Using implicit differentiation, verify that $u=f(x-tu)$ satisfies $\frac{\partial u(x,t)}{\partial t}+u(x,t)\frac{\partial u(x,t)}{\partial x}$.

Question

Using implicit differentiation, verify that $u=f(x-tu)$ satisfies $\frac{\partial u(x,t)}{\partial t}+u(x,t)\frac{\partial u(x,t)}{\partial x}$.

Could someone explain how implicit differentiation works with a pde? Thanks!

Answer 1

Differentiating $u=f(x-t\,u)$ first with respect to $t$ and then with respect to $x$ we get: $$ u_t=f'(x-t\,u)(-u-t\,u_t), $$ $$ u_x=f'(x-t\,u)(1-t\,u_x). $$ Multiplying the second equation by $u$ and adding both equations we get $$ u_t+u\,u_x=-t\,f'(x-t\,u)(u_t+u\,u_x)\implies (1+t\,f')(u_t+u\,u_x)=0. $$ If $f'\ge0$ (i.e. $f$ is increasing) then $1+t\,f'>1$ for all $t\ge0$ and $u_t+u\,u_x=0$ follows.

If $f'<0$ at some point, there is a $T_{\text{max}}$ such that $1+t\,f'>0$ on $[0,T_{\text{max}})$,