Checking the solution of a first order pde

Question

I need some help with this exercise.

Given the following pde: $ \begin{cases} u_t + b(u)\cdot u_x=0\\[6pt] u(x, 0) = u_0(x) \end{cases} $

I have to check that its solution is $u(x, t)=u_0(\psi(x,t))$ where $x=b(u_0(\psi(x,t)))\cdot t +\psi(x,t)$.

It should not be difficult, but i'm making a mess with implicit derivation, and getting nowhere. Thanks a lot for any indication.

Answer 1

Differentiating with repect to $x$ and $t$ we get $$1=b'(u_0(\psi(x,t)))\cdot u'_0(\psi(x,t)\cdot \psi_x(x,t)\cdot t +\psi_x(x,t) \\ 0=b'(u_0(\psi(x,t)))\cdot u'_0(\psi(x,t)\cdot \psi_t(x,t)\cdot t+b(u_0(\psi(x,t))) +\psi_t(x,t)$$ so we have $\psi_t(x,t)=-b\psi_x(x,t)$ equation 1
Moreover,

$$u_x(x, t)=u'_0(\psi(x,t))\cdot \psi_x(x,t) \\ u_t(x, t)=u'_0(\psi(x,t))\cdot \psi_t(x,t)$$

Then $$RHS=u'_0(\psi(x,t))\cdot \psi_t(x,t)+b\cdot u'_0(\psi(x,t))\cdot \psi_x(x,t) \\=0$$ using equation 1