Math 325:  Differential Equations,  Assignment 1,    February 11,  1998
Due to March 4, 1998.

Names (up to 3 students):

Problem 1

In applying elasticity theory to study the transverse vibrations of a beam one encounters
the equation
	(1) E*I*y''''[x] - l*b* y[x] == 0
where y[x] is related to the displacement of the beam at position x, the constant E is Young's modulus,
I is the area moment of inertia which we assume is constant, l is the constant mass per unit length of
the beam, and b is a positive parameter to be determined. We can simplify the equation by letting 
             (2)  K^4 = (l*b)/(E*I),
that is,  we consider 
             (3)  y''''[x] - K^4 y[x] == 0.    
When the beam is clamped at each end, we seek a solution to equation  (3) that satisfies the boundary conditions 
             (4)  y[0] = y'[0] = 0  and  y[L] = y'[L] = 0   
where L is the length of the beam. The problem is to determine those values of K for which equation (3)
has a non trivial solution that satisfies (4). To do this proceed as follows:

(a) Represent the general solution to (3) in terms of sines, cosines and exponentials.

(b) Substitute the general solution obtained in part (a) into the equation in (4) to obtain four equations
in four unknowns.

(c) Determine those values of K for which the system of equations in part (b) has nontrivial solutions.

Solution

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Problem 2

Find the exact solution of the given initial value problem:

 (5)  y'[x] = 1/2 (y - 1)^2,  y[0] = 1; x from [0,1].  

 Then use Mathematica program for the Euler's method twice to approximate this solution on the given interval, first with step size h = 0.01, then with step size h = 0.005. Then approximate the solution by using Mathematica function: NDSolve[]. Finally,  make a table showing for each point  x two approximate values and the actual value of y[x], together with the percentage error in the more accurate approximation, for x an integral multiple of 0.2.

Solution

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Problem 3

A skydiver jumps from an aircraft at an initial altitude of 10,000 ft. Assume that she falls vertically with
initial velocity zero v(0) = 0, weights m = 128 lb, and experiences an upward force F of air resistance given in terms of her velocity v(t)  (in feet per second) by:

(6)  F = (0.01)v + (0.001)v^2 + (0.0001)v^3 

(in pounds, and with the coordinate axis directed downward so that v > 0). 

a) Find a differential equation for v = v(t) and formulate an initial value problem. 
Hint: Use Newtons law: m a(t) = mg - F, where a(t) is an acceleration of the skydiver and
g is acceleration of gravity.
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The equation is as follows: v'[t] = f[v] where f[v] = 32 - F/4. Also v(0) = 0.
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b) The skydiver reaches her terminal velocity when the forces of gravity and air resistance balance, 
so f(v) = 0. (After that v[t] remains constant.) If she does not open her parachute, what will be her terminal velocity in this case? 
Hint: Use Mathematica function: NSolve[a x^3 + b x^2 + c x + d == 0, x] for given coefficients
a,b,c and d. Calculate Vterminal from f[v] = 0. Calculate .95*Vtermminal.

c) Use Mathematica program for the Runge-Kutta method to approximate solution to the initial
value problem obtained in part (a). 

    1) How fast will a skydiver be falling after 5s have alapsed? After 10s? 
    2) How lond will it take to attain the terminal velocity?
        Hint:  Use Runge-Kutta program and choose from the table of values  t's for which v[t]'s
                  are the closest to Vterminal  and .95*Vtermminal.
    3) How long will take to get to the 95% of the terminal velocity?
    4) Sketch graph of the solution.   ******** (By hand!) ************

Solution

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Problem 4

a) Use Mathematica program for the Runge-Kutta method to approximate solution to the following initial
value problem :

(7) y'[x] = 5y[x] - 6 Exp[-x],  y[0] = 1,  x from [0,4]

with step sizes h = 0.2, h = 0.1, and h = 0.05. Notice that the solution of (7) is:  y[x] = Exp[-x]. 
Make a table showing for each point three approximate values and the actual value, for x an integral multiple of 0.4.

It is obvious from the table that our numerical approximation fails. The explanation lies in the fact that the general solution of the equation (7) is y[x] = Exp[-x] + C Exp[5x]. The particular solution satisfying the initial condition y[0] = 1 is obtained when C = 0. But any departure, however small, from the exact solution y[x] = Exp[-x] -- even if due only to roundoff error -- introduces (in effect) a nonzero value of C. And all solution curves with nonzero C diverge strongly away from the one with C = 0, even if their initial values are close to 1.

b) Use Mathematica function Plot [] to plot the solutions:  y[x] = Exp[-x] + C Exp[5x], 
 for C = 0, 0.05, 0.1, 0.2, 0.6 and demonstrate divergence of the solutions explained in part (a).

Solution

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