Page 324 - Schaum's Outline of Differential Equations

CHAP. 31] AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 307

We assume that u(x, t) can be written as a product of functions. That is, u(x, t) = X(x)T(t). Finding the appro-
priate derivatives, we have u xx = X" (x)T(t) and u t = X(x)T'(t). Substitution of these derivatives into the PDE yields
the following.

Equation (1) can be rewritten as

We note that the left-hand side of Eq. (2) is solely a function of x, while the right-hand side of this equation
contains only the independent variable t. This necessarily implies that both ratios must be a constant, because there
are no other alternatives. We denote this constant by c:

We now separate Eq. (3) into two ODEs:

and

We note that the Eq. (4) is a "spatial" equation, while Eq. (5) is a "temporal" equation. To solve for u(x, t), we
must solve these two resulting ODEs.
We first turn our attention to the spatial equation, X"(x) - cX(x) = 0. To solve this ODE, we must consider our
insulated boundary conditions; this will give rise to a "boundary value problem" (see Chapter 32). We note that
u(0, t) = 0 implies that X(0) = 0, since T(t) cannot be identically 0, since this would produce a trivial solution;
similarly, X(n) = 0. The nature of the solutions to this ODE depends on whether c is positive, zero or negative.
If c> 0, then by techniques presented in Chapter 9, we have X(x) = c^e + C 2e~ , where Cj and c 2 are
determined by the boundary conditions. X(0) = Cie° + c 2e° = Cj + c 2 = 0 and X(n) = c le " + c 2e ". These two
equations necessarily imply that Cj = c 2 = 0, which means that X(x) = 0 which renders u(x, t) trivial.
If c = 0, then X(x) = c\x + c 2, where c l and c 2 are determined by the boundary conditions. Here again,
X(0) = X(n) = 0 force c l = c 2 = 0, and we have u(x, t) = 0 once more.
2
Let us assume c < 0, writing c = - X , X > 0 for convenience. Our ODE becomes X"(x) + X?X(x) = 0, which
leads to X(x) = Cj sin Xx + c 2 cos Xx. Our first boundary condition, X(0) = 0 implies c 2 = 0. Imposing X(n) = 0, we
have Ci sin 'kn= 0.
If we let A,= 1, 2, 3, ..., then we have a non-trivial solution for X(x). That is, X(x) = Cj sin nx, where n is a
positive integer. Note that these values can termed "eigenvalues" and the corresponding functions are called "eigen-
functions" (see Chapter 33).
2
2
We now turn our attention to Eq. (5), letting c = -X = -n , where n is a positive integer. That is,
2
T'(t) + n kT(t) = 0. This type of ODE was discussed in Chapter 4 and has T(t) = c 3e" ' as a solution, where c 3 is
an arbitrary constant.
Since u(x, t) =X(x)T(t), we have u(x, t) = Cj sin nx c 3e~" kl = a ne~" kl sin nx, where a n = CjC 3. Not only does
u(x, t) = a ne~" kl sin nx satisfy the PDE in conjunction with the boundary conditions, but any linear combination of
these for different values of n. That is,

where N is any positive integer, is also a solution. This is due to the linearity of the PDE. (In fact, we can even have
our sum ranging from 1 to °°).
We finally impose the initial condition, u(x, 0) = 2 sin 4x - 11 sin Ix, to Eq. (6). Hence, u(x, 0) = sin nx.
Letting n = 4, a 4 = 2 and n = 7, a 7 = -11, we arrive at the desired solution,

It can easily be shown that Eq. (7) does indeed solve the heat equation, while satisfying both boundary
conditions and the initial condition.