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Solutions to Midterm II
Problem 1: (20pts)
Find the most general solution u ( x, y) of the following equation consistent with the boundary
condition stated
y
u
u
x
 3x ,
x
y
u  x 2 on the line y  0
Solution 1: Since the system (1) is linear, the solution is given as a superposition of a
homogenous and a particular solution
ux, y   u H x, y   u P x, y 
We start by computing the homogenous solution;
Characteristic Equation:
dy
x

dx
y
Characteristic Curves:
C  x2  y2
Homogenous solution:

u H ( x, y)  f x 2  y 2

Next, we solve for a particular solution to (1)
We assume
u P x, y   ax  by
If we substitute the particular solution into (1),
We obtain
ay  xb  3x
Hence,
General solution:
 a  0, b  3
u P x, y   3 y


u( x, y)  f x 2  y 2  3 y
On y  0 :
 
u( x,0)  f x 2  x 2
Hence,
u( x, y)  x 2  y 2  3 y
(1)
Problem 2: (30pts)
 2u
 2u  2u
6 2 5

0
xy y 2
x
Solve for u ( x, y) ,
u
 4  6 x both on the line y  0
y
Subject to u  2 x  1 and
Solution 2:
Characteristic equations:
dy  5 

dx
 52  4(6)(1)
2(6)

 5 1
12
Characteristic Curves:
x  3 y  C1
General solution:
and
x  2 y  C2
ux, y   f x  3 y   g x  2 y 
Boundary conditions:
ux,0  f x   g x   2 x  1
Differentiate (1),
u y x,0  3 f x   2 g x   4  6 x
and
f x   g x   2
(1,2)
3 f x   2 g x   4  6 x
(3,4)
g x   2  6 x
(5,6)
We solve equations (3) and (4) simultaneously
f x   6 x
Integrate (5,6)
g x   k1  2 x  3x 2
f x   k o  3x 2
So that
f x  3 y   k o  3x  3 y 
2
g x  2 y   k1  2x  2 y   3x  2 y 
2
Hence,
ux, y   f x  3 y   g x  2 y   k o  3x  3 y   k1  2x  2 y   3x  2 y 
2
Simplify
Boundary condition
So that
2
ux, y   2 x  4 y  6 xy  15 y 2  k o  k1
ux,0  2 x  k o  k1   2 x  1 ,
ux, y   2 x  4 y  6 xy  15 y 2  1
 k o  k1  1
Problem 3: (5pts)
Show that there is no solution of
u
 g on D
n
 2 u  f in D
(1,2)
In three dimensions, unless
 fdV   gdS
(3)
D
D
Hints: Use the divergence theorem




F
dV

F

  nˆdS
D
D
Solution 3:
We integrate (1)
 fdV     u dV
D
(4)
D
We now apply the divergence theorem on the right side of (4)
u
 fdV     u dV   u  nˆdS   n dS   gdS
D
D
D
D
Q.E.D
D
Problem 4: (45pts)
A thin sheet of metal coincides with a unit square in the xy  plane . Initially, the temperature in the
sheet is T x, y,0  hx, y  . If there are no sources of heat in the sheet, find the temperature at any
point at any subsequent time, given that the right and left faces of the sheet are insulated, the
temperature at the lower edge is maintained at zero, and the temperature at the upper edge is
prescribed by T x,1, t   T0 cosx  .
Solution 4:
The governing equation to be solved is
1 T  2T  2T


