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 1

x 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 < π

2 and for π

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

2 DNE (infinite).

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

f (x)∼ a0

2+

∞

∑n=1

an cosnπ

Lx+bn sinn

π

Lx

where for n≥ 0

an =1L

∫ L

−Lf (x)cosn

π

Lx dx

and for n≥ 1

bn =1L

∫ L

−Lf (x)sinn

π

Lx 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) 1

2( 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 < 2L

For which values of f (x) at the points of discontinuity the Fourier series is convergentfor all x?

1

2

Answer:an = 0 for all n≥ 0, and bn =

2nπ(1− cosnπ) for all n≥ 1. Thus

f (x) = ∑n=odd

4nπ

sinnπ

Lx

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

• Letting x = L2 in the Fourier Series representation of the Square-Wave Function, obtain

the following relation:∞

∑k=0

(−1)k

2k+1= 1− 1

3+

15− 1

7+ · · ·= π

4

Answer:Note that if we let x = L

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

1 = f(

L2

)=

4π

∞

∑k=0

sin(2k+1)π

22k+1

=4π

∞

∑k=0

(−1)k

2k+1

Hence∞

∑k=0

(−1)k

2k+1= 1− 1

3+

15− 1

7+ · · ·= π

4

3

Section 9.3 - Fourier Sine and Cosine Series

• Recall f (x) is odd if f (−x) =− f (x), i.e. its graph is symmetric w.r.t y-axis.Example: x2n+1,sinnx for all integers n 6= 0, the square-wave functions

f (x) =

{−1 −L < x < 01 0 < x < L

• Recall f (x) is even if f (−x) = f (x), i.e. its graph is symmetric w.r.t origin.Example: x2n,cosnx for all integers n.

• Remarks:1) If f (x) is odd, then

∫ L−L f (x)dx = 0 for any L

2) If f (x) is even, then∫ L−L f (x)dx = 2

∫ L0 f (x)dx for any L

3) If f and g are odd, then f g is even.4) If f and g are even, then f g is even.3) If f is odd and g is even, then f g is odd.

• Fourier series of odd functions with period 2L:a0 =

1L∫ L−L f (x)dx = 0, an =

1L∫ L−L f (x)cosnπ

Lxdx = 0 since f (x)cosnπ

Lx is odd.bn =

1L∫ L−L f (x)sinnπ

Lxdx = 2L∫ L

0 f (x)sinnπ

Lxdx since f (x)sinnπ

Lx is even.In this case, if f is piecewise smooth, f (x) = ∑bn sinnπ

Lx only involves sine.

Example: f (x) = x2 on (−π,π), then f (x) = ∑

∞n=1

(−1)n+1

n sinnx

• Fourier series of even functions with period 2L:an =

1L∫ L−L f (x)cosnπ

Lxdx = 2L∫ L

0 f (x)cosnπ

Lxdx since f (x)cosnπ

Lx is even.bn =

1L∫ L−L f (x)sinnπ

Lxdx = 0 since f (x)sinnπ

Lx is odd.In this case, if f is piecewise smooth, f (x) = a0

2 +∑an cosnπ

Lx only involves cosine.Example: Try f (x) = x2 on (−π,π).

• Even and odd extensions of a function:Suppose f is a piecewise continuous function defined on interval (0,L).Even extension of f to the interval (−L,0) is

fE(x) =

{f (x) 0 < x < Lf (−x) −L < x < 0

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

fE(x) =

{x2 + x+1 0 < x < Lx2− x+1 −L < x < 0

Odd extension of f to the interval (−L,0) is

fO(x) =

{f (x) 0 < x < L− f (−x) −L < x < 0

4

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

fO(x) =

{x2 + x+1 0 < x < L−x2 + x−1 −L < x < 0

Remark:

1) fE(x) is an even function with Fourier Series of the form a02 +∑an cosnπ

LxHence f (x) = a0

2 +∑an cosnπ

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 sinnπ

LxHence f (x) = ∑bn sinnπ

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,π). Compute the Fourier sine and cosine series and graph thetwo extensions.

