Math 447/547 Partial Differential Equations Prof. Carlson Homework 7
Text section 4.2
1. Solve the diffusion problem
ut = kuxx, 0< x < l, with the mixed boundary conditions
u(t,0) = 0 = ux(t, l).
Solution The analysis proceeds initially as on p. 85. The problem for
X(x) is
−X′′ = λX, X(0) = 0 = X′(l).
For positive eigenvalues λ= β2 solutions must have the form
Xn(x) = Ansin(βx)
because of the boundary condition atx = 0. The condition atx = l becomes cos(βl) = 0
so that
β = (n+ 1/2)π/l, n = 0,1,2, . . . .
It is easy to see that 0 is not an eigenvalue. For general complex eigenvalues λ 6= 0 we have
X(x) = Cei√λx +De−i√λx.
The boundary conditions give
C +D = 0, i√λCe √
λl
−i√λDe−√λl = 0. Together we find that
or e−i√λl[e2i√λl + 1] = 0. This gives 2√λ= (π + 2nπ)/l, or √ λ = (π/2 +nπ)/l n = 0,1,2, . . . .
Thus the positive eigenvalues are the only ones. The general solution then has the form
u(t, x) = X n Xn(x)Tn(t) = ∞ X n=0 Anexp(−(n+ 1/2)2π2kt/l2) sin((n+ 1/2)πx/l).
2. Consider the equation utt =c2uxx for 0< x < l with the boundary conditions ux(t,0) = 0 =u(t, l).
a) Show that the eigenfunctions are cos((n+ 1/2)πx/l). b) Write the series expansion for a solution u(t, x). Solution
a) The analysis proceeds pretty much as in the previous problem. One checks easily that 0 cannot be an eigenvalue. For other complex λ we have
X(x) = Cei√λx +De−i√λx.
The boundary conditions give
i√λC −i√λD = 0, Cei√λl+De−i√λl = 0.
Together we find that
ei√λl+e−i√λl = 0,
or
e−i√λl[e2i√λl + 1] = 0.
This gives √
as before. We now have Xn(x) = Cncos( (n+ 1/2)π l x) +Dnsin( (n+ 1/2)π l x).
To satisfy the boundary condition at 0 we must have Dn = 0, so eigenfunc-tions are nonzero multiples of cos((n+1/2)πl x).
b) As in p. 83 (5) the equation for Tn is
T′′ n +λc2Tn = 0. This gives Tn(t) = Ancos((n+ 1/2)πct/l) +Bnsin((n+ 1/2)πct/l), and u(t, x) = ∞ X n=0 [Ancos((n+ 1 2)πct/l) +Bnsin((n+ 1 2)πct/l)] cos((n+ 1/2)πx). 3. Solve the Schr¨odinger equation
ut = ikuxx, ux(t,0) = 0, u(t, l) = 0, 0 ≤ x ≤ l. As in problem 2,
Xn(x) = cos((n+ 1/2)π
l x).
The equation for Tn is
T′ n(t) +ik[ (n+ 1/2)π l ] 2Tn(t) = 0, so Tn(t) = Cnexp(− ik(n+ 1/2)2π2 l2 t),
and u(t, x) = ∞ X n=0 Cnexp(−ik(n+ 1/2) 2π2 l2 t) cos((n+ 1/2)πx).
4. (NOT ASSIGNED) Consider diffusion inside an enclosed circular tube. Let its length (circumference) be 2l. Let x denote the arc length parameter where −l ≤ x ≤ l. Then the concentration of the diffusing substance satisfies
ut =kuxx, −l ≤x ≤l,
u(t,−l) = u(t, l), ux(t,−l) = ux(t, l).
a) Show that the eigenvalues are λ = (nπ/l)2 for n = 0,1,2, . . .. Solution
There are constant eigenfunctions for eigenvalue λ= 0. For other λ we have
X(x) = Ce√λx +De−√λx.
