# 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:

$$u(x,y) = \xi(y+x) + \eta(y-x) + \int x^2y^yd(y+x)d(y-x)$$ ??

## 2 Answers

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

• it's great. thank doraemonpaul. I think it's the exact solution!! – Lotusquantum Apr 25 '13 at 4:53

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.

• hi, thank you very much for your great answer. I don't know whether this way is acceptable or not, because it's not a formal way in solving. anyway, many thanks for your effort. – Lotusquantum Apr 23 '13 at 2:04
• but when i substitute u(x,y) into the equation, it does not =0? – Lotusquantum Apr 23 '13 at 2:06
• @Lotusquantum in solving a problem, you first guess the answer and then prove the answer is correct. It didn't matter how you get the function. What matters is whether the final function solves the PDE. What did you get? I does get $0$. – achille hui Apr 23 '13 at 5:48
• thank you achille hui, i will check it again. – Lotusquantum Apr 24 '13 at 3:39