Mathematics & Science Learning Center
Computer Laboratory

Applications of Differential Equations

A Suspended Wire

Let's quickly recap the problem, just to remind ourselves what we're working with here:

A suspended wire hangs under its own weight. The differential equation describing the shape it assumes is:

where w represents the linear density of the wire, and |T_P| represents the horizontal component of tension in the wire at its lowest point.

A Preliminary Example

Suppose a wire has linear density of 0.5 lbs/ft and is subjected to a horizontal tension of 40 lbs. A coordinate system is imposed on the hanging wire such that the two axes have units marked off in feet, and the lowest point on the wire passes through the origin. You are given the following instructions:

Instructions

Produce the graph of the wire on the interval -5 ≤ x ≤ 5. Name your plot wire1plot, and force the graph to come out in red.

Setting Things Up

In order to produce a graph we need a function to plot. Unfortunately, all that we have to work with is a differential equation and a short description of how the wire fits into a coordinate system. So what should we do?

Well, the differential equation might be able to give us a function, provided we can solve it. Hold on a second, though! Even if we could solve it, the differential equation all by itself wouldn't give us a function—it would give us a whole family of functions. That's a bit more than we're looking for, or even expect if we think about the reality behind the problem. So, how could we narrow down the family to a single answer?

You should be so used to the idea by now that you should have blurted out the answer to my last question without even thinking about it. A unique solution will be found if we have an initial value problem.

We already have half of the initial value problem in that we have our differential equation. The bit that we are missing is the initial condition. Actually, in this case, since the differential equation is second order, we need not only one initial condition, but two. Where will these come from?

We expect one of the initial conditions to be in the form of a known point through which the solution is guaranteed to pass. Do we have such a point? Of course we do! It was so important that a little earlier I even colored the statement in red—see it? The lowest point on the wire passes through the origin. So there's our first initial condition—the solution passes through (0,0), or to put it another way:

y(0) = 0.

One down, one to go! What about the other initial condition? Well, for a second order initial value problem, the second initial condition should involve a known value of the derivative, i.e. a known slope. Is there any place on the function where we do know its slope?

Think about the statement I made in red again! The lowest point on the wire passes through the origin. If it's the lowest point, then it's a minimum value, correct? And what happens to the derivative at the local minima of a function? It's zero, right? (Assuming the function is continuous, the derivative exists, blah, blah, blah...) This even makes sense if you consider the physical reality behind the problem—at the bottom of the loop, the wire is horizontal. So we now have our second initial condition:

y′(0) = 0.

Summary

Here's what we know so far:

y(0) = 0.

y'(0) = 0.

w = 0.5 and |T_P| = 40.

It looks like we have all of the information we could possibly need to make Mathematica solve the problem for us.

Converting the Problem to Mathematica Format

First, let's substitute the known values of w and |T_P| into the differential equation. We get:

Now, converting the Leibniz notation we've been using so far into Newton's notation (primes) and we get:

Which is just about ready for Mathematica input. Recall that when entering differential equations into Mathematica you must include the argument of the function (so you enter y[x], not just y, remember?) Also, recall that the square root function is denoted by Sqrt[...], and equations require a double equals sign. Using all of these reminders, the Mathematica version of our differential equation would be:

y''[x]==(0.5/40)*Sqrt[1+(y'[x])^2]

But that's just the differential equation. We still haven't instructed Mathematica to solve it for us. Since we actually have an initial value problem, not just a differential equation, we have two options open to us: we could solve the problem analytically, or numerically. Mathematica can handle doing it either way, so we might as well have it do it both ways, just for the sake of comparison. First we'll try the analytic approach.

The Analytical Solution

ansol=DSolve[{y''[x]==(0.5/40)*Sqrt[1+(y'[x])^2], y[0]==0, y'[0]==0}, y[x], x]

(Notice the braces enclosing the list of equations, the double equals signs used on all of the equations, and the argument [x] attached to every appearance of y. Also notice the way that the initial conditions we found earlier are included in the command.)

Well, that was a bit disappointing. Let's go discuss the result you should have just gotten...

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.