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

2nπ(1− (−1)n), so the cosine series is

fO(t) = ∑n=odd

4nπ

sinnx

• 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 = 21−cosnπ

n2π2 ,bn = 0, so the cosine series is

f (t) =12+ ∑

n=odd

4n2π2 cosnπx

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 =

2nπ

, so the cosine series is

f (t) =∞

∑n=1

2nπ

sinnπx

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

5

Piecewise Smooth for all t. If

f (x) =a0

2+

∞

∑n=1

an cosnπ

Lx+bn sinn

π

Lx

then

f ′(x) =∞

∑n=1

(−annπ

L)sinn

π

Lx+(bn

nπ

L)cosn

π

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∫ L

−Lf ′(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 = ∑2Lnπ

(−1)n+1 sinnπ

Lx

if we differentiate

1 6= ∑2π(−1)n+1 cosn

π

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 =

2Ln2π2 ((−1)n−1). Thus

f (x) =12

L+ ∑n=odd

− 4Ln2π2 cosn

π

Lx

Then term wise derivative gives

f ′(x) = ∑n=odd

4nπ

sinnπ

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 =

2nπ(1− cosnπ) which gives

the same Fourier Series as in above.

• Applications to BVP’sConsider 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(π) = 0

Here f (t) = 1, restrict to the interval (0,π) as in Boundary Values.The idea is to find a formal Fourier Series solution of the equation.Since we want x(0) = x(π) = 0, we prefer to consider the Fourier sine series of x(t) onthe interval (0,π),

x(t) = ∑bn sinnt

6

substitute in the equation and compare the coefficients with that of the sine FourierSeries of f (t) = 1 which is

f (t) = ∑n=odd

4nπ

sinnt

Note that by term wise differentiation,

x′′(t)+2x(t) = (2−n2)bn sinnt

Thus bn = 0 for n even and bn =4

πn(2−n2)for n odd. Hence a formal power series of x(t)

is

x(t) = ∑n=odd

4πn(2−n2)

sinnt

• Example:x′′+2x = t, x′(0) = x′(π) = 0

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

x(t) =a0

2+∑an cosnt

substitute in the equation and compare the coefficients with that of the Cosine FS off (t) = t which is

f (t) =π

2− ∑

n=odd

4πn2 cosnt

Note that by termwise differentiation,

x′′(t)+2x(t) = a0 +(2−n2)an cosnt

Thus a0 =π

2 , an = 0 for n even and an = − 4πn2(2−n2)

for n odd. Hence a formal powerseries of x(t) is

x(t) =π

4− ∑

n=odd

4πn2(2−n2)

cosnt

• 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)∼ a0

2+

∞

∑n=1

an cosnπ

Lt +bn sinn

π

Lt

Then we can integrate term by term as∫ t0 f (x)dx =

∫ t0

a02 +∑

∞n=1 an

∫ t0 cosnπ

Lxdx+bn∫ t

0 sinnπ

Lxdx= a0

2 t +∑∞n=1 an

Lnπ

sinnπ

Lx−bnLnπ(cosnπ

Lx−1)

Note that the RHS is not the Fourier series of LHS unless a0 = 0.

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

f (t) = ∑n=odd

4πn

sinnt

7

Then F(t) =∫ t

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

F(t) =12

π− 4π

∑n=odd

1n2 cosnt

Term by term integration gives the same thing!

4π

∑n=odd

1n2 (1− cosnt) =

4π

(∑

n=odd

1n2 =

π2

8

)− 4

π∑

n=odd

1n2 cosnt

8

Section 9.4 - Applications of Fourier Series

• Finding general solutions of 2nd order linear DE’s with constant coefficients:

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

We obtain a particular solution in the following way:

Since L = π and the FS of F(t) is ∑n odd12nπ

sinnt,we may assume x(t) is odd and we consider its Fourier sine series F(t) = ∑

∞n=1 bn sinnt.

Substitute in the equation:

x′′+5x = (5−n2)bn sinnt = F(t) = ∑n odd

12nπ

sinnt

Hence, comparing the coefficients of sinnt on both sides we getbn =

12n(5−n2)π

if n is odd and zero otherwise.

