AMS 212A Applied Mathematical Methods I
Exercise #12
1. Let D =
{( x, y ) x > 0} in 2
and ex be the unit vector along the x-axis.
Consider
1
1
u(x) =
log x x0 +
log x x0( 2 ) ,
x0( 2 ) = x0 2 ex ( x0 ex )
2
2
Verify that for x0 in D , u ( x ) given above satisfies
2u = ( x x0 ) in D
u ( x ) x=0 = 0
Solution:
As a function in R2, u ( x ) satisfies
(
2u = ( x x0 ) + x x0( 2 )
)
For x0 = ( x0 , y0 ) in D , the image x0( 2 ) = ( x0 , y0 ) in not in D. So we have
2u = ( x x0 ) in D
At the boundary x = 0, we have
2
x x0 = x0 2 + ( y y0 ) = x x0( 2 )
u ( x ) x=0 = 0
==>
2. Let D =
{( x, y, z ) x > 0} in 3
and ex be the unit vector along the x-axis.
Consider
u(x) =
1
1
(2) ,
4 x x0 4 x x0
x0( 2 ) = x0 2 ex ( x0 ex )
Verify that for x0 in D , u ( x ) given above satisfies
-1-
AMS 212A Applied Mathematical Methods I
2u = ( x x0 ) in D
u ( x ) x=0 = 0
Solution:
As a function in R3, u ( x ) satisfies
(
2u = ( x x0 ) + x x0( 2 )
)
For x0 = ( x0 , y0 , z0 ) in D , the image x0( 2 ) = ( x0 , y0 , z0 ) in not in D. So we have
2u = ( x x0 ) in D
At the boundary x = 0, we have
2
2
x x0 = x0 2 + ( y y0 ) + ( z z0 ) = x x0( 2 )
u ( x ) x=0 = 0
==>
3. The image of complex number z0 with respect to the circle z = R is defined as
R2
(2)
z0 = z0
z0
2
R2
=
z0
Verify that
R ei z0( 2 ) =
R
R ei z0
z0
Solution:
R ei z0( 2 ) = R ei =
(
R2
R i
=
e z0 Rei
z0
z0
)
R
R
R
z0 R ei =
z0 R ei =
R ei z0
z0
z0
z0
4. For x0 D = B ( 0, R ) in 2 , consider
x0 ( 2 ) 1
1
u(x) =
log x x0 +
log x x0 ,
2
2
R
-2-
(2) R2
x0 = x0 2
x0
AMS 212A Applied Mathematical Methods I
Verify that u ( x ) satisfies
2u = ( x x0 ) in D
u ( x ) = 0 on D
Solution:
As a function in R2, u ( x ) satisfies
(
2u = ( x x0 ) + x x0( 2 )
)
R2
For x0 D = B ( 0, R ) , the image x0( 2 ) = x0 2 in not in D. So we have
x0
2u = ( x x0 ) in D
At the boundary x = 0, we have
x0 ( 2 )
x x0 =
x x0
R
==> u ( x ) = 0 on D
5. For x0 D = B ( 0, R ) in 3 , consider
u(x) =
1
4 x x0
1
,
x0 ( 2 )
4
x x0
R
R2
x0( 2 ) = x0 2
x0
Verify that u ( x ) satisfies
2u = ( x x0 ) in D
u ( x ) = 0 on D
Solution:
As a function in R3, u ( x ) satisfies
(
2u = ( x x0 ) + x x0( 2 )
)
1
x0
R
(2) R2
For x0 D = B ( 0, R ) , the image x0 = x0 2 in not in D. So we have
x0
-3-
AMS 212A Applied Mathematical Methods I
2u = ( x x0 ) in D
At the boundary x = 0, we have
x0 ( 2 )
x x0 =
x x0
R
==> u ( x ) = 0 on D
If we are more careful, we need to discuss the special case of x0 = 0 .
As x0 0 ,
x0 ( 2 )
x0
x0
x x0 = x
R R
R
R
x0
u(x) 1
1
4 x 4 R
-4-