Pde and Fourier

7/30/2019 Pde and Fourier

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Eigenvalue Problems

Consider the problem consisting of the differential equation

P(x)y + Q(x)y + y = 0,

together with some zero-value boundary conditions, e.g.

y() = 0, y() = 0.

The values of for which nontrivial solutions occur are called eigenval-

ues, and the nontrivial solutions are called eigenfunctions.

Examples.

3. In the problem, either solve the given boundary value problem or else

show that it has no solution.

y + y = 0, y(0) = 0, y(L) = 0

Solution.

y + y = 0

r2 + 1 = 0

r1 = i , r2 = iy = c1 cos t + c2 sin t

y = c1 sin t + c2 cos t

0 = c1

0 = c1 sin L + c2 cos L

c1 = 0cos L c2 = 0

y = 0 for all L;y = c2 sin x if sin L = 0

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Either solve the gien boundary problem or else show that it has no solution.

8. y + 4y = sin x, y(0) = 0, y() = 0

Solution.

y + 4y = 0

r2 + 4 = 0

r1 = 2i , r2 =

2i

y1 = cos2x , y2 = sin 2x

Assume Y = A cos x + B sin x,

Y = A sin x + B cos xY = A cos x B sin x

sin x = A cos x B sin x + 4(A cos x + B sin x)= A cos x B sin x + 4A cos x + 4B sin x

= 3A cos x + 3B sin x

3A = 03B = 1

A = 0B = 13

Y =1

3sin x.

Thus

y = c1 cos2x + c2 sin2x +

1

3 sin xChecking it with the boundary conditions, we obtain c1 = 0, that is the

solution has the form

y = c2 sin2x +1

3sin x.

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20. In the problem, find the eigenvalues and eigenfunctions of the given

boundary value problem. Assume that all eigenvalues are real.

x2y xy + y = 0, y(1) = 0, y(L) = 0, L > 1

Solution.

x2y xy + y = 0

r(r 1) r + = 0r 2r + = 0

r1 =2 +

4 42

, r2 =2 4 4

2

If < 1,we have two distinct real roots.

r1 =2 +

4 42

, r2 =2 4 4

2

y1 = x2+

442 , y2 = x

2442

Thus

y = c1x2+

442 + c2x

2442

Checking with the boundary conditions, 0 = c1 + c20 = c1L2+442 + c2L2442

c1 = 0

c2 = 0

So we only have a trivial solution, and there is no eigenvalue.

If = 1, we have two equal roots r1 = r2 = 1. Thus

y = (c1 + c2 ln x)x.

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Checking with the boundary conditions,

0 = c10 = (c1 + c2 ln L)L

c1 = 0c2 = 0

So we only have a trivial solution, and there is no eigenvalue.

If > 1, we have two complex roots

r1 = 1 +

1i, r2 = 1

1i.

Thusy = c1x cos(

1 ln x) + c2x sin(

1 ln x).

Checking with the boundary conditions, 0 = c10 = c1L cos( 1 ln L) + c2L sin( 1 ln L)

c2L sin(

1 ln L) = 0In order to find nontrivial functions, c2 = 0, thus

sin( 1 ln L) = 0 1 ln L = n, n = 1, 2, 3

1 = nln L

1 =

n

ln L

2

=

n

ln L

2+ 1

Thefore we have the eigenvalues and the corresponding eigenfunctions

n = n

ln L2

+ 1,

fn = x sin

n

ln Lln x

;

n = 1, 2, 3,

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10.2 Fourier Series

From now, we will study in detil the Fourier series:

a02

+

m=1

am cos

mx

L+ bm sin

mx

L

.

Periodicity

A function f is said to be periodic with period T > 0 if1) the domain of f contains x + T whenever it contains x;

2) f(x + T) = f(x) for every value of x.

The smallest value of T is called the fundamental period of f.

Orthogonality

The standard inner product (u, v) of two real-valued functions u and v

on the interval x is defined by

(u, v) =

u(x)v(x)dx.

