Method of Characteristics : Other Examples

Example 1. .10

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

with the initial condition

$$u(x, 0) = x,$$

The characteristic equations are

$${dt\over dr} = 1, \qquad \Rightarrow \qquad t = r,$$

and

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

and rearranging we obtain

$$x = A e^{t^{2}} \qquad \Rightarrow \qquad x = x_{0}e^{t^{2}}.$$

Thus,

$$x_{0} = x e^{-t^{2}}.$$

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

$${du\over dr} = u \qquad \Rightarrow \qquad u = A e^{r},$$

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

$$u = F(x_{0}) e^{t},$$

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

$$u(x,t) = x_{0}e^{t} = x e^{-t^{2}}e^{t} = x e^{t - t^{2}}.$$

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

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

with

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

Characteristic equations are

$${dx\over dr} = u^{2}$$ (1.17)

and

$${dt\over dr} = 1, \qquad \Rightarrow \qquad t = r,$$

and (1.16) gives

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

Hence,

$$x = u^{2}r + x_{0} = u^{2}t + x_{0},$$

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

$$x_{0} = x - u^{2}t.$$

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

$$u(x,t) = F(x - u^{2}t),$$

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

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

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

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

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

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

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

Therefore,

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