Thus a particular solution is x(t) = ∑n odd12

n(5−n2)πsinnt

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

Thus, if x1(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(5−n2)π

sinnt

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

x′′+9x = (9−n2)bn sinnt = F(t) = ∑n odd

12nπ

sinnt

We can not find b3 as 9−n2 = 0 for n = 3.In this case we need to use the method of undetermined coefficients to find a functiony(t) such that

y′′+9y =123π

sin3t

Take y = At sin3t +Bt cos3t, and substitute to find A = 0 and B =− 23π

.Therefore, the general solution is

x(t) = c1 cos(3t)+ c2 sin(3t)+ ∑n odd,n6=3

12n(9−n2)π

sinnt− 23π

t cos3t

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

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

Lt is a solution of the associated homogeneous equation.

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 ω0 =√

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

x′′+ω20x =

1m

F(t)

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

F(t) =∞

∑n=1

Fn sinnπ

Lt

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

x(t) =∞

∑n=1

bn sinnπ

Lt

substituting in the equation we get

x′′+ω20x =

(ω

20− (

nπ

L)2)

bn sinnπ

Lt =

1m

F(t) =∞

∑n=1

1m

Fn sinnπ

Lt

Thus (ω

20− (

nπ

L)2)

bn =1m

Fn

If for all n≥ 1, ω20− (nπ

L )2 6= 0, or equivalently√

km ·

π

L is not a positive integer, then wecan solve for bn as

bn =1mFn

ω20− (nπ

L )2

Hence we have a steady periodic solution

xsp(t) =∞

∑n=1

1mFn

ω20− (nπ

L )2sinn

π

Lt

Example: If m = 1,k = 5,L = π,

x′′+5x = F(t)then ω2

0− (nπ

L )2 = 5−n2 6= 0 for all positive integers n≥ 1.

On the other hand, if n0 =√

km ·

π

L is a positive integer, then a particular solution is

x(t) =∞

∑n=1,n6=n0

1mFn

ω20− (nπ

L )2sinn

π

Lt− Fn0

2mω0t cosω0t

10

Example: m = 1,k = 9,L = π,

x′′+9x = F(t)

then ω20− (nπ

L )2 = 9−n2 = 0 for n = 3. There will be a pure resonance.

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

x′′+9x = F(t)

then ω20− (nπ

L )2 = 9− (nπ)2 6= 0 for all positive integers n. We have a steady periodicsolution.

Case 2) Damped Forced Mass-Spring Systems: c 6= 0

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

F(t) =∞

∑n=1

Fn sinnπ

Lt

For each n≥ 1, we seek a function xn(t) such that

mx′′n + cx′n + kxn = Fn sinnπ

Lt = Fn sinωnt

where ωn = nπ

L . Note that there will never be a duplicate solution as c 6= 0.Hence using the method of undetermined coefficients, we can show

xn(t) =Fn√

(k−mω2n)

2 +(cωn)2sin(ωnt−αn)

where

αn = tan−1(

cωn

k−mω2n

)0≤ α≤ π

Therefore, we have a steady periodic solution

xsp(t) =∞

∑n=1

xn(t) =∞

∑n=1

Fn√(k−mω2

n)2 +(cωn)2

sin(ωnt−αn)

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 sinnπ

Lt = ∑

n=odd

8n3π3 sinnπt

So, we have

xsp(t) = ∑∞n=1

Fn√(k−mω2

n)2+(cωn)2

sin(ωnt−αn)

= ∑n= odd

8n3π3√

(30−3n2π2)2+n2π2sin(nπt−αn)

αn = tan−1(

nπ

30−n2π2

)0≤ α≤ π

The fist two terms are

0..0815sin(πt−1.44692)+0.00004sin(3π3−3.10176)+ · · ·

11

Section 9.5 - Heat Conduction and Separation of Variables

• Until now, we studied ODE’s - which involved single variable functions.In this section will consider some special PDE’s - Differential Equations of several vari-able functions involving their Partial Derivatives - and we will apply Fourier Seriesmethod 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 k is a constant - thermal diffiusivity of the material - that depends on the materialof 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 byputting two ice cubes! Then u(0, t) = u(L, t) = 0 for all t ≥ 0.

