I What method can be used to solve this pde?

1. Apr 4, 2017

MAGNIBORO

hi, i know a little bit of ODE but not much about PDE,Some math programs give me the solution but I would like to know what methods they use.

The problem is the following:
$$I(a,b) = \int_{0}^{\infty} e^{-ax^{2}-\frac{b}{x^2}}$$

through differentiation under the integral sign, substitution and integration by parts, we can find this properties.

$$I(a,b) = -\sqrt{\frac{b}{a}}\, \left ( \frac{\partial }{\partial b}I(a,b) \right )=-\frac{2a}{1+2\sqrt{ab}} \left ( \frac{\partial }{\partial a} I(a,b)\right )$$

and the condition

$$I(a,0) = \frac{1}{2}\sqrt{\frac{\pi }{a}}$$

then using a softfware:

$$I(a,b) = -\sqrt{\frac{b}{a}}\, \left ( \frac{\partial }{\partial b}I(a,b) \right )$$
$$I(a,b) = f(a)\, e^{-2\sqrt{ab}}$$

now with the other equation

$$I(a,b) = -\frac{2a}{1+2\sqrt{ab}} \left ( \frac{\partial }{\partial a} I(a,b)\right )$$
$$I(a,b) = g(b)\, \frac{e^{-2\sqrt{ab}}}{\sqrt{a}}$$

comparing the 2 equations and considering the condition I(a,0) we get

$$I(a,b) = \frac{\sqrt{\pi}}{2} \frac{e^{-2\sqrt{ab}}}{\sqrt{a}}$$

To fully understand the development, I would like to know what methods use the program to solve the 2 pde
thanks.

2. Apr 4, 2017

haruspex

Having got to this:
$I(a,b) = -\sqrt{\frac{b}{a}}\, \left ( \frac{\partial }{\partial b}I(a,b) \right )$
we can treat it as an ODE in Ia(b).
$I_a=-I_a'\sqrt{\frac ba}$
$\frac{dI_a}{I_a}=-\sqrt{\frac ab}db$

3. Apr 4, 2017

MAGNIBORO

wow, if we suppose "fix variables" we can "transform" pde into ode, very impressive

4. Apr 4, 2017

haruspex

PDEs are only significantly tougher than ODEs when derivatives wrt different independent variables occur in the same equation.

5. Apr 5, 2017

MAGNIBORO

ok, I find this quite useful To solve the "camouflaged pde".
the function $I(a,b)$ also satisfies the equation
$$I(a,b) = \frac{\partial^2 }{\partial a\, \partial b} I(a,b)$$
If I had tried to solve the problem by this equation, It would be a difficult problem or there is some method to solve it?
thanks

6. Apr 5, 2017

haruspex

I think there should be a minus sign in there, but that's beside the point.
Although the equation is true, it is only one equation instead of two, so may well have extra solutions. Indeed, it is symmetric in a and b, whereas your original equation pair is not, and neither is the solution you already have.

I tried separation of variables... I did get solutions like $k_ne^{\lambda_n a-b/\lambda_n}$, but it is not obvious how a sum of those with different parameters would recreate your original solution.

7. Apr 8, 2017

pasmith

One can obtain solutions of the form $(Ae^a + Be^b)f(a + b)$ where $f'' + f' - f = 0$ so that $$f(z) = Ce^{\frac12 z}\cosh\left(\frac{\sqrt{5}}{2}z\right) + De^{\frac12 z}\sinh\left(\frac{\sqrt{5}}{2}z\right)$$
or $(Ae^a + Be^b)g(a-b)$ where $g'' + g' + g = 0$ so that $$g(z) = Ce^{\frac12 z}\cos\left(\frac{\sqrt{3}}{2}z\right) + De^{\frac12 z}\sin\left(\frac{\sqrt{3}}{2}z\right).$$

8. Apr 8, 2017

MAGNIBORO

For what pde are this solutions?