The functions u and v are said to be orthogonal on x if

(u, v) =

u(x)v(x)dx = 0.

On the intervalL x L, the functions sin(mx/L) and cos(mx/L),m = 1, 2,

form a mutually orthogonal set of functions. In fact, we have

the following

LL

cosmx

Lcos

mx

Ldx =

0, m = nL, m = n

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L

Lcos

mx

L

sinmx

L

dx = 0, all m, n

LL

sinmx

Lsin

mx

Ldx =

0, m = nL, m = n

To prove these equalities, we need the following formulae:

sin A + sin B = 2 sin

A + B

2

cos

A B

2

;

sin A sin B = 2 cos

A + B

2

sin

A B

2

;

cos A + cos B = 2 cosA + B

2

cosA

B

2

;

cos A cos B = 2sin

A + B

2

sin

A B

2

.

Remark. Notice these sin and cos functions are actually the eigenfunctions

of

y + y = 0, on[L, L].The Euler-Fourier Formulas

We can find the Fourier series correponding to a given function f by

an =1

L

LL

f(x)cosnx

Ldx, n = 0, 1, 2,

bn =1

L

LL

f(x)sinnx

Ldx, n = 1, 2, 3,

Remark. It is easy to see that the Fourier series is unique. So if a trigono-

metric polynomial consists of the appropriate sine and cosine functions, it is

a Fourier series.

But this does not guarantee the convergence of the Fourier series, and that

the corresponding Fourier series is equal to the given function. In order to

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obtain these two points, we need more knowledge, which will be studied later.

Examples.

Problem 17 in section 10.2.

(a) Sketch the graph of the given function for three periods.

(b) Find the Fourier series for the given function.

f(x) =

x + L, L x 0,L, 0 < x < L;

f(x + 2L) = f(x)

Solution.

a0 =1

L

LL

f(x)dx

=1

L

L0

Ldx +1

L

0L

(x + L)dx

=1

L

LL

Ldx +1

L

0L

xdx

= 2L L

2

=3L

2

an =1

L

LL

f(x)cosnx

Ldx

=1

L

L0

L cosnx

Ldx +

1

L

0L

(x + L)cosnx

Ldx

=1

L

LL

L cosnx

Ldx +

1

L

0L

x cosnx

Ldx

= 1L

L Ln

sin nxL

LL

+ 1L

Ln

x sin nxL

0L 1

LL

n

0

Lsin nx

Ldx

=1

L

L

n

L

ncos

nx

L

0

L

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= 2L

(2k1)22, n = 2k 1

0, n = 2k

bn =1

L

LL

f(x)sinnx

Ldx

=1

L

L0

L sinnx

Ldx +

1

L

0L

(x + L)sinnx

Ldx

=1

L

LL

L sinnx

Ldx +

1

L

0L

x sinnx

Ldx

= 1L

LL

ncos

nx

L

L

L 1

L

L

nx cos

nx

L

0

L+

1

L

L

n

0L

cosnx

Ldx

= 1L

Ln

x cos nxL

0

L+ 1

LL

nL

nsin nx

L

0

L

= 1L

L

nx cos

nx

L

0

L

=1

n(L) cos(n)

= Ln

(1)n

=L

n(1)n+1

Therefore, we have

f(x) =3L

4+

n=1

2L cos[(2n 1)x/L]

(2n 1)22 +(1)n+1L sin(nx/L)

n

.

Problem 28.

If f is differentiable and is periodic with period T, show that f is also peri-

odic with period T. Determine whether F(x) = x0 f(t)dt is always periodic.Proof.

f(x + T) = limh0

f(x + T + h) f(x + T)h

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= limh0

f(x + h) f(x)

h= f(x)

So we can see that

1) the domain of f contains x + T whenever it contains x;

2) f(x + T) = f(x) for every value of x.

For the second part, we claim F(x) is not always periodic. Here we just

need a counterexample. Just consider any positive periodic function f, and

thus it is easy to know that F is increasing for the whole real line, which con-tradicts the property of periodicity. For example f(t) = 1 and thus F(x) = x.

