285 Exam3 ReviewProblems 2012 Spring

7/31/2019 285 Exam3 ReviewProblems 2012 Spring

1/23

Math 285 - Spring 2012 - Review Material - Exam 3

Section 9.2 - Fourier Series and Convergence

State the definition of a Piecewise Continuous function.Answer: f is Piecewise Continuous if the following to conditions are satisfied:

1) It is continuous except possibly at some isolated points.2) The left and right limits f(x+) and f(x) exist (finite) at the points of discontinuity.

For example, Square-Wave functions are piecewise continuous.

But f(x) = sin 1x for 0 < x < 1, extended to R periodically, is not piecewise continuous,since the right limit at zero DNE.

Also f(x) = tanx for 0 < x < 2

and for 2

< x < , extended to R periodically, is notpiecewise continuous, since the right and left limits at

2DNE (infinite).

Define Fourier Series for functions for a Piecewise Continuous periodic function withperiod 2L.

Answer:

f(x) a02

+

n=1

an cos n

Lx + bn sin n

Lx

where for n 0an =

1

L

LL

f(x) cos n

Lx dx

and for n 1bn =

1

L

LL

f(x) sin n

L

x dx

State the definition of a Piecewise Smooth function.Answer:

f(x) is Piecewise Smooth if both f(x) and f(x) are piecewise continuous.For example, Square-Wave functions are piecewise smooth.

State the Convergence Theorem for Fourier Series.Answer:

If f(x) is periodic and piecewise smooth, then its Fourier Series converges to1) f(x) at each point x where f is continuous.

2) 12 (f(x+) + f(x)) at each point where f is NOT continuous.

Compute the Fourier Series of Square-Wave Function with period 2L:

f(x) =

+1 0 < x < L

1 L < x < 2LFor which values of f(x) at the points of discontinuity the Fourier series is convergentfor all x?

1


2/23

2

Answer:

an = 0 for all n 0, and bn = 2n(1cos n) for all n 1. Thus

f(x) = n=odd

4

nsin n

Lx

Note that f(x) is discontinuous at x = integer multiples ofL at which the average of leftand right limit of f(x) is zero. Thus by Convergence Theorem f(x) must be 0 at those

points.

Letting x = L2

in the Fourier Series representation of the Square-Wave Function, obtain

the following relation:

k=0

(1)k2k+ 1

= 1 13

+1

5 1

7+ =

4

Answer:

Note that if we let x = L2

, then the Fourier series is convergent to f(x) = 1, thus we have

1 = fL2= 4

k=0

sin(2k+ 1)2

2k+ 1

=4

k=0

(1)k

2k+ 1Hence

k=0

(1)k2k+ 1

= 1 13

+1

5 1

7+ =

4


3/23


4/23

4

Example: f(x) = x2 +x + 1 on (0,L)Then

fO(x) =

x2 +x + 1 0 < x < L

x2 +x1 L < x < 0Remark:

1) fE(x) is an even function with Fourier Series of the form a02 +an cos nLxHence f(x) = a0

2+an cos n

Lx for x in (0,L).

This is called the Fourier cosine series of f

2) fO(x) is an odd function with Fourier Series of the form bn sin n

Lx

Hence f(x) = bn sin n

Lx for x in (0,L).

This is called the Fourier sine series of f

Remark: Note that for x in (0,L), f(x) = fE(x) = fO(x).In many cases we are not concerned about f(x) on (L, 0), so the choice between (1)and (2) depends on our need for representing f by sine or cosine.

Example: f(t) = 1 on (0,). Compute the Fourier sine and cosine series and graph thetwo extensions.

1) The Even extension is fE(t) = 1 on (,), period 2.Then a0 = 2, an = 0, bn = 0, so the cosine series is just

f(t) =a0

2= 1

2) The Odd extension is fO(t) = 1 on (0,) and 1 on (, 0), period 2.Then a0 = an = 0, bn =

2

n

(1

(

1)n), so the cosine series is

fO(t) = n=odd

4

nsin nx

Example: f(t) = 1 t on (0, 1). Compute the Fourier sine and cosine series and graphto the to extensions.

