Math 462: HW4 Solutions
Due on August 8, 2014
Jacky Chong
1
Jacky Chong
Remark: We are working in the context of Riemann Integrals.
Problem 1
4.1.3 A quantum-mechanical particle on the line with an infinite potential outside the interval (0, l) (“particle
in a box”) is given by Schrödinger’s equation ut = iuxx on (0, l) with Dirichlet conditions at the ends.
Separate the variables and use (8) to find its representation as a series.
Solution: Consider the problem
(
ut = iuxx
for
0<x<l
u(0, t) = u(l, t) = 0
.
Suppose
u(x, t) = X(x)T (t)
is a solution to the above problem, then we obtain the following decoupled system of ODEs
(
X 00 (x) = λX(x) T 0 (t) = iλT (t)
X(0) = X(l) = 0
.
Case 1. Suppose λ = 0, then we see that
X(x) = C1 + C2 x
is the general solution to X 00 = 0. Applying the boundary condition, we get the trivial solution X ≡ 0.
Case 2. Suppose λ ∈ R>0 , a positive real number, then we see that
√
X(x) = C1 e
λx
+ C2 e−
√
λx
is the general solution to the ode X 00 = λX. Applying the first boundary condition, we get that
X(0) = C1 + C2 = 0
which means X(x) = C sinh
√
i.e.
C2 = −C1
λx. However, applying the second boundary condition, we get that
√
X(l) = C sinh λl = 0
√
which means C = 0 since λl > 0. Again, we get the trivial solution X(x) ≡ 0.
Case 3. Suppose ±(a + ib) are the two square-root of λ where b 6= 0, then we see that
X(x) = C1 e(a+ib)x + C2 e−(a+ib)x
is the general solution that solves X 00 = λX. By the two boundary conditions we get that
C(e(a+ib)l − e−(a+ib)l ) = 0.
Assume C 6= 0, otherwise we get the trivial solution again. Then, we get that
eal = |e(a+ib)l | = |e−(a+ib)l | = e−al
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Jacky Chong
Problem 1
which means a = 0. Therefore, we may assume λ = −b2 for some b > 0. Then it follows
X(x) = C sin bx and X(l) = C sin bl = 0
which means b =
nπ
where n is any positive integer. Thus,
l
X(x) = C sin
nπx
.
l
Next, solving for T (t), we get that
T (t) = Ke−in
which means
un (x, t) = An sin
2
π 2 t/l2
nπx −in2 π2 t/l2
e
l
is a solution to the Schrödinger’s equation with Dirichlet conditions. Suppose v(x, t) solves the Schrödinger’s
equation with Dirichlet conditions, then we see that
v(x, t) =
∞
X
An sin
n=1
nπx −iπ2 n2 t/l2
e
l
is a series representation of the solution.
Problem 2
4.1.4 Consider waves in a resistant medium that satisfy the problem
utt = c2 uxx − rut
u =0
for
0<x<l
at both ends
u(x, 0) = φ(x) ut (x, 0) = ψ(x),
where r is a constant, 0 < r < 2πc/l. Write down the series expansion of the solution.
Solution: Let us begin by finding a solution to utt = c2 uxx − rut of the form
u(x, t) = X(x)T (t)
that has the boundary conditions u(0, t) = u(l, t) = 0. Then we obtain the following system of odes
(
c2 X 00 (x) = λX(x), T 00 (t) + rT 0 (t) = λT (t),
X(0) = X(l) = 0
.
Case 1. Suppose λ = 0, then we see that
X(x) = C1 + C2 x
is the general solution to X 00 = 0. Applying the boundary condition, we get the trivial solution X ≡ 0.
Case 2. Suppose λ ∈ R>0 , a positive real number, then we see that
√
X(x) = C1 e
λx/c
+ C2 e−
√
λx/c
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Jacky Chong
Problem 2
is the general solution to the ode c2 X 00 = λX. Applying the first boundary condition, we get that
X(0) = C1 + C2 = 0
i.e.
C2 = −C1
√
which means X(x) = C sinh
λx
c .
However, applying the second boundary condition, we get that
√
X(l) = C sinh
λl
=0
c
√
which means C = 0 since cλl > 0. Again, we get the trivial solution X(x) ≡ 0.
Case 3. Suppose ±(a + ib) are the two square-root of λ/c2 where b 6= 0, then we see that
X(x) = C1 e(a+ib)x + C2 e−(a+ib)x
is the general solution that solves c2 X 00 = λX. By the two boundary conditions we get that
C(e(a+ib)l − e−(a+ib)l ) = 0.
