The course will only consider linear partial differential equations
of second order. Examples of important linear partial differential
equations are:

Equation (2.1) is the one dimensional wave equation, equation (2.2) is the one dimensional heat (or diffusion) equation and equation (2.3) is the two dimensional Laplace equation. They are all examples of homogeneous, linear partial differential equations.

An example of an inhomogeneous equation is the
two dimensional Poisson equation, namely

The solutions to the above equations are numerous. For example, if one
considers equation (2.3) then it is easily verified that all
of the functions

are solutions. So how do we determine the actual solution we are looking for? The answer, of course, lies in the application of boundary conditions. Once the initial conditions (conditions at ) and the boundary conditions (conditions at specific values of ), where appropriate, are specified, there will be a unique solution to the linear partial differential equation.

Using our knowledge of linear, ordinary differential equations, we
expect that it should be possible to linearly superimpose solutions to
the equations to obtain the most general solution. This is indeed the
case. Thus, if and are both solutions to
(2.1), then

is also a solution if and are constants. However, unlike second order ordinary differential equations where there are two linearly independent solutions and two arbitrary constants, linear partial differential equations may well have an infinte number of linearly independent solutions and we may have to add together solutions involving an infinite number of constants. We will return to this later on.

__Example 2. .12__Obtain a solution, , to the equation

Note that there are only derivatives with respect to and none with respect to . Thus, we can treat the equation like an ordinary differential equation and use first year work to write the solution.

However, unlike the ordinary differential equation case and are not constants but are, in fact, functions of the other independent variable. Hence, the actual solution is

as can be verified by differenitating and substituting into the differential equation.

__Example 2. .13__Consider the equation

To solve this we set so that the equation becomes

Thus, the solution is

Now we must integrate with respect to to get the solution . Therefore, we get

where and is an arbitrary function of integration when we integrate with respect to .