Classical solution of Pde's

Question

How to find a classical solution $u=u(x_{1},u_{2})$ to $$(x_{2}^{2}+u)u_{x_{1}} + x_{2}u_{x_{2}}=0$$ with the initial value $u(x_{2}^{2}/2 , x_{2})=0$.

I worked on this problem but as $g=0$ , I don't know how to find the solution. Please guide me.

Answer 1

Supposing that the PDE is : $$\left(y^2+u(x,y)\right)u_x+y\:u_y=0$$ Chapit-Lagrange characteristic ODEs : $$\frac{dx}{y^2+u}=\frac{dy}{y}=\frac{du}{0}$$ See https://en.wikipedia.org/wiki/Method_of_characteristics

A first characteristic equation comes from $0\:dx=du$ $$u=c_1$$ This was obvious at first sight to the PDE that $u(x,y)=$constant is a characteristic equation.

A second characteristic equation comes from solving $\frac{dx}{y^2+c_1}=\frac{dy}{y}$ : $$y^{c_1}e^{\frac12 y^2-x}=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_1=F(c_2)$ is : $$\boxed{u=F\big(y^{u}e^{\frac12 y^2-x}\big)}$$ $F$ is an arbitrary function.

Condition : $u(\frac12 y^2\:,\:y)=0$

Note that $u(x,y)=$constant is a characteristic equation. This is the present case with constant$=0$. As we know if the condition is specified on a characteristic curve generaly there is not an unique solution. So we don't expect a unique solution.

$$u(\frac12 y^2\:,\:y)=0=F\big(y^{0}e^{\frac12 y^2-\frac12 y^2}\big)$$ $$0=F\big(y^{0}e^{\frac12 y^2-\frac12 y^2}\big)$$ $$F(1)=0$$ Thus any function $F(X)$ is convenient if $F(1)=0$

They are an infinity of functions $F(X)$ which are equal to $1$ when $X=1$. Thus they are an infinity of solutions $u(x,y)$.

$$\boxed{u(x,y)=F\big(y^{u(x,y)}e^{\frac12 y^2-x}\big)\qquad \text{with } F(1)=0.}$$

This is an implicit equation. Depending of the kind of function $F$ one can or not solve it analytically for $u$. Choosing some nice functions $F$ allows to express $u(x,y)$ explicitly.