Chapt_13 - 13 13 Boundary-Value Problems in Rectangular...

13
13
Boundary-Value Problems
in Rectangular Coordinates
EXERCISES 13.1
Separable Partial Differential Equations
1. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y = XY . Separating variables
and using the separation constant − λ , where λ = 0, we obtain
X
X
=
Y
Y
= − λ.
When λ = 0
X + λX = 0 and Y + λY = 0
so that
X = c
1
e
− λx
and Y = c
2
e
− λy
.
A particular product solution of the partial differential equation is
u = XY = c
3
e
− λ ( x + y )
, λ = 0 .
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c
4
, Y = c
5
, and
u = XY = c
6
.
2. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y + 3 XY = 0. Separating
variables and using the separation constant − λ we obtain
X
− 3 X
=
Y
Y
= − λ.
When λ = 0
X − 3 λX = 0 and Y + λY = 0
so that
X = c
1
e
3 λx
and Y = c
2
e
− λy
.
A particular product solution of the partial differential equation is
u = XY = c
3
e
λ (3 x − y )
.
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c
4
, Y = c
5
, and
u = XY = c
6
.
3. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y + XY = XY . Separating
variables and using the separation constant − λ we obtain
X
X
=
Y − Y
Y
= − λ.
680

13.1
Separable Partial Differential Equations
Then
X + λX = 0 and Y − (1 + λ ) Y = 0
so that
X = c
1
e
− λx
and Y = c
2
e
(1+ λ ) y
.
A particular product solution of the partial differential equation is
u = XY = c
3
e
y + λ ( y − x )
.
4. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y = XY + XY . Separating
variables and using the separation constant − λ we obtain
X
X
=
Y + Y
Y
= − λ.
Then
X + λX = 0 and y + (1 + λ ) Y = 0
so that
X = c
1
e
− λx
and Y = c
2
e
− (1+ λ ) y
= 0 .
A particular product solution of the partial differential equation is
u = XY = c
3
e
− y − λ ( x + y )
.
5. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields xX Y = yXY . Separating vari-
ables and using the separation constant − λ we obtain
xX
X
=
yY
Y
= − λ.
When λ = 0
xX + λX = 0 and yY + λY = 0
so that
X = c
1
x
− λ
and Y = c
2
y
− λ
.
A particular product solution of the partial differential equation is
u = XY = c
3
( xy )
− λ
.
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c
4
, Y = c
5
, and
u = XY = c
6
.
6. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields yX Y + xXY = 0. Separating
variables and using the separation constant − λ we obtain
X
xX
= −
Y
yY
= − λ.
When λ = 0
X + λxX = 0 and Y − λyY = 0
so that
X = c
1
e
λx
2
/ 2
and Y = c
2
e
− λy
2
/ 2
.
A particular product solution of the partial differential equation is
u = XY = c
3
e
λ ( x
2
− y
2
) / 2
.
681

13.1
Separable Partial Differential Equations
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c
4
, Y = c
5
, and
u = XY = c
6
.
7. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y + X Y + XY = 0 , which
is not separable.