Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

At Georgia Tech, this was used as the final exam after a short term (less than 10 weeks). Students were allowed 2 hours, 50 minutes, which was adequate. Few students got problem 3 completely correct.

The announced standard for the test was:

A = 2 problems completely solved, a correct idea on the third. B = 1 problem completely solved, correct ideas on the other two. C = good ideas on the all problems, but no complete solutions.

**THERE MAY BE HELPFUL INFORMATION ON THE PAGES AT THE END OF THE TEST!**

1.

a) Solve the following problem:

PDE u_{t} = 4 u_{xx}, for 0 < t, 0 < x < 1

BC u_{x}(t,0) = 0, u_{x}(t,1) = 0, for 0 < t

IC u(0,x) = 2 x - x^{2}, for 0 < x < 1

ANSWER:

u(t,x) = _______________________________________

b) Find the maximum value of u(t,x) for 0 <= t <= 1, 0 <= x <= 1:

ANSWER:

The maximum temperature occurs at x = _____, t = _____

The maximum temperature is u_{max} = _________________

**THERE MAY BE HELPFUL INFORMATION ON THE PAGES AT THE END OF THE TEST!**

2. Some background information:

There is a complete, orthonormal set of functions denoted \phi_{n}(x), for
-infinity < x < infinity, which are eigenfunctions for the
ordinary differential equation

- \phi_{n}'' + x^{2} \phi_{n} = (2n+1) \phi_{n}, n = **0**, 1, 2,
....

You may use the notation \phi n in the answer to this problem.

Consider the following PDE:

PDE u_{tt} = grad^{2} u - x^{2} u, for 0 < t, 0 < x < \pi

BC u(t,x,0) = u(t,x, \pi ) = 0, for 0 < t

Find the normal mode with the lowest frequency of vibration (include the time dependence):

ANSWER:

u(t,x,y) = __________________________________________________

Find the general solution:

ANSWER:

u(t,x,y) = __________________________________________________

3. A slice of pizza is shaped like a sector in cylindrical coordinates,

0 < r < 20 cm,

0 < \theta < \pi /3 radians

0 < z < 1 cm.

It has come to thermal equilibrium while sitting on a student's computer monitor, so that the temperature on its surface is

u(r, \theta , 0) = 30

u(r, \theta , 1) = 20

u(r,0,z) = u(r, \pi /3,z) = 30 - 10 z

u(20, \theta ,z) = 30 - 10 z - 5

This is a low quality pizza consisting of a homogeneous material (independent of position)

a) The partial differential equation for a homogeneous material at thermal equilibrium is:

_______________________________________

b) Answer the following questions.

Are there useful simplifications involving the boundary conditions? If so, what are they?

Be specific and put the answer here:_____________________________________

Are there useful separated solutions? If so, write the specific ordinary differential equations that the separated solutions satisfy below. Include boundary conditions.

____________________________________________________

____________________________________________________

____________________________________________________

____________________________________________________

c) Solve the differential equation with the given boundary conditions.

Possibly helpful information:

The **Laplace operator** in various coordinate systems:

**Bessel functions**

Two independent solutions of

r^{2} R'' + r R' + (\mu r^{2} - m^{2}) R = 0

are J_{+/-m}(\mu^{1/2} r), when m= 1,2,...; For small s,
J_{m}(s) ~ C
s^{m}. When m=0, two independent solutions are J_{0}(u^{1/2} r)
and Y_{0}(u^{1/2} r); J_{0} is bounded at r=0, but
Y_{0} is not.

For each fixed integer m> 0, the set of functions J_{m}(j_{mn} r/A) is complete
for 0 < r < A, with the orthogonality relationship

**An eigenvalue problem. **There is a complete, orthonormal set of
functions denoted \phi n(x), for -infinity < x < infinity,
which are eigenfunctions for the ordinary differential equation

- \phi_{n}'' + x^{2} \phi_{n} = (2n+1) \phi_{n}, n = **0**, 1, 2,
....

In fact,

\phi_{n} = exp(-x^{2}/2) h_{n}(x),

where hn is a polynomial in x of degree n, and that \phi_{n} is even when n is
even, and odd when n is odd.

**Some help from Waterloo, Ontario**:

> int(sin(a*x) * (sin(b*x))^2, x); cos(a x) cos((a + 2 b) x) cos((a - 2 b) x) - 1/2 -------- + 1/4 ---------------- + 1/4 ---------------- a a + 2 b a - 2 b > int(cos(a*x) * (sin(b*x))^2, x); sin(a x) sin((a - 2 b) x) sin((a + 2 b) x) 1/2 -------- - 1/4 ---------------- - 1/4 ---------------- a a - 2 b a + 2 b > int(sin(2*b*x) * (sin(b*x))^2, x); cos(2 b x) cos(4 b x) - 1/4 ---------- + 1/16 ---------- b b > int(cos(2*b*x) * (sin(b*x))^2, x); sin(2 b x) sin(4 b x) 1/4 ---------- - 1/16 ---------- - 1/4 x b b > int(sin(a*x) * x, x); sin(a x) - a x cos(a x) ----------------------- 2 a > int(cos(a*x) * x, x); cos(a x) + sin(a x) a x ----------------------- 2 a > int(sin(a*x) * x^2, x); 2 2 - a x cos(a x) + 2 cos(a x) + 2 sin(a x) a x ---------------------------------------------- 3 a > int(cos(a*x) * x^2, x); 2 2 sin(a x) a x - 2 sin(a x) + 2 a x cos(a x) -------------------------------------------- 3 a

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