I have the following partial differential equation:
\begin{equation}
\frac{\partial \phi}{\partial x_{1}} \left(\frac{2+x_{3}^{2}}{1+x_{3}^{2}}\right) + \frac{\partial \phi}{\partial x_{3}} = 0
\end{equation}
Whose solution is:
\begin{equation}
\phi(x)=x_{1} – x_{3} – \tan^{-1}(x_{3})
\end{equation}
I've been trying, but I can't figure out what steps were followed to arrive to the solution. Unfortunately, I only have experience solving ordinary differential equations and no experience solving partial differential equations.
Can anyone give me an idea? Any help would be appreciated.
PD: The original problem says that it can be solved by separation of variables.
Best Answer
To facilitate writing let $\quad x_1=x$ and $x_3=y$ . \begin{equation} \frac{\partial \phi}{\partial x} \left(\frac{2+y^{2}}{1+y^{2}}\right) + \frac{\partial \phi}{\partial y} = 0 \end{equation} The Charpit-Lagrange characteristic ODEs are : $$\frac{dx}{\left(\frac{2+y^{2}}{1+y^{2}}\right)}=\frac{dy}{1}=\frac{d\phi}{0}$$ A first characteristic equation comes from solving $\frac{dx}{\left(\frac{2+y^{2}}{1+y^{2}}\right)}=\frac{dy}{1}\quad\implies\quad \int \frac{2+y^{2}}{1+y^{2}}dy=x$+constant. $$y-x+\tan^{-1}(y)=c_1$$ A second characteristic equation comes from $d\phi=0$ : $$\phi=c_2$$ This was obvious since $\phi=$constant is obviously a trivial solution of the PDE.
The general solution of the PDE on form of implicit equation $c_2=F(c_1)$ is : $$\boxed{\phi(x,y)=F\big(y-x+\tan^{-1}(y)\big)}$$ $F$ is an arbitrary function until some valid boundary condition be specified.
CONDITION :
If the condition is $$\phi(0,0)=0,$$ then $0=F\big(0-0+\tan^{-1}(0)\big)$ $$F(0)=0$$ They are infinity many functions which are equal to zero when the argument is zero.
Thus the PDE with condition $\phi(0,0)=0$ as infinity many solutions.
The specified condition $\phi(0,0)=0$ is not sufficient to determine a unique solution. This is consistent with the judicious comment from HK Tan.