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The Method of Separation of Variables

This method can be applied to linear partial differential equations, especially those with constant coefficients in the equation. To illustrate the method we consider the one dimensional wave equation,

\begin{displaymath}
{\partial^{2}u\over \partial t^{2}} = c^{2}{\partial^{2}u\over \partial
x^{2}},
\end{displaymath}

where $u(x,t)$ is the displacement (or deflection) of the stretched string. To solve this we need to include both boundary conditions and initial conditions. As an illustration of the method, we consider the case for which the string is fixed at the ends so that
\begin{displaymath}
u(0,t) = 0, \qquad u(L,t) = 0 \qquad \hbox{ for all }t.
\end{displaymath} (2.17)

Other boundary conditions will be considered later. The initial conditions are taken as
\begin{displaymath}
u(x,0) = f(x),
\end{displaymath} (2.18)

and
\begin{displaymath}
{\partial u\over \partial t}(x,0) = g(x).
\end{displaymath} (2.19)

The basic idea is to:
  1. Apply the method of separation to obtain two ordinary differential equations.

  2. Determine the solutions that satisfy the boundary conditions.

  3. Use Fourier series to superimpose the solutions to get the final solution that satisfies both the wave equation and the given initial conditions.
We assume that $u(x,t)$ can be expressed as a product of a function of $x$ and a function of $t$. This is an assumption and if we end up with a contradiction along the line then the assumption was wrong. If there is no contradiction then the assumption is valid. Thus, we seek a solution of the form
\begin{displaymath}
u(x,t) = X(x)T(t).
\end{displaymath} (2.20)

Differentiating, we get

\begin{displaymath}
{\partial^{2}u\over \partial t^{2}} = X(x)\ddot{T}(t),
\end{displaymath}

and

\begin{displaymath}
{\partial^{2}u\over \partial x^{2}} = X^{\prime\prime}(x)T(t),
\end{displaymath}

where the prime denotes differentiation with respect to $x$ and the dot with respect to $t$. Thus, the wave equation becomes,
\begin{displaymath}
X^{\prime\prime}(x)T(t) = {1\over c^{2}}X(x)\ddot{T}(t),
\end{displaymath} (2.21)

and dividing by the product $X(x)T(t)$ (2.21) becomes
\begin{displaymath}
{X^{\prime\prime}\over X} = {\ddot{T}\over c^{2}T}.
\end{displaymath} (2.22)

Since the left hand side of (2.22) is a function of $x$ alone and the right hand side is a function of $t$ alone, this is impossible (we could change the value of $x$, at a fixed time, and the left hand side would have a different value but the right hand side would remain the same) unless they are both equal to a constant. Thus, we have
\begin{displaymath}
{X^{\prime\prime}\over X} = {\ddot{T}\over c^{2}T} =
\hbox{constant} = C.
\end{displaymath} (2.23)

Thus, we obtain
$\displaystyle X^{\prime\prime}$ $\textstyle =$ $\displaystyle C X,$ (2.24)
$\displaystyle \ddot{T}$ $\textstyle =$ $\displaystyle C T.$ (2.25)

We allow the constant to take any value and then show that only certain values allow you to satisfy the boundary conditions, i.e. the conditions in the $x$ direction. We need to consider the three possible cases for $C$, namely $C = p^{2}$ positive, $C=0$ zero and $C = - p^{2}$ negative. These give us three distinct types of solutions that are restricted by the initial and boundary conditions.

Subsections
next up previous
Next: Up: Separation of Variables Previous: Half-Range Expansions
Prof. Alan Hood
2000-10-30