Solutions

MAP 4305/5304 Section 0099/3234
Intermediate Differential Equations
Assignment 3 – Solutions
1. For each of the following regular boundary value problems, (i) convert it
to a regular Sturm-Liouville problem, i.e. one of the form L[u] + λru = 0 with
boundary conditions. Then (ii) find the eigenvalues and a set of normalized
eigenfunctions (normalized, that is, for the inner product with weight r).
(i) u00 + 2u0 + λu = 0,
(ii) u00 + 2u0 + λu = 0,
u(0) = u(π) = 0
u(0) + u0 (0) = 0, u(π) = 0.
Solution. (i) The integrating factor is e2x , so that the equation becomes
(e2x u0 )0 + λe2x u = 0.
so that r(x) = e2x . There are no eigenvalues for λ ≤ 1, and for λ = 1 + a2 the
equation has solutions
u = Ce−x sin ax + De−x cos ax.
Substuting the boundary conditions shows that D = 0 and that sin πa = 0, so
that a is a positive integer and the n-th eigenvalue is λ = 1 + n2 ; the corresponding eigenfunction is
un = Cn e−x sin nx.
The normalization condition is
Z π
Z
2 2x
2
un e dx = Cn
0
π
sin2 nx dx = 1
0
and doing the integral shows that
r
un =
2
sin nx
π
(ii) Once again we find that there are no solutions for λ ≤ 1, and for λ > 1
we must have
2. (i) Find the formal adjoint and the adjoint of the boundary value problem
y 00 + 4y 0 + 5y = 0,
y(0) = y(π) = 0.
(ii) Use the Fredholm alternative to give a criterion for the boundary value
problem
y 00 + 4y 0 + 5y = f (x), y(0) = y(π) = 0.
Solution. (i) The formal adjoint of L[u] = u00 + 4u0 + 5u is L† [v] = v 00 − 4v 0 + 5v.
The bilinear concomittant is
P (u, v) = u0 v − uv 0 + 4uv
and
π
P (u, v) = u0 (π)v(π) − u(π)v 0 (π) + 4u(π)v(π) − u0 (0)v(0) + u(0)v 0 (0) − 4u(0)v(0)
0
= u0 (π)v(π) − u0 (0)v(0)
when we substitute the conditions for u. The adjoint conditions are then
v(0) = v(π) = 0.
(ii) The solutions of the adjoint problem
v 00 − 4v 0 + 5v = 0,
v(0) = v(π) = 0
are constant multiples of
v = e2x sin x
and so the original equation has a solution if and only if
Z π
f (x)e2x sin x dx = 0.
0
3. (i) Find the formal adjoint and the adjoint of the boundary value problem
x2 y 00 + xy 0 + y = 0,
y(1) + y 0 (eπ ) = 0, y 0 (1) = 0.
(ii) Use the Fredholm alternative to give a criterion for the boundary value
problem
x2 y 00 + xy 0 + y = f (x),
y(1) + y 0 (eπ ) = 0, y 0 (1) = 0.
to have a solution.
4. Find a solution of
y 00 + y = x2 − 2πx,
y(0) = y 0 (π) = 0
using the method of eigenfunction expansions.
Solution. We take L[u] = u00 , so that p(x) = 1, q(x) = 0 and r(x) = 1 and for
the inhomogenous problem we must have µ = 1. The eigenfunctions are then
the solutions of
u00 + λu = 0, u(0) = u0 (π) = 0
and the usual method shows that these are λ = (n + 1/2)2 , n a natural number,
p
and un = Cn sin(n + 1/2)x. The normalization condition leads to Cn = 2/π.
We now expand x2 − 2πx in a series of eigenfunctions:
Z π
X
2
x − 2πx =
dn un
dn =
(x2 − 2πx) sin(n + 1/2)x dx.
0
n
A little computation shows that
r
dn = −
and we find
y(x) =
X
n≥0
2
16
π (2n + 1)3
dn
un
1 − (n + 1/2)2
with dn as above.
