Description

Haoran Sun (haoransun@link.cuhk.edu.cn)
Problem 1 (P89 Q3). Easy to obtain
1 T′(t) X′′(x)
= = −λ
i T(t) X(x)
and λ = β2 ≥ 0. Then we have
T(t) = e−iλt
X(x) = Acosβx + B sinβx Applying X(0) = X(l) = 0, we have B = 0 and βn = nπ/l, n ∈ N. Hence
X∞ nπx −i(nπ)2t/l2 u(x,t) = An sin e l
n=1 Problem 2 (P92 Q4).
(a) Easy to obtain
1 T′(t) X′′(x)
= = −λ
k T(t) X(x)
and we can show that λ ≥ 0 given the periodic boundary condition. Hence, we have λ = β2
X(x) = Acosβx + B sinβx, X′(x) = −Aβ sinβx + Bβ cosβx
Plug the P.B.C into it, and we get
X(l) = Acosβl + B sinβl = X(−l) = Acosβl − B sinβl
X′(l) = −Aβ sinβl + Bβ cosβl = X′(−l) = Aβ sinβl + Bβ cosβl
⇒ B sinβl = Aβ sinβl = 0
Hence we have βn = nπ/l and λn = (nπ/l)2.
(b) Using the result from (a), it should be obvious that
1 nπx nπx (nπ/l)2t u(x,t) = AAn cos + Bn sin e−
2l l
Problem 3 (P101 Q9).
(a) Let λ = 0, X(x) = ax + b, X′(x) = a. Plug in the boundary condition, we have
X0(x) = x − 1
(b) Let λ = β2, then
X(x) = Acosβx + B sinβx
X′(x) = A(−β sinβx) + Bβ cosβx
Plug in the boundary condition, we have
= 0
1 β A cosβ sinβ B
The determinant should be zero to get non-trivial solutions
sinβ − β cosβ = 0 ⇒ tanβ = β
HW04 Haoran Sun

(c) Sketch:

Thus there are infinitely many solutions.
(d) Suppose we have negative eigenvalues where λ = (iβ)2, then
X(x) = Acoshβx + B sinhβx
X′(x) = Aβ sinhβx + Bβ coshβx
Since
X(0) = X′(x) ⇒ A = B
X(1) = Ae + e−β + Be − e−β −β > 0
= Ae
2 2
thus we cannot have negative eigenvalues.
Problem 4 (P111 Q2).
(a) For sine series
2 l nπ
An = Z x2 sin dx
l 0 l
···2 l − 2 nπ l Z l nπ !
= Z x cos x + 2 xcos xdx nπ 0 l 0 0 l
=
2 8 2
= + l nπ (nπ)3
Then
2 8
xAnnπx with An = nπ + (nπ)3
HW04 Haoran Sun

(b) For cosine series
A0 =Z x2 dx = l2 l l 2
203
An =Z x2 cosxdx ll nπ
2 0 l
2 2 nπ l − Z l nπ !
= x sin x 2 xsin xdx nπ l 0 0 l
= ···
4l2
=
(nπ)2
Then
2 1 ∞ nπ
x = A0 + X An cos x
2 l
n=1
1nπ nπc nπ
u(x,t) = A An cos t + Bn sin tcos x
2l l l
1 n nπ
u(x,0) = A0 + X An cos x = 0
2 l
n=1
1 ∞ nπc nπ 2
ut(x,0) = B0 + X Bn cos x = cos x
2 l l
n=1
We can conclude that An = 0 and
2 π π 0
B0 = Z cos2 xdx = 1
2 π π 0
Bnnc = Z cos2 xcosnxdx
2 π 1 π 0 2
= Z (cos2x + 1)cosnxdx
1 π 1 π
= Z cos2xcosnx + Z cosnxdx
0 π 0
=δ2n
which means the only two non-zero terms are B0 and B2, therefore
1 1
u(x,t) = t + sin2ctcos2x
2 4c