640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 0
Most of the following exercises are from Strauss [Str] and Rogawski [Rog]. For
every exerc
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 10
Exercises from Haberman [Hab] and Strauss [Str]. Be sure to justify your answers complete
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 7
Exercises are from Haberman [Hab] and Strauss [Str]. Be sure to justify your
answer comple
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 a
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
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
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
=
=
=
=
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
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
=
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 =
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.
x
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 9
Exercises from Haberman [Hab] and Strauss [Str]. Be sure to justify your answers completel
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 8
The following problem set consists of optional additional exercises from Haberman [Hab] an
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 1
Most of the following exercises are from Strauss [Str], Rogawski [Rog] and Haberman [Hab].
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 2
Exercises are from Strauss [Str], Rogawski [Rog] and Haberman [Hab]. For
every exercise, b
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 3
Exercises are from Rogawski [Rog] and Haberman [Hab]. For every exercise,
be sure to justi
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 4
Exercises are from Haberman [Hab] and Strauss [Str]. Be sure to justify your
answer comple
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 5
Exercises are from Haberman [Hab] and Strauss [Str]. Be sure to justify your
answer comple
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 6
Exercises are from Haberman [Hab] and Strauss [Str]. Be sure to justify your
answer comple
640:423 Partial Dierential Equations, Autumn 2012
Department of Mathematics, Rutgers University
Problem Set 11
Exercises from Haberman [Hab] and Strauss [Str]. Be sure to justify your answers complete