5. Convert the problem
u00 + 2u0 + u = 0,
u(0) + u0 (0) = 0, u(π) = 0
to a regular Sturm-Liouville problem. Then find the Green’s function for the
resulting boundary value problem, and use it to give a formula for the solution
of the boundary value problem
u00 + 2u0 + u = −f (x),
u(0) + u0 (0) = 0, u(π) = 0.
Solution. The integrating factor is e2x , and the S-L problem is
(e2x u0 )0 + e2x u = 0
for which p(x) = e2x . The general solution of the equation is
u = Ce−x + Dxe−x
since −1 is a double eigenvalue. The solutions u1 , u2 satisfying the condition
at 0 and π respectively are
u1 = e−x
u2 = (x − π)e−x .
The Wronskian is
−x
e
W (u1 , u2 ) = −x
−e
(x − π)e−x = e−2x
e−x − (x − π)e−x and thus
C = pW (u1 , u2 ) = e2x e−2x = 1.
Therefore
(
G(x, t) =
−e−t (x − π)e−x
−e−x (t − π)e−t
0≤t≤x≤π
0≤x≤t≤π
and the formula for the solution is
Z π
Z
u(x) = −e−x
(t − π)e−t f (t) dt + −(x − π)e−x
x
x
e−t f (t) dt.
0
6. Same as the previous problem, for the boundary value problem
x2 y 00 + xy 0 + y = −f (x),
y(1) = 0, y 0 (eπ ) = 0
7. Suppose G(x, s) is the Green’s function for the regular Sturm-Liouville
boundary value problem
Ly + µy = −f, B(y) = 0
where Ly = (y 0 p)0 + qy. Let f = δ(x − s) (the Dirac delta function; see chapter
7 §8). Show that if un is a normalized system of eigenfunctions with eigenvalues
λn and µ is not an eigenvalue, then
G(x, s) =
∞
X
un (x)un (s)
.
λn − µ
n=1
Solution. The idea is to use the eigenfunction expansion technique on the “function” δ(x − s) and hope this this makes sense. In fact if the un are normalized
eigenfunctions with eigenvalues λn , the expansion of δ(x − s) is
X
δ(x − s) =
un (x)un (s)
n
since
b
Z
un (x)δ(x − s) = un (s)
a
by definition. The formula for the solution then yields
∞
X
un (x)un (s)
G(x, s) =
.
λn − µ
n=1
8. (i) Estimate the spacing of successive zeroes of any nonzero solution of
u00 + (2 + cos x)u = 0
on the interval (−∞, ∞). (ii) Do the same for the equation
u00 + x2 u = 0
on intervals of the form [a, b] and [−b, −a], assuming 0 < a < b.
Solution. (i) With Q(x) = 2 + cos x we have qmin = 1, qmax = 3 (recall that
these are the minimum and maximum values of Q(x) on the interval) and thus
the spacing s of successive zeroes satisfies
π
√ ≤ s ≤ π.
3
(ii) This time, with Q(x) = x2 we have qmin = a2 , qmax = b2 and the spacing
satisfies
π
π
≤s≤ .
b
a
9. (i) Find intervals of infinite extent for which nonzero solutions of
u00 + (E − x2 )u = 0
are not oscillatory (here E
constant). (ii) Estimate the number of
p
pis a positive
zeroes of a solution in [− E/2, E/2].
2
Solution. (i)
√The solutions
√ are not oscillatory when Q(x) = E − x ≤ 0, which
is for x ≤ − E and x ≥ pE. p
(ii) In the interval [− E/2, E/2] we have qmin = E/2, qmax = E. The
successive spacing of zeroes of a solution is bounded by
π
π
√ ≤s≤ p
.
E
E/2
√
The length of the interval is 2E, so the number n of zeroes is bounded by
√
2E
E
≤n≤
π
π
since this number is at least the length divided by maximum spacing, and at
most the length divided by the minimum spacing.
Remark: Up to constants, this is the Schroedinger equation for a onedimensional harmonic oscillator. We see that the wavelength increases linearly
with the energy E, as is expected on physical grounds.

Download Report