1) The Even extension is fE(t) = 1 t on (0, 1) and 1 + t on (1, 0), period 2L, L = 1.Then a0 = 1, an = 2

1cos nn22

, bn = 0, so the cosine series is

f(t) =1

2

+ n=odd

4

n22cos nx

2) The Odd extension is fO(t) = 1 t on (0, 1) and 1 t on (1, 0), period 2L, L = 1.Then an = 0 for all n 0 and bn = 2n , so the cosine series is

f(t) =

n=1

2

nsinnx

Termwise differentiation of a Fourier seriesTheorem: Suppose f(x) is Continuous for all x, Periodic with period 2L, and f is


5/23

5

Piecewise Smooth for all t. If

f(x) =a0

2+

n=1

an cos n

Lx + bn sin n

Lx

then

f(x) =

n=1

(an nL

) sin n

Lx + (bn

n

L) cos n

Lx

Remarks:

1) RHS it the Fourier series of f(x).2) It is obtained by termwise differentiation of the RHS for f(x).3) Note that the constant term in the FS of f(x) is zero as

LL

f(x)dx = f(L) f(L) = 0

4) The Theorem fails if f is not continuous!

For example: consider f(x) = x on (L,L)Then

f(x) = x =

2L

n(

1)n+1 sin n

Lx

if we differentiate

1 =2

(1)n+1 cos n

Lx

For example equality fails at t = 0 or L!

Example: Verify the above Theorem for f(x) = x on (0,L) and f(x) = x on (L, 0).Answer:

bn = 0 as f is even, a0 = L, an =2L

n22((1)n1). Thus

f(x) =1

2L +

n=odd

4Ln22

cos n

Lx

Then term wise derivative gives

f(x) = n=odd

4

nsin n

Lx

On the other hand, directly computing the derivative of f(x) we have f(x) = 1 on (0,L)and f(x) = 1 on (L, 0). Thus an = 0 for all n and bn = 2n(1 cos n) which givesthe same Fourier Series as in above.

Applications to BVPsConsider the BVP of the form

ax + bx + cx = f(t), x(0) = x(L) = 0 or x(0) = x(L) = 0

Example:x + 2x = 1, x(0) = x() = 0

Here f(t) = 1, restrict to the interval (0,) as in Boundary Values.The idea is to find a formal Fourier Series solution of the equation.

Since we want x(0) = x() = 0, we prefer to consider the Fourier sine series ofx(t) onthe interval (0,),

x(t) =bn sin nt


6/23

6

substitute in the equation and compare the coefficients with that of the sine Fourier

Series of f(t) = 1 which is

f(t) = n=odd

4

nsin nt

Note that by term wise differentiation,

x(t) + 2x(t) = (2n2

)bn sin ntThus bn = 0 for n even and bn =

4n(2n2) for n odd. Hence a formal power series ofx(t)

is

x(t) = n=odd

4

n(2n2) sin nt

Example:x + 2x = t, x(0) = x() = 0

Here f(t) = t, restrict to the interval (0,) as in Boundary Values.Since we want x(0) = x() = 0, we prefer to consider the Fourier Cosine Series of

x(t) on (0,),

x(t) =a0

2+an cos nt

substitute in the equation and compare the coefficients with that of the Cosine FS of

f(t) = t which is

f(t) =

2

n=odd

4

n2cos nt

Note that by termwise differentiation,

x(t) + 2x(t) = a0 + (2n2)an cos ntThus a0 =

2

, an = 0 for n even and an =

4

n2(2

n2)for n odd. Hence a formal power

series ofx(t) is

x(t) =

4

n=odd

4

n2(2n2) cos nt

Termwise Integration of the Fourier Series.Theorem: Suppose f(t) is Piecewise Continuous (not necessarily piecewise smooth)periodic with period 2L with FS representation

f(t) a02

+

n=1

an cos n

Lt+ bn sin n

Lt

Then we can integrate term by term ast

0 f(x)dx =t

0a02

+n=1 ant

0 cos n

Lxdx + bnt

0 sin n

Lxdx

= a02

t+n=1 anLn

sin nLxbn Ln(cos nLx1)Note that the RHS is not the Fourier series of LHS unless a0 = 0.

