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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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