Solve $ u_{xx}=x^2 u_{yy}$

Question

I am trying to solve the following PDE: $$ u_{xx}=x^2 u_{yy} $$ I thought it is much easier to consider its canonical form, so I used the change of variable and got my new equation:

$$ u_{vw}=\frac{1}{v-w}(u_v-u_w) $$ But even this seems a bit difficult for me. Can any one help me solving my equation.

Thank you in advance.

Answer 1

I want to preface by saying that all of my PDE solving knowledge comes from the physics courses I took, so hopefully someone more knowledgeable can post a more complete answer.

That said, this seems like a good candidate for solving by separation. Consider solutions of the form $$ u(x,y) = X(x) Y(y) $$ Plugging this into the PDE, we find $$ X''(x)Y(y) = x^2 X(x)Y''(y) $$ Divide through by $X(x)Y(y)$. We have $$ \frac{X''(x)}{X(x)} = \frac{x^2 Y''(y)}{Y(y)} $$ Rearranging a little bit, this means $$ \frac{X''(x)}{x^2X(x)} = \frac{Y''(y)}{Y(y)} $$ The left hand side depends only on $x$ and the right hand side depends only on $y$. This is only possible if both sides are equal to some constant $K$. Thus $$ \frac{X''(x)}{x^2X(x)} = K \quad \quad \frac{Y''(y)}{Y(y)} = K $$ This yields the system of ordinary differential equations $X'' = K x^2 X$ and $Y'' = KY$. At this point we need to know the boundary conditions on $u(x,y)$ to determine the solutions to these differential equations. Suppose we knew the boundary conditions and found families of solutions $x_n$ and $y_n$ for the $X$ and $Y$ equations respectively.

For reasons I don't understand (again, someone please supplement this answer or provide a different more complete answer) any solution $u$ of the PDE will be of the form $$ u= \sum_{n,m} a_{mn} x_n y_n $$ where the $a_{mn}$ are constants.

Answer 2

$$u_{xx}=x^2u_{yy}$$ This is the continuation of the Charles Hudgins's answer. $$ \frac{X''(x)}{x^2X(x)} = \frac{Y''(y)}{Y(y)}=K=\lambda^2 $$ With $\lambda$ real or complex. Note that setting $K=\lambda^2$ doesn't reduce the generality insofar $\lambda$ can be complex. This setting is in interest of simpler typing (avoiding square roots).

$Y''(y)-\lambda^2 Y(y)=0\quad$ gives $$ Y=c_1e^{\pm\lambda y}$$ which includes the sinusoidal cases when $\lambda$ is complex.

$X''(x)-\lambda^2x^2X(x)=0\quad$ gives $$X=c_2x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)$$ This involves the Modified Bessel function of the first kind $I_\nu(z)$ . http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html

A family of particular solutions of the PDE is : $$u=C\,x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)e^{\pm\lambda y}$$ Any linear combination of particular solutions is solution of the PDE. Thus on discret form : $$u(x,y)=\sum_{\forall\lambda} C_\lambda\,x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)e^{\pm\lambda y}$$ where $C_\lambda$ are arbitrary constants, to be determined according to some boundary condition.

Or on integral form : $$u(x,y)=\int C(\lambda)\,x^{1/2}I_{\pm 1/4}\left(\frac12\lambda x^2\right)e^{\pm\lambda y}d\lambda$$ where the function $C(\lambda)$ is arbitrary, to be determined according to some boundary condition.

Of course this is very complicated in general. But simplification might occur in some cases of favourable boundary conditions allowing to reduce to a limited number of standard functions. That is why it is very important that the boundary conditions be known.