Test 3

Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1994,1995,1996 by Evans M. Harrell II and James V. Herod. All rights reserved.


SAMPLE TEST

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 ut = 4 uxx, for 0 < t, 0 < x < 1

   BC ux(t,0) = 0, ux(t,1) = 0, for 0 < t

   IC u(0,x) = 2 x - x2, 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 umax = _________________


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 \phin(x), for -infinity < x < infinity, which are eigenfunctions for the ordinary differential equation

- \phin'' + x2 \phin = (2n+1) \phin, n = 0, 1, 2, ....

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

Consider the following PDE:

    PDE utt = grad2 u - x2 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) = __________________________________________________


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

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

r2 R'' + r R' + (\mu r2 - m2) R = 0

are J+/-m(\mu1/2 r), when m= 1,2,...; For small s, Jm(s) ~ C sm. When m=0, two independent solutions are J0(u1/2 r) and Y0(u1/2 r); J0 is bounded at r=0, but Y0 is not.

For each fixed integer m> 0, the set of functions Jm(jmn 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

- \phin'' + x2 \phin = (2n+1) \phin, n = 0, 1, 2, ....

In fact,

\phin = exp(-x2/2) hn(x),

where hn is a polynomial in x of degree n, and that \phin 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

See the solutions.

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