What method can be used to solve this pde?

  • Context: Undergrad 
  • Thread starter Thread starter MAGNIBORO
  • Start date Start date
  • Tags Tags
    Method Pde
Click For Summary
SUMMARY

The discussion focuses on solving the integral equation $$I(a,b) = \int_{0}^{\infty} e^{-ax^{2}-\frac{b}{x^2}}$$ using methods applicable to partial differential equations (PDEs). Key techniques mentioned include differentiation under the integral sign, substitution, and integration by parts, leading to the relationships $$I(a,b) = -\sqrt{\frac{b}{a}}\, \left ( \frac{\partial }{\partial b}I(a,b) \right )$$ and $$I(a,b) = -\frac{2a}{1+2\sqrt{ab}} \left ( \frac{\partial }{\partial a} I(a,b)\right ).$$ The discussion also highlights the transformation of PDEs into ordinary differential equations (ODEs) by fixing variables, making the problem more manageable.

PREREQUISITES
  • Understanding of integral equations, specifically $$I(a,b) = \int_{0}^{\infty} e^{-ax^{2}-\frac{b}{x^2}}$$
  • Familiarity with differentiation under the integral sign and integration by parts
  • Knowledge of ordinary differential equations (ODEs) and their solutions
  • Basic concepts of partial differential equations (PDEs) and their properties
NEXT STEPS
  • Research the method of separation of variables for solving PDEs
  • Explore the application of differentiation under the integral sign in solving complex integrals
  • Study the properties and solutions of the heat equation and wave equation
  • Examine software tools for symbolic computation, such as Mathematica or Maple, for solving PDEs
USEFUL FOR

Mathematicians, physicists, and engineers interested in solving complex integral equations and understanding the transition from PDEs to ODEs. This discussion is particularly beneficial for those looking to enhance their problem-solving skills in mathematical analysis.

MAGNIBORO
Messages
106
Reaction score
26
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.
 
Physics news on Phys.org
MAGNIBORO said:
I would like to know what methods use the program to solve the 2 pde
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##
 
haruspex said:
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##
wow, if we suppose "fix variables" we can "transform" pde into ode, very impressive
 
MAGNIBORO said:
wow, if we suppose "fix variables" we can "transform" pde into ode, very impressive
PDEs are only significantly tougher than ODEs when derivatives wrt different independent variables occur in the same equation.
 
  • Like
Likes   Reactions: MAGNIBORO
haruspex said:
PDEs are only significantly tougher than ODEs when derivatives wrt different independent variables occur in the same equation.
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
 
MAGNIBORO said:
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
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.
 
  • Like
Likes   Reactions: MAGNIBORO
One can obtain solutions of the form (Ae^a + Be^b)f(a + b) where f&#039;&#039; + f&#039; - f = 0 so that <br /> 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&#039;&#039; + g&#039; + g = 0 so that <br /> g(z) = Ce^{\frac12 z}\cos\left(\frac{\sqrt{3}}{2}z\right) + De^{\frac12 z}\sin\left(\frac{\sqrt{3}}{2}z\right).
 
pasmith said:
One can obtain solutions of the form (Ae^a + Be^b)f(a + b) where f&#039;&#039; + f&#039; - f = 0 so that <br /> 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&#039;&#039; + g&#039; + g = 0 so that <br /> g(z) = Ce^{\frac12 z}\cos\left(\frac{\sqrt{3}}{2}z\right) + De^{\frac12 z}\sin\left(\frac{\sqrt{3}}{2}z\right).

For what pde are this solutions?
 

Similar threads

  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K