Test 2

## Linear Methods of Applied Mathematics Evans M. Harrell II and James V. Herod*

SAMPLE TEST

At Georgia Tech, students took 50 minutes (calculators allowed but no notes). Most students did reasonably well, but few finished early.

1. Find the normal modes (product solutions) for the damped wave equation

utt + ut = uxx

for 0 < x < 1, with boundary conditions:

ux(t,0) = 0

u(t,1) = 0

Make sure to give the specific dependence on x and on t.

THE NORMAL MODES ARE:_____________________________________________

FORMULA OR KEY FACT (FOR POSSIBLE PARTIAL CREDIT):_________

________________________________________________________________

2. Solve the wave equation

utt = uxx

for 0 < x < 1, with boundary conditions

ux(t,0) = ux(t,1) = 0

and initial conditions

u(0,x) = sin2( \pi x)

ut(0,x) = 0

THE GENERAL SOLUTION APPROPRIATE TO THIS PDE WITH THE BC IS:

____________________________________________________________________

SPECIFIC SOLUTION:

u(t,x) = ______________________________________________________________

FORMULA OR KEY FACT (FOR POSSIBLE PARTIAL CREDIT):_________

________________________________________________________________

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

> ODE := diff(y(t), t\$2) + diff(y(t), t) = - mu * y(t);

/   2      \
|  d       |   /  d      \
ODE := |----- y(t)| + |---- y(t)| = - mu y(t)
|   2      |   \ dt      /
\ dt       /

> dsolve( ODE, y(t) );

y(t) =

1/2                                    1/2
_C1 exp(1/2 (- 1 + (1 - 4 mu)   ) t) + _C2 exp(- 1/2 (1 + (1 - 4 mu))   t)

Recall that in this formula, if the square root is imaginary, you can solve instead with


y(t) =
1/2                           1/2
exp(- t/2) (   C1 cos(((4 mu - 1)   /2) t) + C2 sin(((4 mu - 1)   /2) t)   )


by using Euler's formula.