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How may the method of characteristics be applied to solve a second order PDE? For instance, to solve the equation: $u_{tt}=u_{xx}-2u_t$.

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  • $\begingroup$ Your PDE is very odd as both $u$ and $y$ are dependent variables. $\endgroup$ – doraemonpaul Jan 19 '14 at 18:49
  • $\begingroup$ I am terribly sorry! I just emended it. $\endgroup$ – Grtv Jan 19 '14 at 18:55
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Your equation is not suitable for displayng the techniques called  Method of characteristics. Nevertheless, if you really need this equation, you can construct its general solution. To do this, first substitute $\;u=ve^{-t}$,   then change variables $\;\xi=2(t+x)\;$, $\;\eta=2(t-x)\;$, reducing your equation to the simplest possible form $\;v_{\xi\eta}-v=0\;$. General solution of the latter equation looks like $$v(\xi,\eta)=\!\int\limits_0^{\xi}f(s)I_0\Bigl(2\sqrt{\eta(\xi-s)}\Bigr)\,ds +\!\int\limits_0^{\eta}g(s)I_0\Bigl(2\sqrt{\xi(\eta-s)}\Bigr)\,ds \tag{$\ast$}$$ for arbitrarily given functions $f,g\in C^1$ with notation $\;I_0\;$ standing for the zero order modified Bessel function of the first kind.  To obtain general solution $(\ast)$ apply the Laplace transform $$ \hat{f}(p)\overset{\rm def}{=}\mathcal{L}[f(t)]=\int\limits_0^{\infty}f(t)e^{-pt}dt $$ to the equation $v_{\xi\eta}-v=0$   w.r.t.  any of the two variables, say $\eta$, then integrate the equation $p\hat{v}_{\xi}-\hat{v}=v_{\xi}(\xi,0) \overset{\rm not}{=}f(\xi)$   to find that $$ \hat{v}(\xi,p)=\int\limits_0^{\xi}f(s) \frac{e^{\frac{\xi-s}{p}}}{p}\,ds+ c(p)e^{\frac{\xi}{p}}. $$ Since $\mathcal{L}\bigl[I_0\bigl(2\sqrt{\alpha t}\bigr)\bigr]= \frac{1}{p}e^{\frac{\alpha}{p}}$,  the representation $(\ast)$ readily follows by choosing $c(p)=\hat{g}(p)/p$.   Of course,  general solution $(\ast)$ looks rather complicated, but for the last two and a half centuries, nothing better has yet been found to replace it.

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