Showing that an equation is a general solution for a PDE

Question

Consider the following PDE:

$$u_{xx}-\frac{1}{x}u_{x}-x^2u_{yy}=0.$$ Show that $u(x,y)=f(2y+x^2)+g(2y-x^2)$ is a general solution.

To show this would I simply differentiate and substitute in $u(x,y)$.

Answer 1

Just in case the solution is not provided and the PDE has to be solved!

This appears to be a second order PDE.

Before we start, we need to know the domain, which in our case is $V\subset\mathbb{R}^2$. In my case, I consider $V=\{(x,y) \in \mathbb{R}^2/ x>0,y>0\}$.

$\bullet$ Firstly, this PDE is in the form: $A \cdot u_{xx}+2B\cdot u_{xy}+C\cdot u_{yy}+\ldots =0$.

So, we have that $A=1,B=0,C=-x^2$.

$\bullet$ We need to find the discriminant $D=B^2-AC=x^2>0\implies$ we have a second order hyperbolic pde.

$\bullet$ We use the method of characteristic curves, so we have that:

$\dfrac {dy} {dx}=\dfrac{B\pm \sqrt D}A=\pm x$.

Thus, we have to solve 2 different ODEs.

$\begin{array}{c|c} \dfrac{dy} {dx} = -x & \dfrac {dy}{dx}=x\\ dy = -x dx & dy = x dx\\ y =-\dfrac {x^2} {2}+c_1 & y = \dfrac {x^2} {2} +c_2\\ \boxed{c_1=\dfrac{x^2}{2}+y }& \boxed{c_2=-\dfrac{x^2}2+y} \end{array}$

$\bullet$ We define the following transform:

$\begin{array}[t]{l} \xi(x,y)=\dfrac{x^2}2+y\\ \eta(x,y)=-\dfrac{x^2}{2}+y \end{array}$

Moreover, we have to check the Jacobian determinant of the previous transform, which is:

$J(\xi,\eta)=\begin{vmatrix} \dfrac{\partial \xi}{\partial x} & \dfrac{\partial \xi}{\partial y} \\ \dfrac { \partial \eta} {\partial x} & \dfrac {\partial \eta}{ \partial y}\end{vmatrix}=\begin{vmatrix} x & 1 \\ -x & 1\end{vmatrix}=2x>0$ (Jacobian determinant needs to be non-zero).

$\bullet$ Applying the chain rule we take:

$\begin{array}[t]{l} u_x=\xi_x\cdot \bar u_{\xi} + \eta_x \cdot \bar u_{\eta}=x\cdot \bar u_{\xi}+x\cdot \bar u_{\eta}\\ u_y=\xi_y\cdot \bar u_{\xi} + \eta_y \cdot \bar u_{\eta}=-\bar u_{\xi}+ \bar u_{\eta} \end{array}$

We apply chain rule once again to obtain $u_{xx},u_{yy}$ (and $u_{xy}$ if needed).

So, we have:

$\begin{array}[t]{l} u_{xx}=\dfrac{\partial u_x}{\partial x}=\ldots=x^2 \bar u_{\xi\xi}+x^2 \bar u_{\eta\eta}+2x^2\bar u_{\xi\eta}+\bar u_{\xi}+\bar u_{\eta}\\ u_{yy}=\ldots= \bar u_{\xi\xi} + \bar u_{\eta\eta} \end{array}$

After all this procedure, we plug $u_{xx},u_{yy}, u_{x}$ in our initial PDE and we take:

$\begin{array}[t]{l} \bigg(x^2 \bar u_{\xi\xi} +x^2 \bar u_{\eta\eta} +2x^2 \bar u_{\xi\eta}+\bar u_\xi +\bar u_\eta\bigg)-\dfrac1 x \bigg(x\bar u_\xi +x\bar u_\eta\bigg)-x^2 \bigg(u_{\xi\xi}+u_{\eta\eta}\bigg)=0\\ 2x^2 \bar u_{\xi\eta}=0\\ \bar u_{\xi\eta}=0\\ \boxed {\bar u(\xi,\eta)=f(\xi)+g(\eta)} \end{array}$

Finally, we have that $\begin{equation}u(x,y)=f\bigg(\dfrac {x^2}{2}+y\bigg)+g\bigg(-\dfrac{x^2}{2}+y\bigg)\end{equation}$, which is equivalent to the solution you provide.