The boundary conditions give
Ce−√λl +De√λl =Ce√λl +De−√λl,
and √
λ[Ce−√λl
−De√λl] =√λ[Ce√λl −De−√λl]. Some arithmetic leads to
e−√λl = e√λl
or
1 = e2√λl,
so that √
λ= nπ/l.
b) Show that the concentration is
u(t, x) = a0 2 + ∞ X n=1 (ancos( nπx l ) +bnsin( nπx l )) exp(−n 2π2kt/l2).
Solution So far we have identified the eigenvalues, but not the eigen-functions. Notice that for each n = 1,2,3, . . . both functions cos(nπxl ) and
sin(nπxl ) satisfy the periodic boundary conditions. The only exceptional case is n = 0 when the function 1 satisfies the boundary conditions, but x
does not. This results in the given form for u(t, x).
9. On the interval 0 ≤ x ≤ 1 of length 1, consider the eigenvalue problem
−X′′ = λX,
X′(0) +X(0) = 0, X(1) = 0.
a) Find an eigenfunction with eigenvalue 0. Call in X0.
Solution Any multiple of the function 1 − x satisfies the boundary conditions.
b) Find an equation for the positive eigenvalues λ = β2. Solution We may write
Xn(x) = Cncos(βx) +Dnsin(βx). The boundary conditions give
βDn +Cn = 0, Cncos(β) +Dnsin(β) = 0. Together these give
−βcos(β) + sin(β) = 0,
or
tan(β) = β.
c) Show graphically that there are an infinite number of positive eigen-values.
Solution Plot the functions tan(β) andβ and see where they intersect. d) Is there a negative eigenvalue?
Solution Suppose that λ= −γ2, γ > 0. Then we would have
Xn(x) = Cncosh(βx) +Dnsinh(βx) and the boundary conditions would give
The equation for γ is γ = tanh(γ). Since tanh(γ) = e γ −e−γ eγ +e−γ
we see that tanh(0) = 0 and
0 ≤tanh(γ) < 1, 0 ≤ γ < ∞.
Compute
tanh′(γ) = 4
(eγ +e−γ)2.
Notice that eγ +e−γ has a minimum value of 0 which is achieved only at γ = 0. Thus
0 <tanh′(γ) <1, 0 < γ < ∞.
If there were a point γ1 > 0 such that γ = tanh(γ1), then the function
γ −tanh(γ1) would vanish at 0 and at γ1. By Rolle’s theorem the function would have a positive root for the derivative. But the above analysis shows that
1−tanh′(γ) > 0, 0 < γ < ∞.
Thus there are no negative eigenvalues.
11. a) Prove that the total energy is conserved for the wave equation with Dirichlet boundary conditions, where the energy is defined to be
E = 1 2 Z l 0 (c−2u2 t +u2x) dx. Solution We have dE dt = 1 2 Z l 0 (c−22u tutt + 2uxuxt) dx = 1 2 Z l 0 (2utuxx + 2uxuxt) dx.
Integrate the second term by parts to get dE dt = 1 2 Z l 0 (2utuxx +−2uxxut) dx+uxut l 0. The boundary conditions are
u(t,0) = 0 = u(t, l).
Differentiation gives
ut(t,0) = 0 = ut(t, l),
so dEdt = 0.
b) Do the same for Neumann boundary conditions.
Solution The only difference is that the vanishing of the boundary terms comes from ux(t,0) = 0 = ux(t, l).
c) For the Robin boundary conditions, show that
ER = 1 2 Z l 0 (c−2u2 t +u2x) dx+ 1 2alu 2 (t, l) + 1 2a0u 2 (t,0).
Solution Differentiation and integration by parts gives
dER dt = uxut l 0 +alu(t, l)ut(t, l) +a0u(t,0)ut(t,0) = ut(t, l)[ux(t, l) +alu(t, l)]−ut(t,0)[ux(t,0)−a0u(t,0)] = 0.
