Method of characteristics and second order PDE.

[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 u_tt=u_xx-2u_t?

 

[HallsofIvy]

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.