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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,
\begin{displaymath}
c{\partial u\over \partial x} + {\partial u \over \partial t} = 0,
\end{displaymath} (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 $t$.
    \includegraphics [scale=0.7]{figpde1.ps}


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


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


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

    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 $t$.
    \includegraphics [scale=0.7]{figpde2.ps}

  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

    \begin{displaymath}
{\partial u \over \partial x} = F^{\prime}(z) . 1, \qquad {\partial
u \over \partial t} = F^{\prime}(z) . (-c).
\end{displaymath}

    Therefore,

    \begin{displaymath}
c{\partial u\over \partial x} + {\partial u \over \partial t} = 0,
\end{displaymath}

    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

    \begin{displaymath}
u(x,t) = F(x - ct) \qquad \Rightarrow \qquad u(x,0) = F(x).
\end{displaymath}

    Hence, the unknown function, $F$ is the prescribed function, $f$ but the argument is replaced by $x - ct$. Hence, the solution is

    \begin{displaymath}
u(x,t) = f(x - ct).
\end{displaymath}

    As an illustration, if $f(x) = e^{-x^{2}}$, then

    \begin{displaymath}
u(x,t) = e^{-(x -ct)^{2}},
\end{displaymath}

    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

    \begin{displaymath}
c{\partial u \over \partial x} + {\partial u \over \partial t} = - u.
\end{displaymath}

    The solution is $u = e^{-t}F(x - ct)$, where $F$ is an arbitrary function.

    \begin{displaymath}
{\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).
\end{displaymath}

    Hence,

    \begin{displaymath}
{\partial u \over \partial t} = - u - c{\partial u \over \partial x}
\end{displaymath}

    and the equation is satisfied.

  2. Consider the linear equation without non-constant coefficients,

    \begin{displaymath}
x{\partial u \over \partial x} + {\partial u \over \partial t} = 0,
\qquad x>0.
\end{displaymath}

    The solution is given by $u = F(x e^{-t})$, where $F$ is again an arbitrary function but this time $z = x e^{-t}$.

    \begin{displaymath}
{\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}),
\end{displaymath}

    and so the equation is clearly satisfied.
How do we decide on the argument of the arbitrary functions?
next up previous
Next: Method of Characteristics Up: Partial Differential Equations of Previous: Revision of Partial Differentiation
Prof. Alan Hood
2000-10-30