



Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Material Type: Exam; Professor: Fulling; Class: ADV ENGINEERING MATH; Subject: MATHEMATICS; University: Texas A&M University; Term: Unknown 1989;
Typology: Exams
1 / 6
This page cannot be seen from the preview
Don't miss anything!




Alternatively, define normalized eigenfunctions
φ
(x) =
y
(x)
√
∫
y
(x)
dx
,
and then
f (x) =
∑
c
φ
(x), c
=
∫
φ
(x)f (x) dx.
Hidden extra credit:
(1) Evaluate the integral:
∫
sin
(ω
x) dx =
1
2
∫
(1 − cos(2ω
x)) dx
=
1
2
[
x −
1
2 ω
sin(2ω
x)
]
=
1
2
(
π −
1
2 ω
sin(2ω
π)
)
.
As n → ∞ this approaches
, as it should.
(2) Check that there are no zero or negative eigenvalues. This can be done by integration by parts (the
Rayleigh quotient is positive), or directly like this: For λ = 0 we have y(x) = Ax + B = Ax (by
the first boundary condition), and now the second boundary condition forces A = 0. For λ = −κ
we have similarly y(x) = C sinh(κx) and the second boundary condition can’t be satisfied (for
C nonzero). (If the second boundary condition did not contain a minus sign, there could be a
negative eigenvalue.)
of Laplace’s equation,
∂
2
u
∂x
2
∂
2
u
∂y
2
= 0. (It should not be necessary to do extensive
calculations. Use your background knowledge and common sense.)
Problems: (In the rectangular cases, the origin of Cartesian coordinates is at the lower
left corner. In the cylindrical case (f), x is the angular variable and y is the variable along the
axis of the cylinder with y = 0 at the left end.)
(a)
↓ u = 0
↑ u = 0
u = f →
→ ∞
(b)
↓ u = f
↑ u = 0
u = 0 →
→ ∞
(c)
↑ u = 0
u = f →
↗
∞
(d)
↑ u = f
∂u
∂x
= 0 →
↗
∞
The zeroth-order problem is thus
y
= 0, y
(0) = 0, y
(0) = 1,
whose solution is
y
(t) =
1
2
sin(2t).
The first-order problem then becomes
y
= −y
= − cos(2t). y
(0) = 0, y
(0) = 0.
The forcing term is on resonance, so y
will contain a secular term (proportional to t sin(2t)) that
grows with t. This problem will get worse in higher orders, so wherever we stop the perturbation
series, it will give an approximation that is not uniform in t (is not accurate except for rather small t).
(b) Solve the problem by the method of two time scales. (Find a one-term approximation
including the first-order correction.)
See Question 1 in Peter Howard’s Practice Exam 2, Version A:
http://www.math.tamu.edu/~phoward/m401/pexam2asol.pdf
∂
2
u
∂t
2
=
∂
2
u
∂x
2
(−∞ < x < ∞, −∞ < t < ∞), u(0, x) = h(x),
∂u
∂t
(0, x) = j(x).
(If you know the relevant formulas, it is not necessary to rederive them.)
First note that the conventional notation is h = f , j = g.
D’Alembert method: u(t, x) = B(x − t) + C(x + t). Matching the initial conditions yields
B(z) =
[h(z) + G(z)], C(z) =
[h(z) − G(z)],
where G is any antiderivative of j; equivalently,
u(t, x) =
[h(x − t) + h(x + t)] +
∫
j(z) dz.
Fourier transform method: See solution to Homework 14, Exercise 4. Separation-of-variables
method is the same as the transform method except for starting out in a more wordy way.
u(t, x) =
1
2 π
∫
ˆ
h(ω)e
cos(ωt) +
1
2 π
∫
1
ω
ˆ(ω)e
sin ωt
if
ˆ
h(ω) =
∫
e
h(x) dx, etc.
x
3
Find all real roots up through order
2
and all complex roots up through order .
Let x ∼ x
x
. Then
x
∼ x
(x
x
) + 3x
(x
)
= x
x
) +
(3x
x
x
) + O(
).
Therefore, the equation is
0 ∼ x
x
) +
(3x
x
x
) + x
x
− 27.
So the equation of order
is x
= 27, or x
=
√
27 e
for n = 0, 1 , 2. (In fact, n = −1 is
equivalent to n = 2 and leads to a numerical result faster.) That is, the real root is x
= 3, and there
are two complex roots,
x
= 3
[
cos
2 π
3
± i sin
2 π
3
]
=
3
2
(− 1 ± i
√
3).
The equations of orders
and
are
0 = 3x
x
, 0 = 3(x
x
x
) + x
.
For the real case, with x
= 9, we find x
= − 3 /27 = −
, then
0 = 3
(
3
9
)
−
1
9
= 27x
,
so x
= 0.
For the complex case,
x
= −
x
3 x
= −
1
3
2
3
1
− 1 ± i
√
3
= −
2
9
− 1 ∓ i
√
3
1 + 3
=
1
18
(1 ± i
√
3).
Summary:
x ∼ 3 −
9
) is the real root,
x ∼
3
2
(− 1 ± i
√
18
(1 ± i
√
) are the complex roots.
By substituting these into the original equation one can check that that equation is satisfied through
the expected order in each case. (I did that, but I will not type it out.)