fourier series of odd functions with period 2 l
DESCRIPTION
FOURIERTRANSCRIPT
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
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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.
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• 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.