Definitions of Partial Differential Equations

A partial differential equation is an equation that involves an unknown function and its partial derivatives. The order of the highest derivative defines the order of the equation. The equation is called linear if the unknown function only appears in a linear form. Finally, the equation is homogeneous if every term involves the unknown function or its partial derivatives and inhomogeneous if it does not.

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

$${\partial^{2}u\over \partial t^{2}} = c^{2}{\partial^{2}u\over \partial x^{2}},$$ (2.1)

$${\partial u\over \partial t} = k^{2}{\partial^{2}u\over \partial x^{2}},$$ (2.2)

$${\partial^{2}u\over \partial x^{2}} + {\partial^{2}u\over \partial y^{2}} = 0.$$ (2.3)

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

$${\partial^{2}u\over \partial x^{2}} + {\partial^{2}u\over \partial y^{2}} = f(x,y).$$ (2.4)

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

$$u = x^{2} - y^{2}, \qquad u = e^{x}\cos y, \qquad \ln (x^{2} + y^{2}),$$

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 $t=0$) and the boundary conditions (conditions at specific values of $x$), 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 $u_{1}(x,t)$ and $u_{2}(x,t)$ are both solutions to (2.1), then

$$u = c_{1}u_{1}(x,t) + c_{2}u_{2}(x,t),$$

is also a solution if $c_{1}$ and $c_{2}$ 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. .12Obtain a solution, $u(x,t)$, to the equation

$${\partial^{2}u\over \partial t^{2}} + u = 0.$$

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

$$u(x,t) = A \cos t + B \sin t .$$

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

$$u(x,t) = A(x) \cos t + B(x) \sin t ,$$

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

Example 2. .13Consider the equation

$${\partial ^{2} u\over \partial t\partial x} \equiv {\partia... ...\over \partial x}\right ) = - {\partial u \over \partial x}.$$

To solve this we set $p = \partial u /\partial x$ so that the equation becomes

$${\partial p \over \partial t} = - p.$$

Thus, the solution is

$$p = c(x) e^{-t}.$$

Now we must integrate with respect to $x$ to get the solution $u(x,t)$. Therefore, we get

$$u(x,t) = f(x)e^{-t} + g(t),$$

where $f(x) = \int c(x) dx$ and $g(t)$ is an arbitrary function of integration when we integrate with respect to $x$.