Example: Consider f(x) = 1 on (0,) and f(x) = 1 on (, 0), then

f(t) = n=odd

4

nsin nt


7/23

7

Then F(t) =t

0 f(x)dx = t for t (0, 1) and t for t (, 0), whose Fourier Series is

F(t) =1

2 4

n=odd

1

n2cos nt

Term by term integration gives the same thing!

4

n=odd

1

n2

(1

cos nt) =

4

n=odd

1

n2

=2

8

4

n=odd

1

n2

cos nt


8/23

8

Section 9.4 - Applications of Fourier Series

Finding general solutions of 2nd order linear DEs with constant coefficients:

Example: x + 5x = F(t), where F(t) = 3 on (0,) and 3 on (, 0), is odd withperiod 2.

We obtain a particular solution in the following way:

Since L = and the FS ofF(t) is n odd12n

sin nt,

we may assume x(t) is odd and we consider its Fourier sine series F(t) =n=1 bn sin nt.Substitute in the equation:

x + 5x = (5n2)bn sin nt= F(t) = n odd

12

nsin nt

Hence, comparing the coefficients of sin nt on both sides we get

bn =12

n(5

n2)

ifn is odd and zero otherwise.

Thus a particular solution is x(t) = n odd 12n(5n2) sin nt

Definition: We call this a steady periodic solution, denoted by xsp (t).

Thus, ifx1(t),x2(t) are the solutions of the associated homogeneous equation, then thegeneral solution is

x(t) = c1x1(t) + c2x2(t) +xsp (t)= c1 cos(

5t) + c2 sin(

5t) +n odd

12n(5n2) sin nt

Remark: Consider the equation x + 9x = F(t). Then when trying to find a particularsolution we get

x + 9x = (9n2)bn sin nt= F(t) = n odd

12

nsin nt

We can not find b3 as 9n2 = 0 for n = 3.In this case we need to use the method of undetermined coefficients to find a function

y(t) such that

y + 9y =12

3sin3t

Take y = Atsin3t+Btcos3t, and substitute to find A = 0 and B = 23

.

Therefore, the general solution is

x(t) = c1 cos(3t) + c2 sin(3t) + n odd,n=3

12

n(9n2) sin nt2

3tcos3t

Definition: We say in this case a pure resonance occurs.

Remark: To determine the occurrence of pure resonance, just check if for some n,

sin nL t is a solution of the associated homogeneous equation.


9/23

9

Application: Forced Mass-Spring SystemsLet m be the mass, c be the damping constant, and k the constant of spring. Then

mx + cx + kx = F(t)

Consider the case that the external force F(t) is odd or even periodic function.Remark: If F(t) is periodic for t 0, it can be arranged to be odd or even by passingto odd or even extension for values of time t.

Case 1) Undamped Forced Mass-Spring Systems: c = 0

mx + kx = F(t)

Let 0 =

km

be the natural frequency of the system, then we can write

x +20x =1

mF(t)

Assume F(t) is periodic odd function with period 2L.Then

F(t) =

n=1 Fn sin n

Lt

Consider the odd extension of x(t), so that

x(t) =

n=1

bn sin n

Lt

substituting in the equation we get

x +20x =20 (

n

L)2

bn sin n

Lt =

1

mF(t) =

n=1

1

mFn sin n

Lt

Thus 20 (

n

L )2

bn =1

m Fn

If for all n 1, 20( nL )2 = 0, or equivalently

km

Lis not a positive integer, then we

can solve for bn as

bn =1m

Fn

20 ( nL )2Hence we have a steady periodic solution

xsp (t) =

n=1

1m

Fn

20 ( nL )2sin n

Lt

Example: Ifm = 1, k= 5,L = ,

x + 5x = F(t)then 20 ( nL )2 = 5n2 = 0 for all positive integers n 1.

On the other hand, if n0 =

km

Lis a positive integer, then a particular solution is

x(t) =

n=1,n=n0

1m

Fn

20 ( nL )2sin n

Lt Fn0

2m0tcos0t


10/23

10

Example: m = 1, k= 9,L = ,

x + 9x = F(t)

then 20 ( nL )2 = 9n2 = 0 for n = 3. There will be a pure resonance.

