# Midterm Exam 11/5/02 ```MAE 192B MIDTERM EXAM
ENGINEERING MATHEMATICS
FALL 2002
November 5, 2002, 10:00-11:50 1.m., Room 5272 Boelter Hall
(TOTAL : 100 points)
Additional Bonus Problem Worth 15 Points- max. Score could be 115/100
________________________________________________________________________
Note: determine all the constants you generate in your solutions
1. (Total Point 35)
Solve Laplace’s equation inside a circular annulus ( a &lt; r &lt; b ) subject to the boundary
conditions:
u (a,θ ) = f (θ ) ,
and
u (b,θ ) = g (θ )
(1)
Hint: you may consider the solution as a superposition of two solutions u1 and u 2 , each
satisfies Laplace’s equation and one homogeneous boundary condition
2.
(Total Points 35)
Consider a slightly damped vibrating string of length L that satisfies the PDE:
∂ 2u
∂ 2u
∂u
ρ 0 2 = T0 2 − β ,
(2)
∂t
∂t
∂x
where β is a fictional coefficient which is relatively small ( β 2 &lt; 4π 2 ρ 0T0 / L2 ).
Determine the solution (using the separation of variable method) which satisfies the
boundary conditions:
u (0, t ) = 0,
and the initial conditions
u ( x,0) = f ( x),
and
u ( L, t ) = 0,
∂u ( x,0)
= g ( x),
∂t
(3)
(4)
3. (Total Points 30)
Consider the heat equation in two dimensional rectangular region 0 &lt; x &lt; L, and
0 &lt; y &lt; H,
 ∂ 2u ∂ 2u 
∂u
= k  2 + 2 
∂t
∂y 
 ∂x
(5)
subject to the initial condition:
u ( x, y,0) = f ( x, y ).
(6)
-1-
Solve the initial value problem and analyze the temperature as t → ∞ if the boundary
conditions are:
∂u
∂u
(0, y, t ) = 0,
( L, y , t ) = 0, u ( x,0, t ) = 0, u ( x, H , t ) = 0,
(7)
∂x
∂x
Bonus Problem : Points from this problem will be added only after seriously
attempting the above problems
4. (Total Points 15)
If a vibrating string of length L , satisfying the following PDE, initial condition (IC),
and boundary conditions(BC’s):
2
∂ 2u
2 ∂ u
,
=
c
∂t 2
∂x 2
BC’s: u (0, t ) = 0,
PDE:
u ( L, t ) = 0,
∂u ( x,0)
IC:
u ( x,0) = f ( x),
= g ( x),
(8)
∂t
is initially unperturbed ( f ( x) = 0) , with the initial velocity g ( x) given , show that:
u ( x, t ) =
1 x + ct
G (x )dx .
2c ∫x −ct
(9)
where G ( x) is the odd periodic extension of g ( x) .
1
Hint: sin a sin b = [cos(a − b) − cos(a + b)]
2
-2-
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