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Applications of Differential Equations

The Simple Pendulum

(continued from last page...)

You just got the following, I hope:

linsol=linsol[[1,1,2]]

Stripped Solution

We have now finished solving the linearized problem. What we're now going to attempt to do is compare this result with what we get when we solve the original, non-linear problem.

Solving the Non-Linear Problem

The solution to the non-linear problem is the actual, reality based solution. The only roadblock to finding it is that the problem is analytically unsolvable—after all, it was to overcome this that we linearized it in the first place.

But no-one said we had to solve the problem analytically. There is always the numerical approach to fall back on at times like this. So we'll use Mathematica's NDSolve command.

As a reminder, the original, non-linear initial value problem was:

Actual Initial Value Problem

y'' + (g/L) sin(y) = 0
y(0) = yo
y'(0) = vo

which quickly becomes:

y'' + 9.8 sin(y) = 0
y(0) = yo
y'(0) = vo

when we replace g and L by the same values that we used for the linear example we just completed. We'll also use the same initial conditions as before. Mathematica still has these entered during the current session, so there's no need to re-enter them. One other thing that we need for the numerical solver is the interval upon which it is to find the solution. We'll use 0≤t≤15, and put our solution into the variable actsol. Our command will be:

actsol=NDSolve[{y''[t]+9.8 Sin[y[t]]==0,
   y[0]==y0, y'[0]==v0}, y[t], {t,0,15}]

Go back to Mathematica and issue it. Returning here when you're done.

Let's go see what you should have gotten...


Compass If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.
 
 

ODE Laboratories: A Sabbatical Project by Christopher A. Barker

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