13. Solution.

(b)

a0 =1

L

LLxdx = 0 (odd)

an =1

L L

Lx cos nx

Ldx

= 0 (odd)

bn =1

L

LLx sin nx

Ldx

= x1

ncos

nx

L

LL

LL

L

ncos

nx

Ldx

=2L

ncos n L

2

n22sin

nx

L

LL

=2L

ncos n

f(x) = 2Ln

n=1

(1)nn

sin nxL

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18. Solution.

(b)

a0 =1

2

22

f(x)dx

=1

2

11

xdx

= 0

an =1

2

22

f(x)cosnx

2dx

=1

2 1

1x cos

nx

2dx

= 0

bn =1

2

22

f(x)sinnx

2dx

=1

2

11

x sinnx

2dx

= 1n

x cosnx

2

11

+1

n

11

cosnx

2dx

= 2n

cosn

2+

4

n22sin

n

2

f(x) =

n=1

2

n cos

n

2 +

4

n22 sin

n

2

sin

nx

2

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10.3 The Fourier Convergence Theorem

In the previous section, we learnt the definition of the corresponding Fourier

series for a given function, and practice how to calculate it. But so far we

do not have any evidence to say that the series is equal to the function in

some mathematical sense, and even worse we do not know whether the series

converges. We need the convergence and the equality.

In order to reach that, some additional conditions are required. From

a practical point of view, piecewise continuity is broad enough and also

simple enough. But there are also some other conditions.

Piecewise Continuous

A function f is said to be piecewise conitnuous on an interval a x b,if the interval can be partitioned by a finite number of points a = x0 < x1 0,

with the initial condition

u(x, 0) = f(x), 0

x

L.

and the boundary conditions

u(0, t) = 0, u(L, t) = 0, t > 0

Assume

u(x, t) = X(x)T(t)

2X(x)T(t) = X(x)T(t)

X

X

=1

2

T

TAssume X

X= 1

2T

T= , we can see that is independent on x and t, so it

is a constant. Therefore

X + X = 0

X(0) = 0 , X(L) = 0

T + X = 0

Solving X, we have the eigenfunctions and the eigenvalues

Xn(x) = sin

nxL

,

n =n22

L2,

n = 1, 2, 3,

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and thus

T +

n222

L2

T = 0

T(t) = cnen222t/L2

un(x, t) = en222t/L2 sin

nx

L

Now we only to satisfy the initial condition u(x, 0) = f(x), set

u(x, t) =n=1

cnun(x, t) =n=1

cnen222t/L2 sin

nx

L

u(x, 0) =n=1

cn sin

nxL

cn =2

L

L0

f(x)sin

nx

L

dx

Examples.

2. Solution.

tuxx + xut = 0

tXT + xXT = 0X

xX= T

tT=

X xX = 0T + tT = 0

22. Solution.

2(XY T + XYT) = XY T

X

X+

Y

Y=

1

2T

T=

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T + 2T = 0

X

X = Y

Y = X + X = 0

Y + ( )Y = 0

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10.6

21. Solution. We assume that there is a decomposition u(x, t) = v(x) +

w(x, t)(hope so and then try our luck), such that, w is the solution to a

homogeneous heat equation

wt = 2wxx

w(0, t) = 0, w(L, t) = 0

Thus

ut = wt

uxx = vxx + wxx

Plug in the equation for u, we obtain the following

2vxx + s(x) = 0

v(0) = T1, v(L) = T2

and therefore

wt = 2wxx

w(0, t) = 0, w(L, t) = 0

w(x, 0) = f(x) v(x)

22. Solution.

(a) Use the formulas we got in 21,

vxx + k = 0

v(0) = T1, v(L) = T2

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Solving it, we have v(x) = T1 +T2T1

L x +kL2 x k2x2.

