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Problem Set 6 Solutions - APMA 4200 Partial Dierential Equations
1. From the principle of linear superposition, we know that the most general solution to a rst order ODE is the
sum of the solution to the homogeneous equation and the particular solution

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Problem Set 2 Solutions - APMA 4200 Partial Dierential Equations
1.
1.4.1.
(c) Q = 0, u (0) = 0, u(L) = T
x
Solution:
The equilibrium temperature satises:
2u
=0
x2
u
(0) = 0
x
u(L) = T
The general solution for the second order ODE is: u = c1 x + c2 , wh

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Problem Set 9 Solutions - APMA 4200 Partial Dierential Equations
1. 10.3.1a,b: very straightforward calculations.
2. 10.3.5 If
1
2
F ( ) =
f (x)eix dx,
then the inverse transform of ei times that is
F ( )ei(x ) d,
but since
F ( )eix d,
f (x) =
we know i

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1. We apply our usual formula to nd the solution for general forcing f (x) given the Greens function and homogeneous boundary conditions
L
G(x, x0 )f (x0 )dx0 .
u(x) =
0
The Greens function is given by 9.3.46, here L = 1, and f (x) = 1. Thus we have, be

Applied Mathematics E4200
Problem Set 8
Due Mon., Dec. 7, 2008
Note, a couple more problems may be added after the lecture of Monday, Nov.
30. I am posting this now in case anyone wants to get started on the weekend.
1. Consider the inhomogeneous problem

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1. 8.2.6
(a) Since Q = 0 our equilibrium problem is
2
uE = 0, uE (0) = A, uE (L) = B.
x2
We can integrate directly and use the boundary conditions to nd
uE =
BA
x + A.
L
Then we let u = uE + v (x, t) and the remaining problem for u (x, t) is that v sati

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1. 5.3.2
(a) Physical interpretation: this is a damped wave equation, as long as and are both negative. Think of
a maximum in the displacement u; the curvature will be negative, so that the undamped wave equation gives
a restoring force for T0 > 0. In t

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1. 4.4.1
(a) Let c =
T0 /. Substituting u(x, t) = (x)h(t) into the PDE gives:
xx
htt
= 2 =
ch
The solution of the above equation is = C1 cos( x) + C2 sin( x). We can determine by applying
the boundary conditions: = ( n )2 . Plugging this in and solving

Applied Mathematics E4200
Problem Set 4 Solutions
3.2.2
c)
f (x) = Sin
x
L
The fourier series of f (x) on the interval L x L looks like:
(1)
We need to nd the fourier coecients:
a0 =
1
an =
L
bn =
1
L
1
2L
Sin
L
L
L
Sin
L
L
L
Sin
x
L
x
L
x
L
Cos
S

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Problem Set 3 Solutions - APMA 4200 Partial Dierential Equations
Q1. (2.4.1)
To solve this, we will use the method of separation of variables, where we express u(x, t) = (x)G(t). When we
substitute this expression into the heat equation, we obtain the f

Applied Mathematics E4200
Problem Set 1
Due Mon., Sept. 21, 2009
1. ODE practice, I. Solve the following ODE:
dx
+ x = 1,
dt
where > 0 is a constant. Graph a x(t) and show the large-time asymptotic
behavior as t .
2. ODE practice, II. Solve the following