I've got this PDE:
$$2u^3u_x+ u_y= x,\\ u(x,0) =\sqrt x$$
I've gotten the first steps out the way but I have the problem of the differential integration:
$$\frac{\partial x}{\partial \tau } = 2u^3 \\ \frac{\partial y}{\partial \tau } = 1\\ \frac{\partial u}{\partial \tau } = x$$
Now it's obvious how we solve the second $y = \tau + c_1$ and then the conditions to get $y = \tau$
But the problem arises from using charisteric method on (1) + (3) and trying to somehow fit those together.
If someone could help it would be really appreciated thank you
$$2u^3u_x+ u_y= x,\\ u(x,0) =\sqrt x \quad\text{supposing}\quad x\geq 0$$
You wrote : $$\begin{cases}\frac{\partial x}{\partial \tau } = 2u^3 \\ \frac{\partial y}{\partial \tau } = 1\\ \frac{\partial u}{\partial \tau } = x \end{cases}\quad\text{or equivalently to the Charpit-Lagrange form :}\quad \frac{dx}{2u^3}=\frac{dy}{1}=\frac{du}{x}=d\tau$$ A first characteristic equation comes from solving $\frac{dx}{2u^3}=\frac{du}{x}$ : $$u^4-x^2=c_1$$ Before looking for a second characteristic equation one observe that the specified condition $u(x,0)=\sqrt{x}$ is defined on a characteristic curve $u^4-x^2=c_1=0$ since $u^4=x^2$. It is wellknown that if the condition is specified and satisfied on a characteristic curve the solution is not unique : they are an infinity of solutions. Among them the particular solution : $$\big(u(x,y)\big)^4=x^2\quad\implies\quad \boxed{u(x,y)=\sqrt{x}}$$ satisfies both the PDE and the condition. This is a trivial solution of the problem.
Fortunately in the wording of the problem it is not asked to find all the solutions. In fact this would be difficult because a second characteristic equation involves a special function (Elliptic integral of first kind). The final result would includes an arbitrary function, thus leading to an infinity many solutions which all satisfies the condition $u(x,0)=\sqrt{x}$.