Example: m = 1, k= 9,L = 1,

x + 9x = F(t)

then 20 ( nL )2 = 9 (n)2 = 0 for all positive integers n. We have a steady periodicsolution.

Case 2) Damped Forced Mass-Spring Systems: c = 0

mx + cx + kx = F(t)Assume F(t) is periodic odd function with period 2L.Then

F(t) =

n=1

Fn sin n

L

t

For each n 1, we seek a function xn(t) such thatmxn + cx

n + kxn = Fn sin n

Lt = Fn sinnt

where n = n

L. Note that there will never be a duplicate solution as c = 0.

Hence using the method of undetermined coefficients, we can show

xn(t) =Fn

(km2n)2 + (cn)2sin(ntn)

where

n = tan1 cn

km2n 0 Therefore, we have a steady periodic solution

xsp (t) =

n=1

xn(t) =

n=1

Fn(km2n)2 + (cn)2

sin(ntn)

Example: m = 3, c = 1, k = 30, F(t) = t t2 for 0 t 1 is odd and periodic withL = 1. Compute the first few terms of the steady periodic solution.

3x +x + 30x = F(t) =

n=1

Fn sin n

Lt =

n=odd

8

n33sin nt

So, we have

xsp (t) = n=1 Fn

(km2n)2+(cn)2sin(ntn)

= n= odd8

n33(303n22)2+n22 sin(ntn)

n = tan1

n

30n22

0 The fist two terms are

0..0815sin(t1.44692) + 0.00004sin(333.10176) +


11/23

11

Section 9.5 - Heat Conduction and Separation of Variables

Until now, we studied ODEs - which involved single variable functions.In this section will consider some special PDEs - Differential Equations of several vari-

able functions involving their Partial Derivatives - and we will apply Fourier Series

method to solve them.

Heat Equations:Let u(x,t) denote the temperature at pint x and time t in an ideal heated rod that extendsalong x-axis. then u satisfies the following equation:

ut = kuxx

where kis a constant - thermal diffiusivity of the material - that depends on the material

of the rod.

Boundary Conditions:

Suppose the rod has a finite length L, then 0 x L.

1) Assume the temperature of the rod at time t = 0 at every point x is given. Thenwe are given a function f(x) such that u(x, 0) = f(x) for all 0 x L.

2) Assume the temperature at the two ends of the rod is fixed (zero) all the time - say by

putting two ice cubes! Then u(0,t) = u(L,t) = 0 for all t 0.

Thus we obtain a BVP,

ut = kuxxu(x, 0) = f(x) for all 0 x L

u(0,t) = u(L,t) = 0 for all t 0Remark: Other possible boundary conditions are, insulating the endpoints of the rod,

so that there is no heat flow. This means

ux(0,t) = ux(L,t) = 0 for all t 0Remark: Geometric Interpretation of the BVP.

We would like to find a function u(x,t) such that on the boundary of the infinite stript 0 and 0 x L satisfies the conditions u = 0 and u = f(x).

Remark: If f(x) is piecewise smooth, then the solution of the BVP is unique.

Important observations:

1) Superposition of solutions: If u1, u2,. .. satisfy the equation ut = kuxx, then so doesany linear combination of the uis. In other words, the equation ut = kuxx is linear!

2) Same is true about the boundary condition u(0,t) = u(L,t) = 0 for all t 0. Wesay this is a linear or homogeneous condition.

3) The condition u(x, 0) = f(x) for all 0 x L is not homogenous, or not linear!


12/23

12

General Strategy: Find solutions that satisfy the linear conditions and then take asuitable linear condition that satisfies the non-linear conditions.

Example: Verify that un(x,t) = en2tsin nx is a solution ofut = uxx (here k= 1) for anypositive integer n. For example, u1(x,t) = e

tsinx and u2(x,t) = e4tsin2x.