 2 t x 2 y 2
Tx 0, y, t   0 , Tx 1, y, t   0 ,
T x,0, t   0 ,
0  x 1
0  y 1
(1)
T x,1, t   T0 cosx  , T x, y,0  hx, y 
We define our solution as the sum of a steady and a transient component
T x, y, t   x, y   ( x, y, t )
We now substitute (2) into (1) and rearrange
(2)
0  x 1
0  y 1
1   2   2    2   2  
 2  2  2  2 
 2 t
x
y
y 
 x
 x 0, y    x (0, y, t )  0 ,
x,1  ( x,1, t )  To cos x ,
(3)
x,0  ( x,0, t )  0
 x 1, y    x (1, y, t )  0 ,
x, y   ( x, y,0)  h( x, y)
We now make the following choices for x, y, t 
0  x 1
0  y 1
1   2   2 
 2  2
 2 t
x
y
 x 0, y, t   0 ,
 x 1, y, t   0 ,
x,0, t   0 ,
(4a)
x,1, t   0 ,
x, y,0  g x, y 
And for x, y 
 2  2

 0,
x 2 y 2
 x 0, y   0 ,  x 1, y   0 ,
0  x 1
0  y 1
x,0  0 ,
(4b)
x,1  T0 cosx 
We solve (4a) by assuming separable solutions of the form,
x, y, t   X ( x)Y ( y)T (t )
(5)
We now substitute (5) into (4a)
1
2
XYT   X YT  XY T
We now divide through by x, y, t   X ( x)Y ( y)T (t )
1 T  X  Y 


X
Y
2 T
(6)
The left side of (6) is a function of t only. The first term on the right is a function of x only and the last
term a function of y only. The only way for both sides to be equal is if they are each equal to some
constant say  2 ,  2 .
X 
  2 ,
X
Y 
  2 ,
Y
1 T
  2   2
2
 T


By inspection, we have chosen the constants to be negative. So that
X ( x)  2 X ( x)  0 ,
Y ( y)   2Y ( y)  0 ,


T t    2 2   2 T t   0
(7a)
For non-trivial solutions, we select our boundary conditions as
X (0)  X 1  Y (0)  Y (1)  0
(7b)
We now solve (7)
Y ( y)  c3 cosy   c4 siny  , T t   c5 exp   2 2   2 t 
X ( x)  c1 cosx   c2 sinx  ,
We now apply the homogenous boundary conditions
X (0)  c2  0 ,
n  n
c2  0 ,
For nontrivial solutions,
X (1)  c1 sin   0
n  1,2,........
X n ( x)  cn cosnx 
So that,
Y (1)  c4 sin   0
Y (0)  c3  0 ,
 m  m
For nontrivial solutions,
(8)
m  1,2,........
Ym ( y)  bm sinmy 
So that



Tmn t   amn exp   2 2 n 2  m 2 t
Hence,
The solutions



 mn ( x, y, t )   mn cosnx sinmy exp   2 2 n 2  m 2 t
Since the system is linear, a complete solution is given as a linear combination of solutions


n
m



( x, y, t )    mn cosnx sin my  exp   2 2 n 2  m 2 t
The constants Amn are given as
1 1
Amn    g ( x, y) sin my  cosnx dxdy
0 0
We now look at solutions to (4b). We assume separable solutions of the form x, y   Gx H  y  so
that
G( x)  2 G( x)  0
and
H ( y)  2 H ( y)  0
X (0)  X 1  Y (0)  0
(9a,b)
We already solved (9a). The solutions are given in (8). So we look for the solutions of (9b)
Y ( y)  c3 exp y   c4 exp  y 
Y (0)  c3  c4  0 ,
n0
n  1,2,...........
y
 Yn ( y )  
bn sinh ny 
The solutions
 n ( x, y)   o y   n cosnx sinhny 
n  1,2,........
Since the system is linear, a complete solution is given as a linear combination of solutions


n 1
n 1
( x, y)    n ( x, y)   o y    n cosnx sinh ny 
At this point, we apply the non homogenous boundary condition

( x,1)  To cos x   o    n cosnx sinh n 
n 1
By comparing coefficients on the two sides of the above equation, we see that only the n  1 term
survives, yielding
1 
To
sinh  
So that
( x, y) 
T0
cosx sinh y 
sinh 
Finally,
T x, y, t  

To
cosx sinh y   
sinh 
n 1


m 1
mn


 
sin my  cosnx  exp   2 m 2  n 2  2 t