Differentiating Integral with Green's function

In summary, the conversation discusses the difficulty of justifying the step of passing the derivative through the integral in a given expression involving the Modified Bessel function. The conversation mentions the properties of the function and its equivalent representation in the Fourier transform domain. The suggestion of using a polar change of variable and applying theorems on inversion under the integral sign is also mentioned.
  • #1
RUber
Homework Helper
1,687
344
Hello, I am having trouble finding the proper justification for being able to pass the derivative through the integral in the following:
## u(x,y) = \frac{\partial}{\partial y} \int_0^\infty\int_{-\infty}^\infty f(x') K_0( \sqrt{ (x - x')^2 + (y-y')^2 } \, dx' dy' ##
##K_0## is the Modified Bessel function of the second kind with properties:
1. ## K_0(z) \approx - \ln z \text{ as } z \to 0 .##
2. ## \mathop {\lim }\limits_{|z| \to \infty} K_\nu (z) = 0 .##
3. ##K_\nu(z)## is real and positive for ##\nu > -1## and ## z \in \mathbb{R} >0 .##
4. ##K_{-\nu}(z) = K_\nu(z) .##
5. ## \frac{\partial K_\nu (z) }{\partial z } = -\frac 12 \left( K_{\nu -1 }(z) + K_{\nu+1}(z)\right).##

I have seen it done, but my advisor asked me to justify the step.
The problem occurs at the singularity ## x= x', y = y'.##

I also know that ##K_0## has an equivalent representation in the Fourier transform domain of:
##c \int_{-\infty}^\infty \frac { e ^{ - \sqrt{ \xi^2+\alpha^2 } |y- y'|}}{\sqrt{ \xi^2+\alpha^2 }} e^{i (x-x') \xi } d\xi##
for some scaling constant c.
In a few articles, they discuss the properties of pseudo-differential operators...which may be a hint for justifying the derivative.

I am really stumped on this one...any insight would be helpful. I feel like there is something intuitive that I am simply missing.
Thanks.
 
Physics news on Phys.org
  • #2
It looks difficult !
I don't know the answer, but have you tried this: assuming that you've shown integrability of your double integral on ##\mathbb{R}_+\times \mathbb{R}##, would a polar change of variable transform this complicated double integral to a single, parameter dependent integral ? Then, would the usual theorems on inversion under the integral sign apply ?
 

1. What is the purpose of using Green's function in differentiating integrals?

The use of Green's function in differentiating integrals allows for the solution of differential equations by expressing them as an integral equation. This method is particularly useful for solving boundary value problems in physics and engineering.

2. How does Green's function help in solving differential equations?

Green's function acts as a kernel that transforms a differential equation into an integral equation. This integral equation can then be solved using standard integration techniques, making it easier to find solutions to complex differential equations.

3. Can Green's function be used to solve any type of differential equation?

Green's function can be used to solve linear, homogeneous differential equations. However, it may not be applicable to nonlinear or non-homogeneous equations.

4. How is Green's function related to the boundary conditions of a differential equation?

The boundary conditions of a differential equation are used to determine the specific form of Green's function that is needed to solve the equation. Green's function is essentially a solution that satisfies the given boundary conditions.

5. Are there any limitations to using Green's function in differentiating integrals?

Green's function is limited to solving linear, homogeneous differential equations with well-defined boundary conditions. It may not be applicable to all types of differential equations and may not always provide a closed-form solution.

Similar threads

Replies
6
Views
2K
Replies
2
Views
891
Replies
1
Views
2K
Replies
6
Views
908
Replies
5
Views
396
Replies
1
Views
944
Replies
1
Views
1K
Replies
19
Views
3K
Replies
3
Views
1K
Replies
3
Views
655
Back
Top