Verification of Solutions to PDEs

In this section we will verify that certain functions are solutions to first order PDEs. Consider the first order wave equation,

$$c{\partial u\over \partial x} + {\partial u \over \partial t} = 0,$$ (1.2)

where $c$ is a constant called the wave speed. We now verify that the following are all solutions to Equation (1.2).

Example 1. .5

1. $$\begin{eqnarray*} u & = & \sin (x - ct), \\ {\partial u \over \partial x} & =... ...rtial u\over \partial x} + {\partial u\over \partial t} & = & 0. \end{eqnarray*}$$
   
   
   
   
   
   The solution is shown in Figure 1.1 at various values of $ct$. It clearly demonstrates that the initial shape of the function is maintained and simply propagates to the right.

   Figure 1.1: The solution, $u(x,t)$ as a function of $x$ at various values of $t$.

   

2. 
   

   $$u = \left\{ \begin{array}{cc} 1 - (x - ct)^{2} & 0 \leq (x - ct)^{2} \leq 1, \\ 0 & (x - ct)^{2} \geq 1, \\ \end{array} \right.$$

   
   
   
   

   $${\partial u \over \partial x} = \left\{ \begin{array}{cc} -2(x - ... ... \leq (x - ct)^{2} \leq 1, \\ 0 & (x - ct)^{2} \geq 1, \\ \end{array} \right.$$

   
   
   
   

   $${\partial u \over \partial t} = \left\{ \begin{array}{cc} 2c(x - ... ... \leq (x - ct)^{2} \leq 1, \\ 0 & (x - ct)^{2} \geq 1, \\ \end{array} \right.$$

   
   
   Hence, the equation is satisfied. Note that $u$ is continuous but the partial derivatives are not continuous. The solution is shown in Figure 1.2 for various values of $ct$.

   Figure 1.2: The solution, $u(x,t)$ as a function of $x$ at various values of $t$.

   

3. $u = F(x - ct)$ where $F$ is an arbitrary function. It is easier to understand what is happening if we define $z = x - ct$, so that $u = F(z)$. Hence, the partial derivatives, on using the chain rule, are
   
   $${\partial u \over \partial x} = F^{\prime}(z) . 1, \qquad {\partial u \over \partial t} = F^{\prime}(z) . (-c).$$
   
   
   Therefore,
   
   $$c{\partial u\over \partial x} + {\partial u \over \partial t} = 0,$$
   
   
   for any functional form for $F$. This illustrates an important point. Ordinary differential equations have arbitrary constants but Partial differential equations have arbitrary functions. The function $F$ is determined by an initial condition.

4. Assume that the initial condition, at $t=0$, is $u(x,0) = f(x)$, where $f$ is a prescribed function. Now if
   
   $$u(x,t) = F(x - ct) \qquad \Rightarrow \qquad u(x,0) = F(x).$$
   
   
   Hence, the unknown function, $F$ is the prescribed function, $f$ but the argument is replaced by $x - ct$. Hence, the solution is
   
   $$u(x,t) = f(x - ct).$$
   
   
   As an illustration, if $f(x) = e^{-x^{2}}$, then
   
   $$u(x,t) = e^{-(x -ct)^{2}},$$
   
   
   satisfies Equation (1.2) and the initial condition.

The next example will verify the solution to other equations, in terms of an arbitrary function with a specific argument.

Example 1. .6

1. Consider the linear, equation
   
   $$c{\partial u \over \partial x} + {\partial u \over \partial t} = - u.$$
   
   
   The solution is $u = e^{-t}F(x - ct)$, where $F$ is an arbitrary function.
   
   $${\partial u \over \partial x} = e^{-t}F^{\prime}(x - ct), \... ...partial t} = - e^{-t}F(x - ct) - c e^{-t}F^{\prime}(x -ct).$$
   
   
   Hence,
   
   $${\partial u \over \partial t} = - u - c{\partial u \over \partial x}$$
   
   
   and the equation is satisfied.

2. Consider the linear equation without non-constant coefficients,
   
   $$x{\partial u \over \partial x} + {\partial u \over \partial t} = 0, \qquad x>0.$$
   
   
   The solution is given by $u = F(x e^{-t})$, where $F$ is again an arbitrary function but this time $z = x e^{-t}$.
   
   $${\partial u \over \partial x} = e^{-t}F^{\prime}(x e^{-t}),... ...rtial u \over \partial t} = - x e^{-t}F^{\prime}(x e^{-t}),$$
   
   
   and so the equation is clearly satisfied.

How do we decide on the argument of the arbitrary functions?