Example: Use the above example to construct a solution of the following BVP.

ut = uxxu(0,t) = u(,t) = 0 for all t 0u(x, 0) = 2sinx + 3sin2x for all 0 x

Answer:

Here L = . Note that un(x,t) = en2tsin nx also satisfy the linear condition u(0,t) =

u(,t) = 0. Thus it is enough to take u(x,t) to be a linear combination of uns thatsatisfies the non homogenous condition u(x, 0) = 2sinx + 3sin2x. Since un(x, 0) =sin nx, we take

u(x,t) = 2u1 + 3u2 = 2etsinx + 3e4tsin2x

so that

u(x, 0) = 2e0 sinx + 3e0 sin2x = 2sinx + 3sin2x

Remark: The above method in the example for ut = uxx works whenever f(x) is a finitelinear combination of sinx, sin2x,. ..

Example: Use the above example to construct a solution of the following BVP.

ut = uxxu(0,t) = u(,t) = 0 for all t 0u(x, 0) = sin4x cosx for all 0

x

L

Solution: Again we have L = . Note that

f(x) = sin4x cosx =1

2sin(4x +x) +

1

2sin(4xx) = 1

2sin5x +

1

2sin3x

Thus we take

u(x,t) =1

2u3 +

1

2u5 =

1

2e9tsin3x +

1

2e25t sin5x

Remark: When f(x) is a not a finite linear combination of the sine functions, then rep-resent it as an infinite sum using Fourier sine series.

Example: Construct a solution of the following BVP.

ut = uxxu(0,t) = u(,t) = 0 for all t 0u(x, 0) = 1 for all 0 x

Answer: Note that f(x) = 1 and L = , so represent f(x) as a Fourier sine series withperiod 2L = 2

f(x) = n= odd

4

nsin nx


13/23

13

Thus we take

u(x,t) = n= odd

4

nun(x,t) =

n= odd

4

nen

2t sin nx

This is a formal series solution of the BVP, one needs to check the convergence, ...

We only take finitely many terms for many applications.

In general, to solve the following BVP (k is anything, not necessarily 1, and L is any-thing, not just )

ut = kuxxu(0,t) = u(,t) = 0 for all t 0u(x, 0) = f(x) for all 0 x L

we observe that

un(x,t) = ek( nL )2tsin n

Lx

satisfies the equation ut = kuxx and the homogenous boundary condition u(0,t) = u(L,t) =0 for all t

0. Thus, represent f(x) as a Fourier sine series with period 2L,

f(x) =

n=1

bn sin n

Lx

Then

u(x,t) :=

n=1

bnun(x,t) =

n=1

bnek( nL )2tsin n

Lx

satisfies the non homogenous condition u(x, 0) = f(x) for all 0 x L.

Case of a rod with insulated endpoints:Consider the BVP corresponding to a heated rod with insulated endpoint,

ut = kuxxux(0,t) = ux(,t) = 0 for all t 0u(x, 0) = f(x) for all 0 x L

we observe that for n 0,un(x,t) = e

k( nL )2t cos n

Lx

satisfies the equation ut = kuxx and the homogenous boundary condition ux(0,t) =ux(L,t) = 0 for all t 0.Remark: for n = 0, we get u0 = 1 which satisfies the linear conditions!Thus, represent f(x) as a Fourier cosine series with period 2L,

f(x) =a0

2+

n=1

an cos n

Lx

Then

u(x,t) :=a0

2+

n=1

anun(x,t) =a0

2+

n=1

anek( nL )2tcos n

Lx

satisfies the non homogenous condition u(x, 0) = f(x) for all 0 x L.


14/23

14

Remarks:1) In the BVP for heated rod with zero temperature in the endpoints, we have

limtu(x,t) = limt

n=1

bnek( nL )2tsin n

Lx = 0

in other words, heat goes away with no insulation, thus temperature is zero at the end!

2) In the BVP for heated rod with insulated endpoints, we have

limtu(x,t) = limt

a0

2+

n=1

bnek( nL )2tcos n

Lx =

a0

2

which means, with insulation heat distributes evenly throughout the rod, which is the

average of the initial temperature as

a0

2=

1

L

L0

f(x)dx


15/23

15

Section 9.6 - Vibrating Strings and the One-Dimensional Wave Equation

Consider a uniform flexible string of length L with fixed endpoints,stretched along x-axis in the xy-plane from x = 0 to x = L.Let y(x,t) denote the displacement of the points x on the string at time t(we assume the points move parallel to y-axis).

