MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
1. Let α be a constant. Find
(a)
d αx
e
dx
(b)
Z
eαx dx
Check your answer (by differentiating it to see if you get the eαx back
again). You will need to use the answer later on.
(c)
d αx
xe
dx
(d)
Z
xeαx dx
Hint: watch for α = 0.
Solution:
(a)
(b)
d αx
e = αeαx
dx
(
Z
1 αx
e
if α 6= 0
αx
e dx = α
x
if α = 0
(c)
(d)
d αx
xe = (αx + 1)eαx
dx
(
Z
1
x − α1 eαx if α 6= 0
αx
α
xe dx = x2
if α = 0
2
Date: October 12, 2000.
1
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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
2.
√
f (x) = α 2eα(x−π)
−π ≤x<π
where α is a constant, and f (x) has 2π period.
(a) Draw the graph of f (x) assuming α = 1.
(b) Calculate the complex Fourier series of f (x). (The amplitudes depend
on α.) Hint: since α is just a constant, your integrals are products of
exponential functions. Use the rule eA eB = eA+B .
(c) Calculate the energy of f (x) (it depends on α). What happens as α → ∞?
Hint: the function gets very wild.
(d) Calculate the energy stored in each amplitude. Approximately how much
energy is stored in the amplitudes ck for k a positive or negative number
which is small (close to zero) compared with α when α is large? Hint:
it better turn out like Heisenberg’s uncertainty principle—a long list of
nearly constant amplitudes.
Solution:
(a) See figure 1 on the facing page.
(b)
√
(−1)k 2α 1 − e−2απ
ck =
2π(α − ik)
(c)
kf k2 = α 1 − e−4απ → ∞
as α → ∞.
(d)
2
1 − e−2απ
1
2
→
2π |ck | =
2
k2
2π
2π 1 + α2
if |k| α: the energy spreads out evenly among the amplitudes ck with
|k| α.
3.
Calculate
Z
2π
3
esin
x
3
− ecos
x
dx
0
(hint: don’t work out any integrals).
Solution: 0 because sin and cos are the same function shifted over, and we are
integrating over a complete period.
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
3
1.4
1.2
1
0.8
0.6
0.4
0.2
–10
–5
0
5
x
10
Figure 1. Exponential-like function
4.
(a) Find the real Fourier series of f (x) = cos2 x − sin2 x (hint: don’t calculate
any integrals).
(b) The same for f (x) = sin x cos x.
Solution:
(a) cos2 x − sin2 x = cos 2x
(b) sin x cos x = 12 sin 2x
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MATH 3150: PDE FOR ENGINEERS
5.
MIDTERM TEST #1
The ordinary differential equation
df
d2 f
+µ +f =0
dt2
dt
(where µ is a real constant) has a periodic solution f (x) which is not constant.
Plug in a complex Fourier series for f (x) to find the possible values of µ.
Solution:
To solve the ODE, the amplitudes of f (x) have to satisfy
2
2πik
2πik
ck + µ
ck + ck = 0
T
T
so that if any ck is not 0, we must have k solving
2
2πik
2πik
+µ
+1=0
T
T
Split into real and imaginary parts:
2
2πk
0=−
+1
T
2πik
0=µ
T
so the first equation says
T
k=±
2π
(in particular, k 6= 0), and then the second forces µ = 0.
6.
Suppose that f (x) has period T . Let
g(x) = f (T x/S)
where S is a positive constant.
(a) g(x) is periodic. With what period?
(b) Calculate the energy of g(x), expressed in terms of the energy of f (x).
(c) Express the complex amplitudes of g(x) in terms of those of f (x).
Solution:
(a) S
(b)
S
kf k2
T
(c) The amplitudes of g(x) are the same as those of f (x).
kgk2 =