Thus we obtain a BVP,ut = kuxx

u(x,0) = f (x) for all 0≤ x≤ Lu(0, t) = u(L, t) = 0 for all t ≥ 0

Remark: 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 ≥ 0

Remark: 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 ui’s. 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

• 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) = e−n2t sinnx is a solution of ut = uxx (here k = 1) for anypositive integer n. For example, u1(x, t) = e−t sinx and u2(x, t) = e−4t sin2x.

• Example: Use the above example to construct a solution of the following BVP.ut = uxx

u(0, t) = u(π, t) = 0 for all t ≥ 0u(x,0) = 2sinx+3sin2x for all 0≤ x≤ π

Answer:Here L = π. Note that un(x, t) = e−n2t sinnx also satisfy the linear condition u(0, t) =u(π, t) = 0. Thus it is enough to take u(x, t) to be a linear combination of un’s thatsatisfies the non homogenous condition u(x,0) = 2sinx + 3sin2x. Since un(x,0) =sinnx, we take

u(x, t) = 2u1 +3u2 = 2e−t sinx+3e−4t sin2x

so thatu(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 = uxx

u(0, t) = u(π, t) = 0 for all t ≥ 0u(x,0) = sin4xcosx for all 0≤ x≤ L

Solution: Again we have L = π. Note that

f (x) = sin4xcosx =12

sin(4x+ x)+12

sin(4x− x) =12

sin5x+12

sin3x

Thus we take

u(x, t) =12

u3 +12

u5 =12

e−9t sin3x+12

e−25t 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 = uxx

u(0, t) = u(π, t) = 0 for all t ≥ 0u(x,0) = 1 for all 0≤ x≤ π

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

f (x) = ∑n= odd

4nπ

sinnx

13

Thus we take

u(x, t) = ∑n= odd

4nπ

un(x, t) = ∑n= odd

4nπ

en2t sinnx

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 π)

ut = kuxx

u(0, t) = u(π, t) = 0 for all t ≥ 0u(x,0) = f (x) for all 0≤ x≤ L

we observe thatun(x, t) = e−k( nπ

L )2t sinnπ

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 sinnπ

Lx

Then

u(x, t) :=∞

∑n=1

bnun(x, t) =∞

∑n=1

bne−k( nπ

L )2t sinnπ

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 = kuxx

ux(0, t) = ux(π, 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( nπ

L )2t cosnπ

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 cosnπ

Lx

Then

u(x, t) :=a0

2+

∞

∑n=1

anun(x, t) =a0

2+

∞

∑n=1

ane−k( nπ

L )2t cosnπ

Lx

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

14

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

limt→∞

u(x, t) = limt→∞

∞

∑n=1

bne−k( nπ

L )2t sinnπ

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

limt→∞

u(x, t) = limt→∞

a0

2+

∞

∑n=1

bne−k( nπ

L )2t cosnπ

Lx =

a0

2

which means, with insulation heat distributes evenly throughout the rod, which is theaverage of the initial temperature as

a0

2=

1L

∫ L

0f (x)dx

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 BVPytt = a2yxx

y(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 = a2yxx 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

If yA(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)

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

• Solving a BVP of type (A)

(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

16

Verify directly that for all positive integers n, the function

yn(x, t) = cosnπ

Lat · sinn

π

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 cosnπ

Lat · sinn

π

Lx

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

f (x) = y(x,0) =∞

∑n=1

bnyn(x,0) =∞

∑n=1

bn sinnπ

Lx

Thus bn’s are the Fourier sine coefficients of f (x).

• Example: Triangle initial position (pulled from the midpoint) with zero initial velocity

Assume a = 1, L = π and f (x) =

{x if 0 < x < π

2π− x if π

2 < x < π. Then the BVP is type (A)