Then y satisfies the One-dimensional wave equation:

ytt = a2yxx

where a is a constant that depend on the material of the string and the tension!

Boundary Conditions:

1) Since the endpoints are fixed, y(0,t) = y(L,t) = 0.2) The initial position of the string y(x, 0) at each x is given as a function y(x, 0) = f(x).3) The solution also depends on the initial velocity yt(x, 0) of the string at each x, givenas a function yt(x, 0) = g(x).

Thus we obtain the following BVP

ytt = a2yxxy(0,t) = y(L,t) = 0 for all t 0y(x, 0) = f(x) for all 0 x Lyt(x, 0) = g(x) for all 0 x L

Important Observations:1) ytt = a

2yxx is a linear equation. Thus the superposition of solutions applies.

2) Condition y(0,t) = y(L,t) = 0 is linear.3) Conditions y(x, 0) = f(x) and yt(x, 0) = g(x) are not linear.

General Strategy: Split the BVP into two problems,

(A)

ytt = a2yxx

y(0,t) = y(L,t) = 0 for all t 0y(x, 0) = f(x) for all 0 x Lyt(x, 0) = 0 for all 0 x L

(B)

ytt = a2yxx

y(0,t) = y(L,t) = 0 for all t 0y(x, 0) = 0 for all 0 x Lyt(x, 0) = g(x) for all 0 x L

IfyA(x,t) and yB(x,t) are the respective solutions, then

y(x,t) = yA(x,t) +yB(x,t)

satisfies the original BVP as

y(x, 0) = yA(x, 0) +yB(x, 0) = f(x) + 0 = f(x)

and

yt(x, 0) = (yA)t(x, 0) + (yB)t(x, 0) = 0 + g(x) = g(x)

Solving a BVP of type (A)

(A)

ytt = a2yxxy(0,t) = y(L,t) = 0 for all t 0y(x, 0) = f(x) for all 0 x Lyt(x, 0) = 0 for all 0 x L


16/23

16

Verify directly that for all positive integers n, the function

yn(x,t) = cos n

Lat sin n

Lx

satisfies the equation ytt = a2yxx and the linear conditions y(0,t) = y(L,t) = 0 andyt(x, 0) = 0. Thus, by superposition law, we need to find coefficients bn such that

y(x,t) =

n=1

bnyn(x,t) =

n=1

bn cos n

Lat sin n

Lx

satisfies y(x, 0) = f(x). Note that

f(x) = y(x, 0) =

n=1

bnyn(x, 0) =

n=1

bn sin n

Lx

Thus bns are the Fourier sine coefficients of f(x).

Example: Triangle initial position (pulled from the midpoint) with zero initial velocityAssume a = 1, L = and f(x) =

x if 0 < x <

2

x if 2

< x < . Then the BVP is type (A)

(A)

ytt = yxxy(0,t) = y(,t) = 0 for all t 0

y(x, 0) = f(x) =

x if 0 < x <

2

x if 2

< x <

yt(x, 0) = 0 for all 0 x

Fourier sine series of f(x) is

n=1

4sin n2

n2sin nx =

n=odd

4(1) n12n2

sin nx

Thus, since a = 1 and L = ,

y(x,t) = n=1 bn cos n

Lat sin n

Lx

= n=14sin n2n2

cos nt sin nx

= n=odd4(

1)n1

2

n2 cos nt sin nx= n=odd

4(1) n12n2

cos nt sin nx= 4

cos tsinx 4

9cos3tsin3x + 4

25cos5tsin5x +

Remark: Using the identity

2sinA cosB = sin(A +B) + sin(AB)

we can write the solution as


17/23

17

y(x,t) = n=odd4(1) n12

n2 cos nt sin nx

= 12 n=odd

4(1) n12n2

2cos nt sin nx= 1

2 n=odd4(1) n12

n2 (sin(nx + nt) + sin(nxnt))

