Substituting we find
u
vw
= 0
Problem 1.12
Write the PDE
u
xx
+ 2
u
xy

3
u
yy
= 0
in the coordinates
v
(
x, y
) =
y

3
x
and
w
(
x, y
) =
x
+
y.
SOLUTIONS TO SECTION 1
9
Solution.
We have
u
x
=

3
u
v
+
u
w
u
xx
=

3(

3
u
v
+
u
w
)
v
+ (

3
u
v
+
u
w
)
w
= 9
u
vv

6
u
vw
+
u
ww
u
xy
=

3
u
vv
+
u
vw

3
u
vw
+
u
ww
=

3
u
vv

2
u
vw
+
u
ww
u
y
=
u
v
+
u
w
u
yy
=(
u
v
+
u
w
)
v
+ (
u
v
+
u
w
)
w
=
u
vv
+ 2
u
vw
+
u
ww
.
Substituting into the PDE we find
u
vw
= 0
Problem 1.13
Write the PDE
au
x
+
bu
y
= 0
in the coordinates
s
(
x, y
) =
ax
+
by
and
t
(
x, y
) =
bx

ay.
Assume
a
2
+
b
2
>
0
.
Solution.
According to the chain rule for the derivative of a composite function, we
have
u
x
=
u
s
s
x
+
u
t
t
x
=
au
s
+
bu
t
u
y
=
u
s
s
y
+
u
t
t
y
=
bu
s

au
t
.
Substituting these into the given equation to obtain
a
2
u
s
+
abu
t
+
b
2
u
s

abu
t
= 0
or
(
a
2
+
b
2
)
u
s
= 0
and since
a
2
+
b
2
>
0 we obtain
u
s
= 0
Problem 1.14
Write the PDE
u
x
+
u
y
= 1
in the coordinates
s
=
x
+
y
and
t
=
x

y.
10
CONTENTS
Solution.
Using the chain rule we find
u
x
=
u
s
s
x
+
u
t
t
x
=
u
s
+
u
t
u
y
=
u
s
s
y
+
u
t
t
y
=
u
s

u
t
.
Substituting these into the PDE to obtain
u
s
=
1
2
Problem 1.15
Write the PDE
au
t
+
bu
x
=
u,
a, b
6
= 0
in the coordinates
v
=
ax

bt
and
w
=
1
a
t.
Solution.
We have
u
t
=

bu
v
+
1
a
u
w
and
u
x
=
au
v
.
Substituting we find
u
w
=
u
SOLUTIONS TO SECTION 2
11
Solutions to Section 2
Problem 2.1
Determine
a
and
b
so that
u
(
x, y
) =
e
ax
+
by
is a solution to the equation
u
xxxx
+
u
yyyy
+ 2
u
xxyy
= 0
.
Solution.
We have
u
xxxx
=
a
4
e
ax
+
by
, u
yyyy
=
b
4
e
ax
+
by
,
and
u
xxyy
=
a
2
b
2
e
ax
+
by
.
Thus,
substituting these into the equation we find
(
a
4
+ 2
a
2
b
2
+
b
4
)
e
ax
+
by
= 0
.
Since
e
ax
+
by
6
= 0, we must have
a
4
+ 2
a
2
b
2
+
b
4
= 0 or (
a
2
+
b
2
) = 0
.
This is
true only when
a
=
b
= 0
.
Thus,
u
(
x, y
) = 1
Problem 2.2
Consider the following differential equation
tu
xx

u
t
= 0
.
Suppose
u
(
t, x
) =
X
(
x
)
T
(
t
)
.
Show that there is a constant
λ
such that
X
00
=
λX
and
T
0
=
λtT.
Solution.
Substituting into the differential equation we find
tX
00
T

XT
0
= 0
or
X
00
X
=
T
0
tT
.
The LHS is a function of
x
only whereas the RHS is a function of
t
only.
This is true only when both sides are constant. That is, there is
λ
such that
X
00
X
=
T
0
tT
=
λ
and this leads to the two ODEs
X
00
=
λX
and
T
0
=
λtT
12
CONTENTS
Problem 2.3
Consider the initial value problem
xu
x
+ (
x
+ 1)
yu
y
= 0
,
x, y >
1
u
(1
,
1) =
e.
Show that
u
(
x, y
) =
xe
x
y
is the solution to this problem.
Solution.
We have
xu
x
+(
x
+1)
yu
y
=
x
y
(
e
x
+
xe
x
)+(
x
+1)
y

xe
x
y
2
= 0 and
u
(1
,
1) =
e
Problem 2.4
Show that
u
(
x, y
) =
e

2
y
sin (
x

y
) is the solution to the initial value prob
lem
u
x
+
u
y
+ 2
u
= 0
,
x, y >
1
u
(
x,
0) = sin
x.
Solution.
We have
u
x
+
u
y
+2
u
=
e

2
y
cos (
x

y
)

2
e

2
y
sin (
x

y
)

e

2
y
cos (
x

y
)+
2
e

2
y
sin (
x

y
) = 0 and
u
(
x,
0) = sin
x
Problem 2.5
Solve each of the following differential equations:
(a)
du
dx
= 0 where
u
=
u
(
x
)
.
(b)
∂u
∂x
= 0 where
u
=
u
(
x, y
)
.
Solution.
(a) The general solution to this equation is
u
(
x
) =
C
where
C
is an arbitrary
constant.
(b) The general solution is
u
(
x, y
) =
f
(
y
) where
f
is an arbitrary function of
y
Problem 2.6
Solve each of the following differential equations:
(a)
d
2
u
dx
2
= 0 where
u
=
u
(
x
)
.
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 Fall '15
 BrodyJohnson
 Differential Equations, Equations, Partial Differential Equations, Boundary value problem