# Method of characteristics and second order PDE.

1. Jan 19, 2014

### peripatein

This isn't a homework question per se. Am merely seeking an explanation how the method of characteristics may be applied to a second order PDE. For instance, how is it used to solve utt=uxx-2ut?

2. Jan 22, 2014

### HallsofIvy

Staff Emeritus
That is a "hyperbolic" equation and, ignoring the lower order, first derivative is $u_{tt}= u_{xx}$ so has "characteristic equation"$t^2= x^2$ or $t= \pm x$ so the "characteristics" are $t- x= constant$ or $t+ x= constant$.

That tells us that we can simplify the equation by taking p= t- x and q= t+ x as variables instead of x and t. By the chain rule, $u_t= u_pp_t+ u_qq_t= u_p+ u_q$ and then $u_{tt}= (u_p+ u_q)_t= (u_p+ u_q)_p+ (u_p+ u_q)_q= u_{pp}+ 2u_{pq}+ u_{qq}$.

Similarly, $u_x= u_pp_x+ u_qq_x= -u_p+ u_q$ and then $u_{xx}= (-u_p+ u_q)_x= -(-u_p+ u_q)_p+ (-u_p+ u_q)_q= u_pp- 2u_{pq}+ u_{qq}$.

So $u_{tt}= u_{xx}+ u_t$ becomes $u_{pp}+ 2u_{pq}+ u_{qq}= u_pp- 2u_{pq}+ u_{qq}+ u_p+ u_q$. The "$u_{pp}$" and "$u_{qq}$" cancel, leaving $4u_{pq}= u_p+ u_q$.