I keep getting stuck and have a hard time understanding my professor, so I'm hoping to get some help here. The question is: Solve the partial diff/eq:
${\partial u}\over{\partial t}$ + $c {{\partial u}\over{\partial x}}$ = $e^{2x}$
of an unknown function of two variables $u=u(t, x)$ where $c \in \mathbb{R}$ is a fixed parameter, and with the additional initial condition $u(0, x) = f(x)$, where $f$ is a given function.
What I have so far:
1st Step: Characteristic Curves
$s \to (t(s), x(s))$
${d \over ds} u(t(s), x(s)) = t'(s){\partial u \over \partial t} + x'(s){\partial u \over \partial x}$
Char. curves:
\begin{cases}
t'(s) = 1 \\
x'(s) = c \\
\end{cases}
so:
\begin{cases}
t(s) = s+d_1 \\
x(s) = cs + d_2 \\
\end{cases}
the characteristics eq's thus become:
\begin{cases}
t(s) = s' \\
x(s) = cs + d \\
\end{cases}
2nd Step:
As long as everything above is correct, this is now where I'm getting stuck: Solve for $u$ along the curve:
$z(s) = u(t(s), x(s))$
$z'(s) + z(s) = e^{2x}$
and then I'm lost…any help would be great…
Best Answer
You can solve for $u$ in a simpler fashion as follows. Since your PDE is linear, it must hold:
$$\frac{dt}{1} = \frac{dx}{c} = \frac{du}{e^{2x}},$$
which gives you the characteristics:
$$ \begin{align} dx - cdt = & 0 \iff x- ct = \eta, \\ e^{2x}dx/c = & du \iff u = e^{2x}/(2c) + \xi. \end{align}$$ Since the original equation is a 1st order PDE, then $\xi$ must be a function of $\eta$, $\xi = g(\eta)$ (or conversely, $\eta = h(\xi)$), so your solution is given by (prove that it satisfies the original PDE):
$$u(t,x) = e^{2x}/(2c) + g(x-ct).$$
Apply now the initial condition to obtain:
$$u(0,x) = f(x) = e^{2x}/(2c) + g(x),$$
so $g(x) = f(x)-e^{2x}/(2c)$, and hence the solution which satisfies the given initial condition is:
$$u(t,x) = e^{2x}/(2c) + f(x-ct) - e^{2(x-ct)}/(2c).$$
Cheers!
Edit
Everything commented above is valid for $c \neq 0$ (non-zero phase velocity). What happens if $c = 0$? Then, the original equation can be rewritten as follows:
$$u_t = e^{2x},$$
so the solution can be easily achieved by integrating "partially" with respect to $t$:
$$u(x,t) = e^{2x} t + h(x),$$
being $h(x)$ an arbitrary function. Apply again the initial condition to obtain $f(x) = h(x)$, so the solution becomes:
$$u(x,t) = e^{2x} t + h(x).$$
(Note that you could have done: $\int^{u(t,x)}_f u_t \, dt = \int^t_0 e^{2x} \, dt$ or write again the characteristics equations, which yields $dx = 0 \Rightarrow x = \eta$).