(b) Use the formulas in 21 and in part (a)

wt = wxx

w(0, t) = 0, w(20, t) = 0

w(x, 0) = 2x + 110

x2

Solving it,

w(x, t) =

n=1

160(cos n 1)

n3

3

en22t/400 sin

nx

20

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The Wave Equation

The one-dimensiongal wave equations

a2uxx = utt

in the domain 0 < x < L, t > 0.

Now Let us consider the following conditions:

the boundary conditions,

u(0, t) = 0, u(L, t) = 0, t

0;

and the initial conditions

u(x, 0) = f(x), ut(x, 0)

We also use the method of separation of variables to find the solution.

Assuming that

u(x, t) = X(x)T(t)

a2(X(x)T(t))xx = (X(x)T(t))tt

a2

X

T = XT

X

X=

1

a2T

T=

X + X = 0

T + a2T = 0

X(0) = 0, X(L) = 0.

T(0) = 0.

Now we have the two equations to solve, and the second equation is

dependent on the result of the first equation

X + X = 0

X(0) = 0

X(T) = 0

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T + a2T = 0

T(0) = 0

The first is just an eigenvalue problem, which we have dealt with in the heat

equations. So it is easy to know that

n =n22

L2, n = 1, 2, ,

Xn = sinnx

L

and thus we have

Tn = k1 cosnat

L+ k2 sin

nat

L

By the initial condition T(0) = 0, k2 = 0. So,

un(x, t) = sinnx

Lcos

nat

L

Therefore, we have a family of functions satisfying the wave equation

with the boudary conditions and the second initial condition. And it is quite

obvious that the linear combination of the functions such as

u(x, t) =n=1

cnun(x, t) =n=1

cn sinnx

Lcos

nat

L

under some convergence conditions.

So far, we only have one inital condition to meet.

u(x, 0) =n=1

cn sinnx

L

= f(x)

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According to the uniqueness of the Fourier series, we have the formula

for the coefficients

cn =2

L

L0

f(x)sinnx

Ldx, m = 1, 2,

Examples.

1a. Solution.

Here we just need to use the formula on the textbook,

u(x, t) =

n=1cn sin

nx

Lcos

nat

L

and

cn =2

L

L0

f(x)sinnx

Ldx.

So

cn =2

L

L/20

2x

Lsin

nx

Ldx +

2

L

LL/2

2(x L)L

sinnx

Ldx

=2

L

L0

2x

Lsin

nx

Ldx 2

L

LL/2

2sinnx

Ldx

Integration by parts

= 82

1n2

sin n2

Thereofore,

u(x, t) =8

2

n=1

1

n2sin

n

2sin

nx

Lcos

nat

L

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Laplace Equations

Dirichlet Problem for a Rectangle

uxx + uyy = 0

u(x, 0) = 0, u(x, b) = 0

u(0, y) = 0, u(a, y) = f(y)

Use separation of variables to find all possible solutions for the homoge-

neous conditions:

u = XY

uxx + uyy = 0 XY + XY = 0 X

X= Y

Y=

X X = 0Y + Y = 0

u(x, 0) = 0 X(x)Y(0) = 0

Y(0) = 0u(x, b) = 0 X(x)Y(b) = 0

Y(b) = 0u(0, y) = 0 X(0)Y(y) = 0

X(0) = 0

In sum,

X

X = 0

X(0) = 0 Y

+ Y = 0

Y(0) = 0, Y(b) = 0

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The second equation is an eigenvalue problem, and we have solved it several

times. The details are skipped here, but we know

n =n22

b2, n = 1, 2,

Yn = sin(n

by)

Xn = sinh(n

bx)

un = sinh(n

bx)sin(

n

by)

u =

n=1

cn sinh(n

b

x)sin(n

b

y)

u(a, y) =n=1

cn sinh(na

b) sin(

n

by) = f(y)

Now we use the Fourier series to find the cn. As this is a sine series, so

we apply the Fourier series of the odd extension of f.

f(y) =n=1

bn sin(n

by)

bn =2

b

b

0

f(y)sin(n

b

y)dy

cn sinh(na

b) = bn

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