how to solve this PDE How to find general solution of the PDEs
$$\frac{∂^2u}{∂x^2}-\frac{∂^2u}{∂y^2}=x^2y^y$$
the problem is the term $y^y$ in the equation. May i solve it by transforming into the canonical form? I have tried but it lead to a complicated equation. Please experts help! many thanks
so i think we can just write the general solution in form:
$$u(x,y) =  \xi(y+x) + \eta(y-x) + \int x^2y^yd(y+x)d(y-x)$$  ?? 
 A: There are several approaches that can solve this inhomogeneous linear ODE.
Approach $1$: classical variables transformations
Let $\begin{cases}p=x+y\\q=x-y\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial y}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\therefore\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}-\left(\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}\right)=\left(\dfrac{p+q}{2}\right)^2\left(\dfrac{p-q}{2}\right)^{\frac{p-q}{2}}$
$4\dfrac{\partial^2u}{\partial pq}=\dfrac{(p+q)^2(p-q)^{\frac{p-q}{2}}}{4\times2^{\frac{p-q}{2}}}$
$\dfrac{\partial^2u}{\partial pq}=\dfrac{(p+q)^2(p-q)^{\frac{p-q}{2}}}{16\times2^{\frac{p-q}{2}}}$
$u(p,q)=f(p)+g(q)+\dfrac{1}{16}\int_b^q\int_a^p\dfrac{(s+t)^2(s-t)^{\frac{s-t}{2}}}{2^{\frac{s-t}{2}}}ds~dt$
$u(x,y)=f(x+y)+g(x-y)+\dfrac{1}{16}\int_b^{x-y}\int_a^{x+y}\dfrac{(s+t)^2(s-t)^{\frac{s-t}{2}}}{2^{\frac{s-t}{2}}}ds~dt$
Approach $2$: Duhamel's principle
With reference to http://en.wikipedia.org/wiki/Duhamel%27s_principle#Wave_equation and http://en.wikipedia.org/wiki/Wave_equation#Inhomogeneous_wave_equation_in_one_dimension, we have $u(x,y)=f(x+y)+g(x-y)+\dfrac{1}{2}\int_0^x\int_{y-x-s}^{y+x-s}s^2t^t~dt~ds$ or $u(x,y)=f(x+y)+g(x-y)-\dfrac{1}{2}\int_0^y\int_{x-y-t}^{x+y-t}s^2t^t~ds~dt$
Approach $3$: See achille hui's answer
A: I don't know how to solve this in a mathematical elegant manner. Here is my sort of ugly solution.
Since the R.H.S of the PDE consists of a single power in $x$, I will rewrite $u(x,y)$ as a former power series in $x$ with functions of $y$ as coefficients:
$$ u(x,y) = \sum_{n=0}^{\infty} a_n(y) x^n$$
and looks for simplification. The PDE becomes:
$$ \sum_{n=0}^{\infty} \left((n+1)(n+2) a_{n+2}(y) - a_{n}''(y) \right) x^n = y^y x^2$$
For the coefficients of even $n$, we have:
$$\begin{align}
2\;a_2(y) - a_0''(y)  &= 0\\
12\;a_4(y) - a_2''(y) &= y^y\tag{*}\\
30\;a_6(y) - a_4''(y) &= 0\\
&\;\vdots
\end{align}$$
Since the PDE is linear, it just suffices for us to find one particular solution for the inhomogeneous case. Looking at $(*)$, the simplest way to achieve this is set $a_n(y) = 0$ for all $n \ne 0, 2$. Let $F$ as the $4^{th}$ antiderivative of $y^y$, i.e.
$$F(y) = \int^{y} dp \int^{p} dq \int^{q} dr \int^{r} ds\;s^s$$
The equation $(*)$ suggest:
$$u(x,y) = -\left(2 F(y) +x^2 F''(y)\right) + \xi(x+y) + \chi(x-y)$$
where $\xi(\cdot), \chi(\cdot)$ are arbitrary $C^2$ functions will be a solution for the PDE. Substitute this back into the PDE, this is indeed the case.
