Separation of variables is a procedure which can turn a partial differential equation into a set of ordinary differential equations. The procedure only works in very special cases involving a high degree of symmetry. Remarkably, the procedure works for many important physics examples. Here, we will use the procedure on the wave equation.

Step 1: Write the partial differential equation in appropriate coordinate system. For the wave equation we have:

\begin{equation} \frac{\partial^2 f}{\partial x^2}-\frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} =0 \label{Wave} \end{equation}

Step 2: Assume that the solution $f(x,t)$ can be written as the product of functions, each of which depends on only one variable, in this case $x$ or $t$, i.e.\ assume \begin{equation} f(x,t)=X(x)T(t) \label{separated} \end{equation} This is a very strong assumption. Not all solutions will be of this form. However, it turns out that all of the solutions can be written as linear combinations of solutions of this form. The study of when and why this works is called Sturm-Liouville theory.

Plug this assumed solution (\ref{separated}) into the partial differential equation (\ref{Wave}). Because of the special form for $f$, the partial derivatives each act on only one of the factors in $f$. \begin{equation} T\frac{d^2 X}{d x^2}-\frac{1}{v^2}X\frac{d^2 T}{d t^2} =0 \label{wavesep} \end{equation} Any partial derivatives that act only on a function of a single variable may be rewritten as total derivatives.

Step 3: Divide by $f$ in the form of (\ref{separated}). Many, many students forget this step. Don't be one of them! The rest of the procedure doesn't work if you do. \begin{eqnarray} \frac{1}{X}\frac{d^2 X}{d x^2}-\frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2} =0 \label{wavedivide} \end{eqnarray}

Step 4: Isolate all of the dependence on one coordinate on one side of the equation. Do as much algebra as you need to do to achieve this. In our example, notice that in (\ref{wavedivide}), all of the $t$ dependence is already in one term while all of the dependence on the $x$ variable is in the other term. In this case, the $t$ dependence is trivially isolated by putting the $t$ term on the other side of the equation, without any other algebra on our part. \begin{eqnarray} \frac{1}{X}\frac{d^2 X}{d x^2}=\frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2} \label{wavedivide2} \end{eqnarray}

Step 5: Now imagine changing the isolated variable $t$ by a small amount. In principle, the right-hand side of (\ref{wavedivide2}) could change, but nothing on the left-hand side would. (This argument is the magic of the separation of variables procedure–compare it to arguments about constants of the motion from classical mechanics.) Therefore, if the equation is to be true for all values of $t$, the particular combination of $t$ dependence on the right-hand side must be constant. We will call this constant $A$. Note that we don't know (yet) whether the constant is positive or negative. \begin{equation} \frac{1}{X}\frac{d^2 X}{d x^2}=\frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2}\equiv A \label{Adefinition} \end{equation} In this way we have broken our original partial differential equation up into a pair of equations, one of which is an ordinary differential equation involving only $t$, the other is an ordinary differential equation involving only $x$. \begin{eqnarray} \frac{1}{X}\frac{d^2 X}{d x^2}=A \label{wavexdivide}\\ \frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2}= A \label{wavetdivide} \end{eqnarray} The separation constant $A$ appears in both equations.

Step 6: Write each equation in standard form by multiplying each equation by its unknown function to clear it from the denominator. \begin{eqnarray} \frac{d^2 X}{d x^2}-AX=0 \label{wavex}\\ \frac{1}{T}\frac{d^2 T}{d t^2}-Av^2T=0 \label{wavet} \end{eqnarray}

Notice that (\ref{wavex}) is an eigenvalue equation for the operator $\frac{d^2}{dx^2}$. At the moment, the eigenvalue $A$ could be anything. We will use the spatial boundary conditions for a problem that we want to solve to find the possible values of $A$. Once we have found the possible eigenvalues $A$, (\ref{wavet}) becomes a second order ordinary differential equation with constant coefficients which you should know how to solve. It is NOT an eigenvalue equation.