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Consider $$u_{tt} - u_{xx} - u = 0$$ with initial condition $u(x, t=0) = c$, $c \in \mathbb{R}.$

How do you solve the problem using characteristics, without separation of variables?

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  • $\begingroup$ en.wikipedia.org/wiki/Wave_equation#Algebraic_approach $\endgroup$ – polfosol Oct 15 '16 at 12:18
  • $\begingroup$ I gave it some thought and it seems my approach didn't get me anywhere. Sorry -_- $\endgroup$ – polfosol Oct 15 '16 at 17:19
  • $\begingroup$ Substitute $u = f(x - ct) = f(z)$ into your PDE, multiply both sides by $f'$ and integrate. The integration constant can be set to zero if you require your wave to be bounded away from the origin. That is, $f, f' \to 0$ as $z \to \infty$ (You haven't set any boundary conditions, so I'm applying some myself that are physically reasonable). You then have a separable ODE to solve. Also note conditions on the magnitude of your wave speed. $\endgroup$ – mattos Oct 16 '16 at 1:19
  • $\begingroup$ @Mattos: I'm not sure how your substitution works; can you add more details (in an answer)? $\endgroup$ – user361459 Oct 16 '16 at 7:34
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Approach $1$: power series method

Similar to PDE - solution with power series:

$u_{tt}-u_{xx}-u=0$

$u_{tt}=u+u_{xx}$

Consider $u_t(x,0)=f(x)$ ,

Let $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{t^n}{n!}\dfrac{\partial^nu(x,0)}{\partial t^n}$ ,

Then $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{t^{2n}}{(2n)!}\dfrac{\partial^{2n}u(x,0)}{\partial t^{2n}}+\sum\limits_{n=0}^\infty\dfrac{t^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(x,0)}{\partial t^{2n+1}}$

$u_{tttt}=u_{tt}+u_{xxtt}=u+u_{xx}+u_{xx}+u_{xxxx}=u+2u_{xx}+u_{xxxx}$

Similarly, $\dfrac{\partial^{2n}u}{\partial t^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}u}{\partial x^{2k}}$

$u_{ttt}=u_t+u_{xxt}$

$u_{ttttt}=u_{ttt}+u_{xxttt}=u_t+u_{xxt}+u_{xxt}+u_{xxxxt}=u_t+2u_{xxt}+u_{xxxxt}$

Similarly, $\dfrac{\partial^{2n+1}u}{\partial t^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}u}{\partial x^{2k}\partial t}$

$\therefore u(x,t)=\sum\limits_{n=0}^\infty\dfrac{ct^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(x)t^{2n+1}}{(2n+1)!}=c\cosh t+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(x)t^{2n+1}}{(2n+1)!}$

Approach $2$:

Let $\begin{cases}p=\dfrac{t+x}{4}\\q=\dfrac{t-x}{4}\end{cases}$ ,

Then $u_t=u_pp_t+u_qq_t=\dfrac{u_p+u_q}{4}$

$u_x=u_pp_x+u_qq_x=\dfrac{u_p-u_q}{4}$

$u_{tt}=\left(\dfrac{u_p+u_q}{4}\right)_t=\left(\dfrac{u_p+u_q}{4}\right)_pp_t+\left(\dfrac{u_p+u_q}{4}\right)_qq_t=\dfrac{u_{pp}+u_{pq}}{4}+\dfrac{u_{pq}+u_{qq}}{4}=\dfrac{u_{pp}+2u_{pq}+u_{qq}}{4}$

$u_{xx}=\left(\dfrac{u_p-u_q}{4}\right)_x=\left(\dfrac{u_p-u_q}{4}\right)_pp_x+\left(\dfrac{u_p-u_q}{4}\right)_qq_x=\dfrac{u_{pp}-u_{pq}}{4}-\dfrac{u_{pq}-u_{qq}}{4}=\dfrac{u_{pp}-2u_{pq}+u_{qq}}{4}$

$\therefore\dfrac{u_{pp}+2u_{pq}+u_{qq}}{4}-\dfrac{u_{pp}-2u_{pq}+u_{qq}}{4}-u=0$

$u_{pq}-u=0$

According to Method of charactersitics and second order PDE.,

$u(p,q)=\int_0^pf(s)I_0\left(2\sqrt{q(p-s)}\right)~ds+\int_0^qg(s)I_0\left(2\sqrt{p(q-s)}\right)~ds$

$u(x,t)=\int_0^\frac{t+x}{4}f(s)I_0\left(2\sqrt{\dfrac{t-x}{4}\left(\dfrac{t+x}{4}-s\right)}\right)~ds+\int_0^\frac{t-x}{4}g(s)I_0\left(2\sqrt{\dfrac{t+x}{4}\left(\dfrac{t-x}{4}-s\right)}\right)~ds$

$u(x,t)=\int_0^\frac{t+x}{4}f(s)I_0\left(\dfrac{\sqrt{(t-x)(t+x-4s)}}{2}\right)~ds+\int_0^\frac{t-x}{4}g(s)I_0\left(\dfrac{\sqrt{(t+x)(t-x-4s)}}{2}\right)~ds$

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