(A)

ytt = yxx

y(0, t) = y(π, t) = 0 for all t ≥ 0

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

{x if 0 < x < π

2π− x if π

2 < x < π

yt(x,0) = 0 for all 0≤ x≤ π

Fourier sine series of f (x) is

∞

∑n=1

4sin nπ

2πn2 sinnx = ∑

n=odd

4(−1)n−1

2

πn2 sinnx

Thus, since a = 1 and L = π,

y(x, t) = ∑∞n=1 bn cosnπ

Lat · sinnπ

Lx

= ∑∞n=1

4sin nπ

2πn2 · cosnt · sinnx

= ∑n=odd4(−1)

n−12

πn2 · cosnt · sinnx

= ∑n=odd4(−1)

n−12

πn2 · cosnt · sinnx= 4

πcos t sinx− 4

9πcos3t sin3x+ 4

25πcos5t sin5x+ · · ·

• Remark: Using the identity

2sinAcosB = sin(A+B)+ sin(A−B)

we can write the solution as

17

y(x, t) = ∑n=odd4(−1)

n−12

πn2 · cosnt · sinnx

= 12 ∑n=odd

4(−1)n−1

2

πn2 ·2cosnt · sinnx

= 12 ∑n=odd

4(−1)n−1

2

πn2 · (sin(nx+nt)+ sin(nx−nt))

= 12 ∑n=odd

4(−1)n−1

2

πn2 · sinn(x+ t)+ 12 ∑n=odd

4(−1)n−1

2

πn2 · sinn(x− t)= 1

2 fO(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) = sinnπ

Lat · sinn

π

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 sinnπ

Lat · sinn

π

Lx

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

yt(x, t) =∞

∑n=1

cn(nπ

La)cosn

π

Lat · sinn

π

Lx

Thus

g(x) = yt(x,0) =∞

∑n=1

cn(nπ

La)sinn

π

Lx

Then cn(nπ

La)’s are the Fourier sine coefficients of g(x).Hence cn =

π

nLa ·n-th Fourier sine coefficients of g(x)

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

(B)

ytt = yxx

y(0, t) = y(π, t) = 0 for all t ≥ 0y(x,0) = 0yt(x,0) = g(x) = 1 for all 0≤ x≤ π

Fourier sine series of g(x) = 1 with L = π is

∑n=odd

4πn

sinnx

18

Thus, since a = 1 and L = π

y(x, t) = ∑∞n=1 cn sinnπ

Lat · sinnπ

Lx= ∑n=odd

π

nπ(1) ·4

πn · sinnt · sinnx= ∑n=odd

4πn2 · sinnt · sinnx

19

Section 9.7 - Steady-State Temperature and Laplace’s 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 of u, then we can write

ut = k∇2u

Remark: The 2-dimensional wave equation is

ztt = a2(zxx + zyy) = a2∇

2z

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 whiletemperature 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 BVP{

uxx +uyy = 0u(x,y) = f (x,y) on the boundary of the plate

This BVP is called a Dirichlet Problem.

• Remark: If the Boundary C is 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 in xy-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

uxx +uyy = 0u(x,0) = f1(x,y),0 < x < au(x,b) = f2(x,y),0 < x < au(0,y) = g1(x,y),0 < y < bu(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 whichonly 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 = 0u(x,0) = f1(x,y)u(x,b) = 0u(0,y) = 0u(a,y) = 0

In this case, one can show

un(x,y) = sinnπ

ax · sinh

nπ

a(b− y)

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

2(ex− e−x) and coshx = 1

2(ex + e−x),

so (sinhx)′ = coshx and (coshx)′ =−sinhx

• Method of separation of variables: To actually find un’sAssume u(x,y) = X(x)Y (y).Then uxx +uyy = 0 implies X ′′Y +XY ′′ = 0.Thus X ′′

X =−Y ′′Y .

Since RHS only depends on y and LHS only depends on x, these fractions must be con-stant, say −λ.Then X ′′

X =−λ and Y ′′Y = λ.

Thus we obtain X and Y are nonzero solutions of X ′′+λX = 0 and Y ′′−λY = 0, suchthat Y (b) = 0, X(0) = X(a) = 0This is an endpoint problem on X ,we seek those values of λ for which there are nonzero solutions X , such that X(0) =X(a) = 0. The eigen values are λn = (nπ

a )2

and the corresponding eigenfunctions are scalar multiples of Xn(x) = sin nπ

a x.Now substitute λn = (nπ

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

a (b− y).Hence un(x,y) = Xn(x)Yn(y) = sin nπ

a x · sinh nπ

a (b− y)

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 · sinnπ

ax · sinh

nπ

a(b)

