# Second order PDE with mixed derivative

In my study of QFT I've faced the following PDE in Peskin&Schroeder textbook:

$$\frac{\partial^2 g(x,\xi)}{\partial w \partial \xi}=A g(x,\xi),$$

where $w=\ln(1/x)$ and $A$ is a constant.

The book provides a solution only in the limit $w\xi\gg1$ for some reason.

$$g=K(Q^2)exp(\left(4w(\xi-\xi_0)\right)^{1/2}),$$

where $\xi=\ln \ln \left(Q^2\right)$, and $K(Q^2)$ is a function determined from initial condition.

So, my questions are:

1. Why dont they use separation of variables for this PDE?
2. Is there any method to obtain general solution? Or can you explain how they get this approximate form?
• Have you tried to separate the variables for this PDE? My guess is that it may not work and that's why they did not do so. – Paichu May 6 '17 at 22:13

The method of separation of variables could be used as shown below: $$\frac{\partial^2 g(x,\xi)}{\partial w \partial \xi}=A g(x,\xi),$$ where $w=\ln(1/x) \quad\to\quad dw=-\frac{1}{x}dx\quad\to\quad \frac{\partial g}{\partial w}=\frac{\partial g}{\partial x}\frac{dx}{dw}=-x\frac{\partial g}{\partial x}$

$$-x\frac{\partial^2 g(x,\xi)}{\partial x \partial \xi}=Ag(x,\xi),$$ $g=F(x)G(\xi) \quad\to\quad -x\frac{F'}{F}\frac{G'}{G}=A \quad\to\quad \begin{cases} -x\frac{F'}{F}=\lambda \quad\to\quad F=x^{-\lambda} \\ \frac{G'}{G}=\frac{A}{\lambda} \quad\to\quad G=e^{A\xi/\lambda} \end{cases}$

This leads to a family of particular solutions of the PDE : $\quad g_{\lambda}(x,\xi)=C_\lambda e^{A\xi/\lambda}x^{-\lambda}$

Combining all these particular solutions leads to a general form expressing the solution of the PDE: $$g(x,\xi)=\int\Phi(\lambda) e^{A\xi/\lambda}x^{-\lambda}d\lambda =\int\Phi(\lambda) e^{A\xi/\lambda-\lambda\ln{x}} d\lambda$$ $$g(x,\xi)=\int\Phi(\lambda) e^{\frac{A}{\lambda}\xi+\lambda w} d\lambda$$ with any function $\Phi$ insofar the integral be convergent. This function $\Phi$ has to be determined according to the boundary conditions. Generally that is the most difficult part of the job, depending a lot of what are the boundary conditions.

To the question : 1.Why don't they use separation of variables for this PDE? , the answer might be : Because this form of function to express the PDE solution is complicated and not easy to approximate according to the context of the problem.

You wrote : "The book provides a solution only in the limit $w\xi≫1$ for some reason". Moreover, the boundary conditions are not explicitly given. All this makes difficult to derive an approximate from the above general formula. That is probably why a simpler approach (may be for didactic reason) was investigated and then presented in the book.

For $A=1$ ,

According to Method of charactersitics and second order PDE.,

The general solution is $g(x,\xi)=\int_0^xf(s)I_0\left(2\sqrt{\xi(x-s)}\right)~ds+\int_0^\xi g(s)I_0\left(2\sqrt{x(\xi-s)}\right)~ds$

And I think for other values of $A$ , the form of general solutions are similar.

Note that using separation of variables for this PDE cannot work well for deduce its general solution, because:

Let $g(x,\xi)=X(x)\Xi(\xi)$ ,

Then $X'(x)\Xi'(\xi)=AX(x)\Xi(\xi)$

$\dfrac{X'(x)}{X(x)}=\dfrac{A\Xi(\xi)}{\Xi'(\xi)}=s$

$\begin{cases}\dfrac{X'(x)}{X(x)}=s\\\dfrac{A\Xi(\xi)}{\Xi'(\xi)}=s\end{cases}$

$\begin{cases}X(x)=c_1(s)e^{xs}\\\Xi(\xi)=c_2(s)e^\frac{A\xi}{s}\end{cases}$

$\therefore g(x,\xi)=\int_sc(s)e^{xs+\frac{A\xi}{s}}~ds$