PDE first order solution

Question

Consider the following PDE $x_2 u_{x_1} + u_{x_2}= u$ in $\{x_1>1\} \times\mathbb{R}$.Can someone give me a hint how to solve this equation ?

Answer 1

Let us rewrite your equation as $$ yu_x+u_y=u, $$ which is a bit more nicely-looking. Also, what it seems necessary to assume is either $y > 0,$ or $y < 0.$ Let's consider the case when $y > 0,$ that is, work to obtain the general solution of the PDE on $\mathbf R \times (0,+\infty).$

We apply the method of characteristics for first-order semilinear equations, equations of the form $$ a(x,y)u_x + b(x,y) u_y =c(x,y,u). $$ The main steps are as follows (I'd recommend my book on PDEs to see how to treat first-order semilinear PDEs.)

1) Solve the characteristic ODE $$ y'= \frac{b(x,y)}{a(x,y)} $$ in your case it is the separable ODE $$ y'=\frac 1y \iff T(x,y)=y^2-2x=C $$ where $C$ is an arbitrary constant.

2) Next, according to the general statement of the method the change of variables $$ \begin{cases} s=x,\\ t=T(x,y) \end{cases} $$ in your case $$ \begin{cases} s=x,\\ t=y^2-2x \end{cases} $$ transforms your PDE to $$ a(x,y) w_s =w $$ or to $$ y w_s =w, $$ which in new terms is \begin{equation*} \tag{1} \sqrt{2s+t} w_s =w. \end{equation*} This PDE is reducible to the ODE $$ \sqrt{2s+t} z'(s)=z(s) \iff z(s)= C e^{\sqrt{2s+t}} $$ which means that the general solution of the PDE (1) is $$ w(s,t)= f(t) e^{\sqrt{2s+t}} $$ where $f$ is a continuously differentiable function.

3) Accordingly, the general solution of the original PDE is $$ u(x,y) =//\, f(t) e^{\sqrt{2s+t}}\,// = f(y^2-2x) e^{\sqrt{y^2}} = f(y^2-2x) e^y $$ (rememebering our assumption on $y$).

Answer 2

You can use the so-called method of characteristics. For example you can check:

https://en.wikipedia.org/wiki/Method_of_characteristics

Answer 3

The solution to this problem is given by $u(x,t)=\frac{x_2}{ln(x_1)+1}$