Assume C 6= 0, otherwise we get the trivial solution again. Then, we get that
eal = |e(a+ib)l | = |e−(a+ib)l | = e−al
which means a = 0. Therefore, we may assume λ = −c2 b2 for some b > 0. Then it follows
X(x) = C sin bx and X(l) = C sin bl = 0
which means b =
nπ
where n is any positive integer. Thus,
l
X(x) = C sin
2
Using the fact that λ = − c
n2 π 2
,
l2
nπx
.
l
we obtain the linear second ode
T 00 (t) + rT 0 (t) = −
c2 n2 π 2
T (t)
l2
2
2
2
with corresponding characteristic polynomial P (z) = z 2 + rz + c nl2 π . Solving for the roots of P (z), we get
that
p
−r ± r2 − 4c2 n2 π 2 /l2
z=
2
which are complex numbers since −(4c2 n2 π 2 /l2 − r2 ) =: −αn2 < 0 for all n > 0. Then it follows
T (t) = C1 e−rt/2 cos αn t + C2 e−rt/2 sin αn t
which mean
un (x, t) = [An e−rt/2 cos αn t + Bn e−rt/2 sin αn t] sin
nπx
l
is a solution to utt = c2 uxx − rut with the zero Dirichlet boundary conditions at the endpoints. Hence any
solution will have the representation
u(x, t) =
∞
X
[An e−rt/2 cos αn t + Bn e−rt/2 sin αn t] sin
n=1
nπx
.
l
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Jacky Chong
Problem 2
Applying the initial conditions, we get that
φ(x) = u(x, 0) =
∞
X
An sin
n=1
and
ψ(x) = ut (x, 0) =
∞ X
n=1
or
nπx
l
r nπx
αn Bn − An sin
2
l
∞
X
r
nπx
ψ(x) + φ(x) =
αn Bn sin
2
l
n=1
Problem 3
4.2.1 Solve the diffusion problem ut = kuxx in 0 < x < l, with the mixed boundary conditions u(0, t) =
ux (l, t) = 0.
Solution: Let us consider solution to the boundary valued diffusion equation of the form
u(x, t) = X(x)T (t)
where both X(x) and T (t) are real-valued functions. Then we get the following system of odes
(
X 00 (x) = λX(x), T 0 (t) = λkT (t)
X(0) = X 0 (l) = 0
.
Case 1. Suppose λ = 0, then we see that
X(x) = C1 + C2 x
is the general solution to X 00 = 0. Applying the boundary condition, we get the trivial solution X ≡ 0.
Case 2. Suppose λ > 0, then we have that
√
X(x) = C1 e
λx
+ C2 e−
√
λx
is the general to X 00 = λX. Applying the first boundary condition, we get that
√
X(x) = C sinh
λx,
then by the second boundary condition we get that
√
√
X 0 (l) = C λ cosh λl = 0
which implies C = 0 since cosh x 6= 0 for all x ∈ R.
Case 3. Suppose λ < 0, then we have that
X(x) = C1 cos
√
−λx + C2 sin
√
−λx.
By the first boundary condition, we have that
X(x) = C sin
√
−λx.
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Jacky Chong
Problem 3
Applying the second boundary condition, we get that
√
√
X 0 (l) = C −λ cos −λl = 0.
Assuming C 6= 0, then we see that
n. Thus, it follows
√
−λ = n +
1
2
π/l or λ = − n +
X(x) = C sin
Since λ = − n +
1 2
2
n+
1
2
πx
l
1 2
2
π 2 /l2 for any nonnegative integer
.
π 2 /l2 , then
T (t) = Ae−(n+ 2 )
1
2
π 2 kt/l2
.
Now, it follows
1
2
n+
πx
2
2
2
e−n π kt/l
l
is a solution to the above problem. Hence any solution u will have the following series representation
∞
X
n + 21 πx −(n+ 1 )2 π2 kt/l2
2
e
u(x, t) =
An sin
l
n=0
un (x, t) = An sin
for some sequence of An .
Problem 4
4.2.3 Solve the Schrödinger equation ut = ikuxx for real k in the interval 0 < x < l with the boundary
conditions ux (0, t) = 0, u(l, t) = 0.
Solution: Let us consider solution to the boundary valued Schrödinger equation of the form
u(x, t) = X(x)T (t).
Then we obtain the following system of odes
(
X 00 (x) = λX(x), T 0 (t) = λikT (t)
X 0 (0) = X(l) = 0
.
Case 1. Suppose λ = 0, then we see that
X(x) = C1 + C2 x
is the general solution to X 00 = 0. Applying the boundary condition, we get the trivial solution X ≡ 0.
Case 2. Suppose λ ∈ R>0 , a positive real number, then we see that
√
X(x) = C1 e
λx
+ C2 e−
√
λx
is the general solution to the ode X 00 = λX. Applying the first boundary condition, we get that
X 0 (0) =
which means X(x) = C cosh
√
√
λ(C1 − C2 ) = 0
i.e.
C2 = C1
λx. However, applying the second boundary condition, we get that
√
X(l) = C cosh
λl = 0
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Jacky Chong
Problem 4
which means C = 0 since cosh x 6= 0 for all x ∈ R. Again, we get the trivial solution X(x) ≡ 0.
