HOMEWORK 2
SHUANGLIN SHAO
1. P5. # 1.
Proof. The process is similar to the derivation of one dimensional wave equation in Example 2. In the transverse direction,
x
T ux (x, t) T ux (x0 , t)
rut dx.
x0
Then dierentiating in x,
T uxx rut = utt .
Then
ut t
HOMEWORK 4
SHUANGLIN SHAO
1. P45. # 1.
Proof. By the maximum principle, u(x, t) = 1 x2 2kt attains the maximum at the bottom or on the two sides. When t = 0, 1 x2 2kt = 1 x2
attains the maximum at x = 0, i.e., 1 x2 = 1.
When x = 0, 1 x2 2kt = 1 2kt attain
HOMEWORK 5
SHUANGLIN SHAO
1. Section 3.1. # 1.
Proof. This is the diusion equation (1) with the function (x) = ex . By
the solution formula (6),
1
4kt
1
=
4kt
1
=
4kt
e
(xy)2
4kt
e
(x+y)2
4kt
(y)dy
e
v(x, t) =
(xy)2
4kt
e
(x+y)2
4kt
ey dy
0
0
e
(xy)2
y
HOMEWORK 7
SHUANGLIN SHAO
1. Section 5.1. # 2.
Proof. (a). The Fourier sine series expansion is
x2 =
An sin nx, 0 x 1,
n=1
where for n 1,
An =
2
1
1
x2 sin nxdx
0
1
x2 sin nxdx
=2
0
1
x2
=2
0
=
=
=
=
=
1
cos nx
n
dx
1
2 2
2
x cos nx|1 +
x cos nxdx
0
n
n 0
HOMEWORK 8
SHUANGLIN SHAO
1. Section 5.3. # 3.
Proof. By the exercise #2 in Section 4.2, we have
u(x, t) =
sin
n=0
(2n + 1)x
2l
(2n + 1)ct
(2n + 1)ct
+ Bn sin
2l
2l
An cos
Since ut (x, 0) = 0, we have
0=
Bn
n=0
(2n + 1)c
2l
sin
(2n + 1)x
.
2l
Thus Bn = 0.
HOMEWORK 10
SHUANGLIN SHAO
1. Section 6.3 # 2.
Proof. By the formula (10) in this section, we have
u(r, ) =
A0
+
2
rn (An cos n + Bn sin n),
n=1
where
2
1
h() cos nd,
an 0
2
1
Bn =
h() sin nd.
an 0
We need to compute An and Bn for specic h(). In this prob
HOMEWORK 9
SHUANGLIN SHAO
1. Section 6.1. # 1.
Proof. We write f in a power series of z: f (z) =
n
n=0 an z .
That is to say,
an (x + iy)n .
u(x + iy) + v(x + iy) =
n=0
Dierentiating both sides in x and y:
v
u
+i
=
x
x
u
v
+i
=
y
y
nan (x + iy)n1 ,
n=0
in
HOMEWORK 6
SHUANGLIN SHAO
1. Section 4.1. # 2.
Proof. By using the solution formula,
An sin
nx
.
l
An sin
u(x, t) =
nx
.
l
n=1
Let t = 0. Then
u(x, 0) =
n=1
By the expansion of ,
1=
An sin
n=1
nx
l
= sin
x
2x
3x
nx
+ A2 sin
+ A3 sin
+ + An sin
+
l
l
l
l