Practice Final Exam, Math 112 C
1 June, 2007
Revision I
Name:
ID:
Instructions: The actual final exam will have about 5 problems that look very much
like a subset of the following 12. Complete these problems with the aid of the book. The
use of group work and external text reference is encouraged. On the exam, you should
be able to state and use any theorem, fact (obtained from an exercise or example), or
definition from the book that you use to solve these practice exam problems. You do not
have to know theorems by name or number only the statements of them.
1. Find a series solution for the problem
!
2
∂2u
1 ∂u
1 ∂2u ∂2u
2 ∂ u
−
c
+
+
+ 2
∂t2
∂r 2
r ∂r r 2 ∂θ2
∂z
u(1, θ, z, t)
u(r, θ, 0, t)
∂u
(r, θ, L, t)
∂z
u(r, θ, z, 0)
∂u
(r, θ, z, 0)
∂t
2. Solve the heat equation with convection:
= 0,
r < 1, 0 < z < L, t > 0,
= 0,
= 0,
= 0,
= f (r, θ, z),
= 0.
∂2u
∂u
∂u
= k 2 −c ,
∂t
∂x
∂x
lim u(x, t) = 0, t ≥ 0,
−∞ < x < ∞, t > 0,
x→±∞
u(x, t = 0) = f (x),
−∞ < x < ∞.
3. Solve the heat equation on a semi-infinite interval:
∂u
∂2u
= k 2,
∂t
∂x
∂u
(x = 0, t) = 0,
∂x
lim u(x, t) = 0,
x→∞
t ≥ 0,
t ≥ 0,
u(x, t = 0) = f (x),
1
0 < x < ∞, t > 0,
0 < x < ∞.
4. Solve the following damped vibrating string problem:
2
∂2u
∂u
2∂ u
−
c
+β
− F (x) cos(ωt)
2
2
∂t
∂x
∂t
u(x = 0, t)
u(x = π, t)
u(x, t = 0)
∂u
(x, t = 0)
∂t
Assume that c > 0, and 0 < β < 2c.
= 0,
0 < x < π, t > 0,
= 0,
= 0,
= 0,
= 0.
5. Solve the heat equation with a heat sink:
∂u
∂2u
= k 2 − γu,
∂t
∂x
lim u(x, t) = 0, t ≥ 0,
−∞ < x < ∞, t > 0,
x→±∞
u(x, t = 0) = f (x),
−∞ < x < ∞,
where γ ≥ 0
6. Solve Laplace’s equation in the half-plane:
∂2u ∂2u
+
= 0,
∂x2 ∂y 2
lim u(x, y) = 0,
0 < y < ∞,
lim u(x, y) = 0,
−∞ < x < ∞,
x→±∞
y→∞
−∞ < x < ∞, 0 < y < ∞,
u(x, y = 0) = f (x),
−∞ < x < ∞.
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7. Show that if f (x) = xe−x /2 and g(x) = x2 /2 then f ∗ g(x) = Cx, where C is a
constant and
Z ∞
f (x − x̄)g(x̄) dx̄.
f ∗ g(x) =
−∞
8. Find the solution of
!
∂ 2 u 2 ∂u
∂2u
∂u
1
1
∂
sin(θ)
+
+
+
= 0,
∂r 2
r ∂r r 2 sin(θ) ∂θ
∂θ
r 2 sin2 (θ) ∂φ2
u(r = 1, θ, φ) = x3 .
9. Show that the right-hand-side of Rodrigues formula
1 dn 2
Pn (x) = n
(x − 1)n
n
2 n! dx
satisfies Legendre’s differential equation
"
#
dy
d
(1 − x2 )
+ n(n + 1)y = 0.
dx
dx
2
r<1
−π ≤ φ ≤ π ,
0≤θ≤π
10. Find the solution of
!
∂ 2 u 2 ∂u
∂u
1
1
∂
∂2u
sin(θ)
+
+
+
= 0,
∂r 2
r ∂r r 2 sin(θ) ∂θ
∂θ
r 2 sin2 (θ) ∂φ2
u(r = 1, θ, φ) =
(
r<1
−π ≤ φ ≤ π ,
0≤θ≤π
1 0 ≤ θ ≤ π/2
.
−1 π/2 < θ ≤ π
Hint: You may find these formulae useful
Z
1
P2n+1 (x) dx = (−1)n
0
Z
(2n)!
22n+1 n!(n
+ 1)!
,
1
P2n (x) dx = 0,
n 6= 0,
0
where n is a non-negative integer.
11. Find a solution for the problem
!
1 ∂2u
∂ 2 u 1 ∂u
∂u
−k
+
+
= 0, r < 1, t > 0,
∂t
∂r 2
r ∂r r 2 ∂θ2
∂u
(1, θ, t) = 0,
∂r
u(r, θ, 0) = f (r, θ).
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