the point x = x0 + x equals

Tx

(x0 + x,t)

so that the amount of heat entering the section, in a short time t equals

T

x(x0 + x,t) T

x(x0, t)

t.

Now the average change of temperature T in the small section of the rod is proportional to the amount of heat introduced (given above) andinversely proportional to the mass x of the section. The constant of proportionality is c1 where c is called the specific heat of the material.Hence

T =1

c 1

x

T

x(x0 + x,t) T

x(x0, t)

t.

If we rearrange this toT

t=

c

T

x(x0 + x,t) T

x(x0, t)

x

and let x and t tend to zero then we find

Tt = c 2

Tx2 . (7.2)

Equation (7.2) is known as the heat equation or diffusion equation. The constant /c is often written as 2 and is called the thermaldiffusivity of the rod so that the heat equation reads

T

t= 2

2T

x2.

More generally, in three dimensions, the heat equation reads

Tt

= 22T.Note that the steady-state solutions of the heat equation, i.e. those solutions that dont depend on time, are the solutions of Laplaces equation

2T = 0.

When modelling the heat flow in such a rod 0 x L there are two natural boundary conditions which might arise.

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An end of the rod may be kept at a particular temperature. For example we might haveT (L, t) = T0 for all t.

An end of the rod may be insulated so that no heat is lost through that end. For example, if the end at x = 0 were insulated it would meanT

x(0, t) = 0 for all t.

Consider the example below. To solve it completely we will have to calculate the separable solutions of the heat equation and calculate anotherFourier series.

Example 98 The temperature T in a rod 0 x L is initially T0. The ends of the rod are kept at temperature T = 0; that is for all t we have

T (0, t) = T (L, t) = 0. (7.3)

Determine the temperature T (x, t) in the rod at time t and at position x.

Solution. We begin by determining the separable solutions to the heat equation that satisfy the boundary conditions (7.3). If we substitute

T (x, t) = A (x) B (t)

into the heat equationT

t= 2

2T

x2

we findA (x) B (t) = 2A (x) B (t)

and separating the variables we get

A

(x)A (x) = B

(t)2B (t) = k (constant),

as only constants are functions of x alone and also of t alone. So we have, depending on the sign of k

A (x) =

P exp

kx

+ Q expkx

k > 0,

P x + Q k = 0,

P cos

kx

+ Q sin

kx

k < 0.

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As we have seen before, it is only the third of these, when k < 0 that yields any non-zero solutions meeting the boundary conditions

A (0) = A (L) = 0.

In this case we have solutions of the form

A (x) = Q sin

nxL

where

k = n/L. Substituting this back into the equationB (t) = k2B (t)

we get separable solutions of the form

T (x, t) = Qn sinnx

L exp

n222tL2

.

A general solution of the heat equation, satisfying the boundary conditions (7.3) is of the form

T (x, t) =n=1

Qn sinnx

L

exp

n222tL2

.

If this is solution is going to meet the initial condition that T (x, 0) = T0 in the rod then we must have

T0 =

n=1Qn sin

nx

L . (7.4)

Recall that a crucial idea behind Fourier analysis is thatL0

sinnx

L

sinmx

L

dx =

0 if n = m,

L/2 if n = m,

though our earlier examples in Section 6.3 we on an interval of length 2 rather than the more general L here. Hence

Qn =2T0

L L

0

sin

nx

L dx =

2T0L

Ln

cos

nx

L L

0

= 2T0

n{(1)n 1} =

4T0n if n is odd,0 if n is even.

Hence

T (x, t) =4T0

n odd

1

nsinnx

L

exp

n222tL2

. (7.5)

Notice that T (x, t) tends to 0 as t tends to at all points of the rod.

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Figure 7-1 The Fourier Series used for T0 in example solution (7.1) above.

Remark 99 Note that the Fourier Seriesn=1

Qn sinnx

L

cannot be equal to T0 whenx = 0 or x = L. This Fourier series in fact converges to the function depicted in Figure 7-1. Analogously to example(89) previously, the Fourier series in not continuous. At x = 0 it converges to the mean of the limits on approaching x = 0 from below andfrom above, and similarly for other discontinuities. Being VERY careful with the calculation of limits for the expression (7.5) one finds

T0 = limx0

limt0

T(x, t) = limt0

limx0

T(x, t) = 0

and such complicated behaviour must be expected given the initial conditions are not consistent with the boundary conditions for the aboveexample.



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Example 100 Suppose now that the temperatures of the ends of the rod are maintained so that

T (0, t) = T1 and T (L, t) = T2.

We cannot immediately use separation of variables, because our boundary conditions are no longer homogenous. If we have two solutions whichsatisfy the heat equation and the above boundary conditions, say TA and TB we do NOT have their sum also satisfies the boundary conditionsas, for example,

TA(0, t) + TB(0, t) = T1 + T1 = 2T1.

Thus, we cannot find separable solutions and sum them to find a general solution whose summation coefficients can then be fined tune to satisfythe initial conditions.

In direct analogy to differential equations, we can treat this as an inhomogeneous version of the previous problem and just look for a particularsolution that satisfies the equation and conditions. One might first consider the steady-state solution that does this as it is easy to find ...

The steady-state heat equation just reads, in this one-dimensional case as

T (x) = 0

which has solutions

T (x) = Ax + B.

Of these the one that meets the boundary conditions is

T (x) = T1 +x (T2 T1)

L.

So if we let

U(x, t) = T (x, t)

T1 +x (T2 T1)

L

then U(x, t) satisfies the heat equation, the boundary condition

U(0, t) = U(L, t) = 0

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and the initial condition

U(x, 0) = T (x, 0)

T1 +x (T2 T1)

L .

Utilising Fourier Analysis as in the previous example this problem is now tractable.


Calculus of Two Variables Lecture Notes 2009 Part II p63to97

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