Function of first integrals pde

Question

Show that if $F$ and $G$ are first integrals of the characteristic system of $u_{t} + c(x,t,u)u_{x} = g(x,t,u)$ then $\Psi(F(x,t,u),G(x,t,u))=0$ defines the solution $u=u(x,t)$ of this equation. It seems like an easy exercise, but I don't know how to start - can anyone give me a hint? Thanks in advance!

Answer 1

$$0=\frac{\partial \Psi }{\partial x}=\frac{\partial \Psi }{\partial F}\cdot \left( \frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\cdot \frac{\partial u}{\partial x} \right)+\frac{\partial \Psi }{\partial G}\cdot \left( \frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\cdot \frac{\partial u}{\partial x} \right).$$

$$0=\frac{\partial \Psi }{\partial t}=\frac{\partial \Psi }{\partial F}\cdot \left( \frac{\partial F}{\partial t}+\frac{\partial F}{\partial u}\cdot \frac{\partial u}{\partial t} \right)+\frac{\partial \Psi }{\partial G}\cdot \left( \frac{\partial G}{\partial t}+\frac{\partial G}{\partial u}\cdot \frac{\partial u}{\partial t} \right).$$

$$\left[ \begin{matrix} \frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\cdot \frac{\partial u}{\partial x} & \frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\cdot \frac{\partial u}{\partial x} \\ \frac{\partial F}{\partial t}+\frac{\partial F}{\partial u}\cdot \frac{\partial u}{\partial t} & \frac{\partial G}{\partial t}+\frac{\partial G}{\partial u}\cdot \frac{\partial u}{\partial t} \\ \end{matrix} \right]\left[ \begin{matrix} \frac{\partial \Psi }{\partial F} \\ \frac{\partial \Psi }{\partial G} \\ \end{matrix} \right]=0\text{ }.$$
Since $\frac{\partial \Psi }{\partial F}$ and $\frac{\partial \Psi }{\partial G}$ do not vanish, from Cramer’s rule we know
$$\Delta = \left| {\begin{array}{*{20}{c}}{\frac{{\partial F}}{{\partial x}} + \frac{{\partial F}}{{\partial u}} \cdot \frac{{\partial u}}{{\partial x}}}&{\frac{{\partial G}}{{\partial x}} + \frac{{\partial G}}{{\partial u}} \cdot \frac{{\partial u}}{{\partial x}}}\\{\frac{{\partial F}}{{\partial t}} + \frac{{\partial F}}{{\partial u}} \cdot \frac{{\partial u}}{{\partial t}}}&{\frac{{\partial G}}{{\partial t}} + \frac{{\partial G}}{{\partial u}} \cdot \frac{{\partial u}}{{\partial t}}}\end{array}} \right| = 0{\rm{ }}.$$

$$\left( \frac{\partial F}{\partial x}+\frac{\partial F}{\partial u}\cdot \frac{\partial u}{\partial x} \right)\left( \frac{\partial G}{\partial t}+\frac{\partial G}{\partial u}\cdot \frac{\partial u}{\partial t} \right)-\left( \frac{\partial G}{\partial x}+\frac{\partial G}{\partial u}\cdot \frac{\partial u}{\partial x} \right)\left( \frac{\partial F}{\partial t}+\frac{\partial F}{\partial u}\cdot \frac{\partial u}{\partial t} \right)=0\text{ }.$$

$$\left( {\frac{{\partial G}}{{\partial x}}\frac{{\partial F}}{{\partial u}} - \frac{{\partial F}}{{\partial x}}\frac{{\partial G}}{{\partial u}}} \right)\frac{{\partial u}}{{\partial t}} + {\rm{ }}\left( {\frac{{\partial F}}{{\partial t}}\frac{{\partial G}}{{\partial u}} - \frac{{\partial F}}{{\partial u}}\frac{{\partial G}}{{\partial t}}} \right)\frac{{\partial u}}{{\partial x}} = \left( {\frac{{\partial F}}{{\partial x}}\frac{{\partial G}}{{\partial t}} - \frac{{\partial F}}{{\partial t}}\frac{{\partial G}}{{\partial x}}} \right).$$

$${u_t} + c\left( {x,t,u} \right){u_x} = g\left( {x,t,u} \right){\rm{ }}.$$