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Di erential Equations

Lecture 40: Separation of Variables and Heat Equation IVPs

2. Separation of Variables

In Section 1, we derived the heat equation (40.1a) for our bar of length L. Suppose we’ve got an initial value problem such as (40.1). How do we proceed? We’ll want to try to build up a general solution out of smaller solutions that are easier to find.

A rather straightforward approach is to start by assuming these “mini”-solutions have some nice form. For example, we might suppose we have a separated solution, where

u(x, t) = X(x)T (t).

That is, our solution is the product of a function that depends only on x and a function that depends only on t. We can then try to write down an equation depending only on x and another depending only on t before using our knowledge of ODEs to try to solve them.

It should be noted that this is very special and should not be expected in general. In fact, this method cannot always be used, and even when it can be used often it’s hard to move past the first step. However, it works for all of the equations we will be considering in this class, and is a valuable starting point.

How does the method work? We begin by plugging our separated solution into the heat equation (40.1a).

∂t

[X(x)T (t)] = k x 2 X ( x ) T 0 ( t ) = k X 0 0 ( x 2 ) T ( t ) [X (x)T (t)]

Now notice that we can move everything depending on x to one side and everything depending on t to the other.

T 0 ( t ) kT (t)

=

X00(x) X(x)

This equation should look a little funny to you. On the left, we have an expression which depends only on t, while on the right, we have an expression that depends only on x. Yet these two sides have to be equal for any choice of x and t we make! The only way this is possible is if both sides of the equation are the same constant. In other words,

T 0 ( t ) kT (t)

=

X00(x) X(x)

=

λ.

We’ve written the minus sign explictly for convenience

: it will turn out that λ > 0, but these

expressions should be negative. The equation above really contains a pair of separate ordinary di erential equations,

X00 + λX = 0

(40.2a)

0 T + λkT = 0.

(40.2b)

Notice that, with these separated equations, our boundary conditions (39.4b) become

X(0)

=

0

and

X(l)

=

0.

Now,

(40.2b)

is

easy

enough

to

solve:

we

have

T 0

=

λkT ,

so

that

T (t) = Ae

kt

.

What about (40.2a)?

This just gives us the boundary value problem

X00 + λX = 0

X(0) = 0

X(l) = 0.

This should look familiar: it’s very similar to Example 38.6. The only di erence is instead of our second condition occuring at x = 2π, it’s at some x = l. As in that example, it will turn out that

2

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