Solving for the Unknown Integral Kernel?

In summary: So, in the discrete case, the number of solutions is (infinitely many). However, in the continuous case, there is only one solution. This is a conclusion I have already come to. However, I do not understand your argument. How can you say h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int h(\mathbf{v}) d\mathbf{v}}{\int d\mathbf{v}} implies \int h'(\mathbf{v}) d\mathbf{v}= 0? Certainly this is true when h(\mathbf{v}) is a constant but how can you say this is so when h(\math
  • #1
vmw
4
0
Consider,

[tex] f(\mathbf{w}) = \int K(\mathbf{w,\mathbf{v}}) g(\mathbf{v}) d\mathbf{v} [/tex]

where [itex]\mathbf{v},\mathbf{w} \in \mathbb{R^3}[/itex].

Is it possible to solve for the integral kernel, [itex] K(\mathbf{w,\mathbf{v}}) [/itex], if [itex] f(\mathbf{w}) [/itex] and [itex] g(\mathbf{v}) [/itex], are known scalar functions and we require [itex] \int K(\mathbf{w,\mathbf{v}}) d\mathbf{v} = 1 [/itex]? These are definite integrals: [itex]\int \rightarrow \int_{a1}^{b1}\int_{a2}^{b2}\int_{a3}^{b3}[/itex]

Thank you for any solution/advice/insight!
 
Last edited:
Physics news on Phys.org
  • #2
Hi vmw,
Suppose you have a K'(w,v) which satisfies the =1 integral constraint. Consider any function h(v), and define h'(v) = h(v)-∫h(v).dv/∫dv. So ∫h'(v).dv = 0.
If it happens that ∫h'(v)g(v).dv is nonzero then set K = K' + h'(v).j(w) where
j(w) = (f(w) - ∫K'(w,v)g(v).dv)/∫h'(v)g(v).dv.
 
Last edited:
  • #3
It seems like you are making an argument that there are infinitely many [itex]K(\mathbf{w},\mathbf{v}) [/itex] so one cannot solve for a unique kernel. This is a conclusion I have already come to. However, I do not understand your argument. How can you say [itex]h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int h(\mathbf{v}) d\mathbf{v}}{\int d\mathbf{v}}[/itex] implies [itex]\int h'(\mathbf{v}) d\mathbf{v}= 0[/itex]? Certainly this is true when [itex]h(\mathbf{v})[/itex] is a constant but how can you say this is so when [itex]h(\mathbf{v})[/itex] is any function of [itex]\mathbf{v}[/itex]?
 
  • #4
vmw said:
It seems like you are making an argument that there are infinitely many [itex]K(\mathbf{w},\mathbf{v}) [/itex] so one cannot solve for a unique kernel. This is a conclusion I have already come to. However, I do not understand your argument. How can you say [itex]h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int h(\mathbf{v}) d\mathbf{v}}{\int d\mathbf{v}}[/itex] implies [itex]\int h'(\mathbf{v}) d\mathbf{v}= 0[/itex]? Certainly this is true when [itex]h(\mathbf{v})[/itex] is a constant but how can you say this is so when [itex]h(\mathbf{v})[/itex] is any function of [itex]\mathbf{v}[/itex]?
Maybe it would have been clearer if I'd written
[itex]h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int h(\mathbf{x}) d\mathbf{x}}{\int d\mathbf{x}}[/itex]
 
  • #5
I still can't see why defining [itex] h'(\mathbf{v}) = h(\mathbf{v}) + \frac{\int h(\mathbf{x}) d \mathbf{x}}{\int d \mathbf{x}}[/itex] implies [itex] \int h'(\mathbf{v}) d \mathbf{v} = 0 [/itex]. I am probably just being dense.

Instead of getting caught up in the details... Is it accurate to say you are showing me that given a kernel K' that satisfies my condition, you can construct a different kernel K from K' that still satisfies my conditions. Thus K' is not unique?
 
  • #6
vmw said:
I still can't see why defining [itex] h'(\mathbf{v}) = h(\mathbf{v}) + \frac{\int h(\mathbf{x}) d \mathbf{x}}{\int d \mathbf{x}}[/itex] implies [itex] \int h'(\mathbf{v}) d \mathbf{v} = 0 [/itex].
It's -, not +:
Write [itex]c = \frac{\int h(\mathbf{x}) d \mathbf{x}}{\int d \mathbf{x}}= \frac{\int h(\mathbf{v}) d \mathbf{v}}{\int d \mathbf{v}}[/itex]
[itex] \int h'(\mathbf{v}) d \mathbf{v} = \int h(\mathbf{v}) d \mathbf{v} - \int c.d \mathbf{v} = c \int d \mathbf{v} - \int c.d \mathbf{v}= 0[/itex].
Instead of getting caught up in the details... Is it accurate to say you are showing me that given a kernel K' that satisfies my condition, you can construct a different kernel K from K' that still satisfies my conditions. Thus K' is not unique?
Not quite. I'm saying that given a K' that satisfies one criterion (the = 1 criterion), it's not hard to construct one that satisfies both. There is the degenrate case where g is a constant. In this case there is no solution unless f happens to be the same (the integral of h'g will always be 0). It would be interesting to prove that an h' can be found which makes h'g integral nonzero when g is not (almost everywhere) constant.
 
Last edited:
  • #7
I see. I was, indeed, being dense. You make an interesting point. Thank you.

Ultimately, I am now uninterested in the problem since I realized I cannot solve for a unique K. If you consider the discrete case of my problem where f and g are length-N vectors, there are infinitely many matrices, K, which satisfy the two relationships I have proposed. The continuous cause can be (not-so-rigorously) thought of as the case where N->infinity.
 

1. What is an unknown integral kernel?

An unknown integral kernel is a mathematical function that represents the relationship between an unknown variable and a known variable in an integral equation. It is typically denoted by a capital letter, such as K(x,y), and is used to solve problems in various fields of science and engineering.

2. How is an unknown integral kernel determined?

The determination of an unknown integral kernel involves solving the integral equation in which it appears. This can be done analytically or numerically, depending on the complexity of the equation and the available tools. In some cases, the unknown kernel may also be estimated through experimental data or simulations.

3. What is the importance of unknown integral kernels?

Unknown integral kernels play a crucial role in solving various types of integral equations, which are widely used in physics, engineering, and other sciences. They also have applications in data analysis, image processing, and signal processing. By understanding and accurately determining these kernels, scientists can better understand and model complex physical systems.

4. What are some common techniques for solving unknown integral kernels?

Some common techniques for solving unknown integral kernels include the method of separation of variables, the method of Laplace transforms, and the method of Fourier transforms. These techniques involve transforming the integral equation into a simpler form, which can then be solved using standard mathematical tools or numerical methods.

5. Are there any challenges associated with determining unknown integral kernels?

Yes, there can be several challenges associated with determining unknown integral kernels. These may include the nonlinearity of the integral equation, the complexity of the kernel itself, and the need for accurate and reliable numerical methods to solve the equation. Additionally, experimental data may be limited or noisy, making it difficult to determine the unknown kernel accurately.

Similar threads

Replies
4
Views
143
  • Calculus
Replies
1
Views
901
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
Replies
2
Views
544
  • Advanced Physics Homework Help
Replies
26
Views
3K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
2
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
14
Views
1K
Back
Top