= 12 n=odd4(

1)n1

2

n2 sin n(x + t) + 12 n=odd4(

1)n1

2

n2 sin n(x t)= 1

2fO(x + t) +

12

fO(x t)

Solving a BVP of type (B)

(A)

ytt = a2yxx

y(0,t) = y(L,t) = 0 for all t 0y(x, 0) = 0 for all 0 x Lyt(x, 0) = g(x) for all 0 x L

Verify directly that for all positive integers n, the function

yn(x,t) = sin n

Lat sin n

Lx

satisfies the equation ytt = a2yxx and the linear conditions y(0,t) = y(L,t) = 0 andy(x, 0) = 0. Thus, by superposition law, we need to find coefficients cn such that

y(x,t) =

n=1

cnyn(x,t) =

n=1

cn sin n

Lat sin n

Lx

satisfies yt(x, 0) = g(x). Note that by termwise differentiation with respect to variable twe have

yt(x,t) =

n=1

cn(n

La) cos n

Lat sin n

Lx

Thus

g(x) = yt(x, 0) =

n=1

cn(n

La) sin n

Lx

Then cn(n

La)s are the Fourier sine coefficients of g(x).

Hence cn =

nLan-th Fourier sine coefficients ofg(x)

Example: Assume a = 1, L = and g(x) = 1. Then the following BVP is type (B)

(B)

ytt = yxxy(0,t) = y(,t) = 0 for all t 0y(x, 0) = 0

yt(x, 0) = g(x) = 1 for all 0 x Fourier sine series of g(x) = 1 with L = is

n=odd

4

nsin nx


18/23

18

Thus, since a = 1 and L =

y(x,t) = n=1 cn sin n

Lat sin n

Lx

= n=odd

n(1) 4n sin nt sin nx= n=odd

4n2

sin nt sin nx


19/23

19

Section 9.7 - Steady-State Temperature and Laplaces Equation

Consider the temperature in a 2-dimensional uniform thin plate in xy-planebounded by a piecewise smooth curve C.

Let u(x,y,t) denote the temperature of the point (x,y) at time t.Then the 2-dimensional eat equation states that

ut = k(uxx + uyy)

where k is a constant that depends on the material of the plate.

If we let 2u = uxx + uyy, which is called the laplacian ofu, then we can write

ut = k2u

Remark: The 2-dimensional wave equation is

ztt = a2(zxx +zyy) = a

22z

where z(x,y,t) is the position of the point (x,y) in a vibration elastic surface at time t.

Case of the steady state temperature: i.e. we consider a 2-dimensional heat equationin which the temperature does not change in time (The assumption is after a while

temperature becomes steady). Thus

ut = 0

Therefore, 2u = uxx + uyy = 0.This equation is called the 2-dimensional Laplace equation.

Boundary problems and Laplace equation:If we know the temperature on the boundary C of a plate, as a function f(x,y), can wedetermine the temp. at every point inside the plate?

In other word, can we solve the BVPuxx + uyy = 0

u(x,y) = f(x,y) on the boundary of the plate

This BVP is called a Dirichlet Problem.

Remark: If the Boundary Cis Piecewise Smooth and the function f(x,y) is Nice!, thenDirichlet Problem has a unique solution.

We will consider the cases that C is rectangular or circular!

Case of Rectangular Plates:Suppose the plate is a rectangle positioned inxy-plane with vertices (0, 0), (0, b), (a, b), (a, 0).Assume we are given the temp. at each side of the rectangle.

Then we have the following type BVP.


20/23

20

uxx + uyy = 0

u(x, 0) = f1(x,y), 0 < x < a

u(x, b) = f2(x,y), 0 < x < a

u(0,y) = g1(x,y), 0 < y < b

u(a,y) = g1(x,y), 0 < y < b

General Strategy:The main equation uxx + uyy = 0 is linear, but all of the boundary conditions are non-linear, the idea is to split this BVP into 4 simpler BVP denotes by A, B, C, D, in which

only one of the boundary conditions in non-linear and apply Fourier series method there,

and and the end,

u(x,y) = uA(x,y) + uB(x,y) + uC(x,y) + uD(x,y)

Solving a type BVP of type (A)For 0 < x < a and 0 < y < b,

uxx + uyy = 0

u(x, 0) = f1(x,y)

u(x, b) = 0

u(0,y) = 0

u(a,y) = 0

In this case, one can show

un(x,y) = sinn

ax sinh n

a(by)

satisfies the linear conditions of the BVP.Recall that sinhx = 1

2(ex ex) and coshx = 1

2(ex + ex),

so (sinhx) = coshx and (coshx) = sinhx

Method of separation of variables: To actually find unsAssume u(x,y) = X(x)Y(y).Then uxx + uyy = 0 implies X

Y+XY = 0.Thus X

X

=YY

.

