# Partial differential equations/Separation of variables method Action required: please create Category:Partial differential equations/Lectures and add it to Category:Lectures.

## Introduction

We often consider partial differential equations such as

$\nabla ^{2}\psi ={\frac {1}{c^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}$ ,

which is recognisable as the wave equation in three dimensions, with $\nabla ^{2}$ being the Laplacian operator, $\psi$ being some function of three spacial dimensions and time, and c being the speed of the wave.

These are often found by considering the physical connotations of a system, but how can we find a form of $\psi$ such that the equation is true?

## Finding General Solutions

One way of doing this is to make the assumption that $\psi$ itself is a product of several other functions, each of which is itself a function of only one variable. In the case of the wave equation shown above, we make the assumption that

$\psi (x,y,z,t)=X(x)\times Y(y)\times Z(z)\times T(t)$ (NB Remember that the upper case characters are functions of the variables denoted by their lower case counterparts, not the variables themselves)

By substituting this form of $\psi$ into the original wave equation and using the three dimensional cartesian form of the Laplacian operator, we find that

$YZT{\frac {d^{2}X}{dx^{2}}}+XZT{\frac {d^{2}Y}{dy^{2}}}+XYT{\frac {d^{2}Z}{dz^{2}}}={\frac {1}{c^{2}}}XZY{\frac {d^{2}T}{dt^{2}}}$ We can then divide this equation through by $\psi$ to produce the following equation:

${\frac {1}{X}}{\frac {d^{2}X}{dx^{2}}}+{\frac {1}{Y}}{\frac {d^{2}Y}{dy^{2}}}+{\frac {1}{Z}}{\frac {d^{2}Z}{dz^{2}}}={\frac {1}{c^{2}}}{\frac {1}{T}}{\frac {d^{2}T}{dt^{2}}}$ Both sides of this equation must be equal for all values of x, y, z and t. This can only be true if both sides are equal to a constant, which can be chosen for convenience, and in this case is -(k2).

The time-dependent part of this equation now becomes an ordinary differential equation of form

${\frac {d^{2}T}{dt^{2}}}=-c^{2}k^{2}T$ This is easily soluble, with general solution

$T(t)=A\cos(ckt)+B\sin(ckt)$ with A and B being arbitrary constants, which are defined by the specific boundary conditions of the physical system. Note that the key to finding the time-dependent part of the original function was to find an ODE in terms of time. This general process of finding ODEs from PDEs is the essence of this method.