Therefore, cn · sinh nπ

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

cn = bn/sinhnπb

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

the temp at the center of the rectangle.Solution: Note that

f (x) = ∑n=odd

4nπ

sinnπ

ax = ∑

n=odd

4nπ

sinnx

Thus

u(x,y) = ∑n=odd

(4

nπ/sinh

nπba

)· sin

nπ

ax · sinh

nπ

a(b− y)

Then

u(x,y) = ∑n=odd

(4

nπsinhnπ

)· sinnx · sinhn(π− y)

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

u(π/2,π/2) = ∑n=odd( 4

nπsinhnπ

)· sinnπ/2 · sinhnπ/2

= ∑n=odd

(2

nπsinhnπ/2

)· sinnπ/2

∼ .25

In 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 =−Y ′′Y .

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

X = λ and Y ′′Y =−λ.

Thus we obtain X and Y are nonzero solutions of X ′′−λX = 0 and Y ′′+λY = 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 λ for which there are nonzero solutions Y , such that Y (0) =Y (b) = 0.The eigen values are λn = (nπ

b )2

22

and the corresponding eigenfunctions are scalar multiples of Yn(y) = sin nπ

b y.Now substitute λn = (nπ

b )2 in X ′′−λX = 0 and solve for X(x).We obtain Xn(x) = Ane

nπ

b x +Bne−nπ

b x

Since un = XnYn and u and Yn = sin nπ

b y are bounded, then Xn must be bounded too.Hence An = 0. Suppress bn.Thus un(x,y) = Xn(x)Yn(y) = e−

nπ

b x · sin nπ

b yNote that un(x,0) = un(x,b) = 0 and un(x,y)< ∞ as x→ ∞.

Now, to solve the BVP:We want u(x,y) = ∑cnun(x,y) such that u(0,y) = g(y).Hence

g(y) = u(0,y) = ∑cnun(0,y) = ∑cn · (1) · sinnπ

by

Therefore, cn is the n-th Fourier sine coefficient bn of g(y) over interval (0,b).

• Example: If b = 1 and g(y) = 1, then compute u(x,y).Solution:

g(y) = ∑n=odd

4nπ

sinnπ

bx = ∑

n=odd

4nπ

sinnx

Thus

u(x,y) = ∑n=odd

4nπ· e−nπx · sinnπy

• Case of a circular disk:Using polar coordinates (r,θ) to represent the points in a disk, on can transform theLaplace equation into

urr +1r

ur +1r2 uθθ = 0

Note that 0 < r < a = the radius of the disk, and 0 < θ < 2π.It turns out that the solution is of the form

u(r,θ) =a0

2+

∞

∑n=1

(an cosnθ+bn sinnθ)rn

Boundary conditions: If we know the temp. on the boundary of the disk, want todetermine the temp. inside. To give the temp. on the boundary means to give a functionf (θ) such that

u(a,θ) = f (θ) for 0 < θ < 2π

Thusf (θ) = u(a,θ) = a0

2 +∑∞n=1(an cosnθ+bn sinnθ)an

= a02 +∑

∞n=1 anan cosnθ+anbn sinnθ

Therefore anan and anbn are the n-th Fourier series coefficient of f (θ) over the interval(0,2π).

23

• Example: If f (θ) = 1 on (0,π) and −1 on (π,2π), and radius r = 1, then computeu(r,θ).

Answer:We compute the Fourier series of f (θ) = ∑n=odd

4nπ

sinnθ.Thus an = 0 for all n. Hence

u(r,θ) =a0

2+

∞

∑n=1

(an cosnθ+bn sinnθ)rn = ∑n=odd

4nπ

rn sinnθ

• Remark: Intuitively, in the above example u(r,0) = 0, which is consistent with thesolution: If θ = 0 or π,

u(r,θ) = ∑n=odd

4nπ

rn sinnθ = 0

• Remark: In the above example, if we change the boundary condition to f (θ) = 1 on(0,2π), then intuitively, u(r,θ) = 1 everywhere by symmetry. This is consistent with theactual solution to BVP: In this case, the Fourier coefficients of f are a0 = 2,an = bn = 0for all n≥ 1. Thus u(r,θ) = 1 for all 0 < r < 1 and 0 < θ < 2π.