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Next: Partial Differential Equations of Up: Method of Characteristics Previous: Initial-boundary-value Problems

Method of Characteristics : Other Examples

Example 1. .10

\begin{displaymath}
2 xt {\partial u \over \partial x} + {\partial u \over \partial t} =
u, \qquad -\infty < x < \infty,
\end{displaymath} (1.15)

with the initial condition

\begin{displaymath}
u(x, 0) = x,
\end{displaymath}

The characteristic equations are

\begin{displaymath}
{dt\over dr} = 1, \qquad \Rightarrow \qquad t = r,
\end{displaymath}

and

\begin{displaymath}
{dx\over dr} = 2xt \qquad \Rightarrow \qquad {dx\over x} = 2t dt
\qquad \Rightarrow \qquad \log x = t^{2} + C,
\end{displaymath}

and rearranging we obtain

\begin{displaymath}
x = A e^{t^{2}} \qquad \Rightarrow \qquad x = x_{0}e^{t^{2}}.
\end{displaymath}

Thus,

\begin{displaymath}
x_{0} = x e^{-t^{2}}.
\end{displaymath}

On using ${du\over dr} = {\partial u \over \partial x}{dx\over dr} +
{\partial u \over \partial t}{dt\over dr}$, (1.15) is now written as

\begin{displaymath}
{du\over dr} = u \qquad \Rightarrow \qquad u = A e^{r},
\end{displaymath}

where the constant $A$ is actually constant along the characteristic defined by the value of $x_{0}$. Thus, $A$ is really a function of $x_{0}$ and we write

\begin{displaymath}
u = F(x_{0}) e^{t},
\end{displaymath}

since $t=r$ and the function $F$ is determined by the initial conditions. At $t=0$, we have $x=x_{0}$ and $u(x_{0},0) = x_{0}$ so that $F(x_{0}) = x_{0}$. Replacing $x_{0}$ by the above expression involving $x$ and $t$, we obtain the final solution

\begin{displaymath}
u(x,t) = x_{0}e^{t} = x e^{-t^{2}}e^{t} = x e^{t - t^{2}}.
\end{displaymath}

Example 1. .11Consider the initial-boundary-value problem

\begin{displaymath}
u^{2}{\partial u \over \partial x} + {\partial u \over \partial t} =
0, \qquad x > 0, t > 0,
\end{displaymath} (1.16)

with

\begin{displaymath}
\begin{array}{cc}
u(x, 0) = \sqrt{x} & x > 0, \\
u(0, t) = 0 & t > 0.
\end{array}\end{displaymath}

Characteristic equations are
\begin{displaymath}
{dx\over dr} = u^{2}
\end{displaymath} (1.17)

and

\begin{displaymath}
{dt\over dr} = 1, \qquad \Rightarrow \qquad t = r,
\end{displaymath}

and (1.16) gives

\begin{displaymath}
{du\over dr} = 0 \qquad \Rightarrow \qquad u = \hbox{constant on
characteristic } = F(x_{0}).
\end{displaymath}

Hence,

\begin{displaymath}
x = u^{2}r + x_{0} = u^{2}t + x_{0},
\end{displaymath}

since $u$ is constant along the characteristic and dx/dr is the derivative of $x$ along the characteristic curve. Therefore, we have

\begin{displaymath}
x_{0} = x - u^{2}t.
\end{displaymath}

Thus, substituting for $x_{0}$ into $F(x_{0})$ we obtain the implicit solution

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

where $F$ is determined by the initial condition, namely, $t=0$, $x=x_{0}$ and $u = \sqrt{x_{0}}$. Thus, $F(x_{0}) = \sqrt{x_{0}}$ and so

\begin{displaymath}
u = \sqrt{x - u^{2}t}, \qquad x - u^{2}t > 0.
\end{displaymath}

Squaring both sides, we can manipulate this solution into an explicit solution for $u$ in terms of $x$ and $t$.

\begin{eqnarray*}
u^{2} & = & x - u^{2}t, \\
u^{2}(1 + t) & = & x, \\
u^{2} & = & {x\over 1 + t},
\end{eqnarray*}



and so the final solution is

\begin{displaymath}
u = \sqrt{{x\over 1+t}}, \qquad x > 0, t > 0.
\end{displaymath}

We can easily check that this is the solution to (1.16) by calculating the partial derivatives,

\begin{displaymath}
{\partial u \over \partial x} = {1\over 2} x^{-1/2}(1 + t)^{-1/2},
\end{displaymath}


\begin{displaymath}
{\partial u \over \partial t} = -{1\over 2}x^{1/2}(1 + t)^{-3/2},
\end{displaymath}

Therefore,

\begin{displaymath}
u^{2}{\partial u\over \partial x} = {x\over 1+t}{1\over 2
...
...x^{1/2}\over (1 + t)^{3/2}} =
-{\partial u \over \partial t}
\end{displaymath}


next up previous
Next: Partial Differential Equations of Up: Method of Characteristics Previous: Initial-boundary-value Problems
Prof. Alan Hood
2000-10-30