Since RHS only depends on y and LHS only depends on x, these fractions must be con-

stant, say .Then X

X =

and Y

Y = .

Thus we obtain X and Y are nonzero solutions of X +X = 0 and YY = 0, suchthat Y(b) = 0, X(0) = X(a) = 0This is an endpoint problem on X,

we seek those values of for which there are nonzero solutions X, such that X(0) =X(a) = 0. The eigen values are n = (

na )

2

and the corresponding eigenfunctions are scalar multiples ofXn(x) = sinna x.

Now substitute n = (na

)2 in YY = 0 with Y(b) = 0, and solve for Y(y).We obtain the solutions are a scalar multiple of Yn(y) = sinh

na (by).

Hence un(x,y) = Xn(x)Yn(y) = sinna x sinh na (by)


21/23

21

Back to solving a type BVP of type (A)We want u(x,y) = cnun(x,y) such that u(x, 0) = f1(x).Hence

f1(x) = u(x, 0) =cnun(x, 0) =cn sinn

ax sinh n

a(b)

Therefore, cn sinh na (b) is the n-th Fourier sine coefficient bn of f1(x) over interval(0, a). Hence

cn = bn/ sinhnb

a Example: Solve the BVP of type (A) ifa = b = and f1(x) = 1. Compute u(/2,/2),

the temp at the center of the rectangle.

Solution: Note that

f(x) = n=odd

4

nsin

n

ax =

n=odd

4

nsin nx

Thus

u(x,y) = n=odd

4

n/ sinh nb

a

sin n

ax sinh n

a(by)

Then

u(x,y) = n=odd

4

nsinh n

sin nx sinh n(y)

Also note that after computing the first few terms and using sinh2t = 2sinh tcosh t,

u(/2,/2) = n=odd

4nsinh n

sin n/2 sinh n/2= n=odd

2

nsinh n/2

sin n/2

.25In fact, one can argue by symmetry that is is exactly .25.

Case of a semi-infinite strip plate!Assume the plate is an infinite plate in the first quadrant whose vertices are (0, 0) and(0, b), and u = 0 along the horizontal sides, u(0,y) = g(y) and u(x,y) < as x .Apply the separation of variable method to solve the corresponding Dirichlet problem.

Answer:

Let u(x,y) = X(x)Y(y).Then uxx + uyy = 0 implies X

Y+XY = 0.Thus X

X =

YY

.

Since RHS only depends on y and LHS only depends on x, these fractions must be con-stant, say .

Then X

X= and Y

Y

=.Thus we obtain X and Y are nonzero solutions of XX = 0 and Y +Y = 0, suchthat Y(0) = Y(b) = 0, and u(x,y) = X(x)Y(y) in bounded as x This is an endpoint problem on Y.

We want those values of for which there are nonzero solutions Y, such that Y(0) =Y(b) = 0.The eigen values are n = (

nb )

2


22/23

22

and the corresponding eigenfunctions are scalar multiples ofYn(y) = sinnb y.

Now substitute n = (nb

)2 in XX = 0 and solve for X(x).We obtain Xn(x) = Ane

nb

x +Bne nb x

Since un = XnYn and u and Yn = sinnb y are bounded, then Xn must be bounded too.

Hence An = 0. Suppress bn.Thus un(x,y) = Xn(x)Yn(y) = e

nb x sin nb y

Note that un(x, 0) = un(x, b) = 0 and un(x,y)

285 Exam3 ReviewProblems 2012 Spring

Documents

Transcript of 285 Exam3 ReviewProblems 2012 Spring