Case 3. Suppose ±(a + ib) are the two square-root of λ where b 6= 0, then we see that
X(x) = C1 e(a+ib)x + C2 e−(a+ib)x
is the general solution that solves X 00 = λX. By the two boundary conditions we get that
C(e(a+ib)l + e−(a+ib)l ) = 0.
Assume C 6= 0, otherwise we get the trivial solution again. Then, we get that
eal = |e(a+ib)l | = |e−(a+ib)l | = e−al
which means a = 0. Therefore, we may assume λ = −b2 for some b > 0. Then it follows
X(x) = C cos bx and X(l) = C cos bl = 0
which means b =
n+
1
2
π/l where n is any nonnegative integer. Thus,
1
2
n+
X(x) = C cos
πx
l
.
Next, solving for T (t), we get that
T (t) = Ke−i(n+ 2 )
1
which means
un (x, t) = An cos
n+
1
2
l
2
πx
π 2 kt/l2
e−i(n+ 2 )
1
2
π 2 kt/l2
is a solution to the Schrödinger’s equation with Dirichlet conditions. Suppose v(x, t) solves the Schrödinger’s
equation with Dirichlet conditions, then we see that
∞
X
n + 21 πx −i(n+ 1 )2 π2 kt/l2
2
e
v(x, t) =
An cos
l
n=0
is a series representation of the solution.
Problem 5
4.3.1 Find the eigenvalues graphically for the boundary conditions
X(0) = 0,
X 0 (l) + aX(l) = 0.
Assume that a 6= 0.
Solution: Considering the eigenvalue problem −X 00 = λX. Let a > 0.
Case 1. Let λ = 0, then we have that
X(x) = C1 + C2 x.
Applying the boundary conditions, we get that X(x) = Cx and
X 0 (l) + aX(l) = C + aCl = C(1 + al) = 0.
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Jacky Chong
Problem 5
Since a, l > 0, then it follows C = 0. Therefore X ≡ 0, the trivial solution.
Case 2. Let λ = β 2 > 0, then we have that
X(x) = C1 cos βx + C2 sin βx.
By the first boundary condition, we have that X(x) = C sin βx. Applying the second boundary condition,
we get that
X 0 (l) + aX(l) = Cβ cos βl + aC sin βl = 0
or
β
tan βl = − .
a
Case 3. Let λ = −β 2 < 0, then we have that
X(x) = C1 eβx + C2 e−βx .
Applying the first boundary condition, we get that X(x) = C sinh βx. Using the second boundary condition,
we get that
X 0 (l) + X(l) = C(β cosh βl + sinh βl) = 0.
But β cosh βl + sinh βl > 0, then it follows C = 0. Thus, there is no nontrivial eigenfunctions corresponding
to λ < 0.
Apply a similar argument when a < 0.
Problem 6
4.3.9 On the interval 0 ≤ x ≤ 1 of length one, consider the eigenvalue problem
−X 00 = λX
X 0 (0) + X(0) = 0
and
X(1) = 0
(absorption at one end and zero at the other).
(a) Find a eigenfunction with eigenvalue zero. Call it X0 (x).
(b) Find an equation for the positive eigenvalues λ = β 2 .
(c) Show graphically from part (b) that there are an infinite number of positive eigenvalues.
(d) Is there a negative eigenvalue?
Solution:
(a) Suppose λ = 0, then we have that
X(x) = C1 + C2 x.
Applying the boundary conditions, we obtain the following linear of equation
C1 + C2 = 0
or C2 = −C1 .
Let X0 = 1 − x then it’s clear that X0 is an eigenfunction with the eigenvalue zero.
(b) Suppose λ = β 2 > 0, then we see that
X(x) = C1 cos βx + C2 sin βx
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Jacky Chong
Problem 6
solves −X 00 = λX. Applying the boundary conditions, we get that
X 0 (0) + X(0) = C1 + βC2 = 0
and X(1) = C1 cos β + C2 sin β = 0.
WLOG, we may assume C1 , C2 6= 0, otherwise we will again arrive at the trivial solution. Hence it
follows
C1
tan β = −
=β
C2
(c) Sketch is left to the reader.
(d) Suppose λ < 0, then we see that
√
X(x) = C1 e
−λx
+ C 2 e−
√
−λx
is the general solution to −X 00 = λX. Applying the boundary conditions, we get
X 0 (0) + X(0) = C1 + C2 +
and
√
X(1) = C1 e
−λ
√
−λ(C1 − C2 ) = 0
√
+ C 2 e−
−λ
= 0.
Again, wlog, we may assume C1 , C2 6= 0, then we get that
−
√
C2
= e2 −λ
C1
which in turn yields
e
√
2 −λ
√
1 + −λ
√ .
=
1 − −λ
1+x
However, the function e2x and 1−x
does not have any positive point of intersection, i.e. there does not
√
exists λ < 0 such that −λ is a point of intersection.
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