Solve Non-Periodic Signal Equation: (1-\nabla^2)B

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SUMMARY

The discussion focuses on solving the non-periodic signal equation S = (1 - ∇²)B, where S and B are non-periodic functions. The user proposes a method involving the decomposition of S into S₀ and S₋, applying Fourier transforms to derive B. The equation B = F⁻¹(Ŝ₋/(1 + k²)) - S₁ is established as a solution, contingent on S₀ being an analytically integrable function that meets boundary conditions. The challenge lies in ensuring S₀ satisfies the correct values on all borders in two dimensions.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with Fourier transforms and their applications
  • Knowledge of boundary value problems in mathematical physics
  • Experience with numerical methods for solving PDEs
NEXT STEPS
  • Study Fourier transform techniques for solving PDEs
  • Research boundary value problem methods in two dimensions
  • Explore numerical algorithms for non-periodic signal processing
  • Investigate analytical solutions for specific forms of S₀
USEFUL FOR

Mathematicians, physicists, and engineers working with signal processing, particularly those dealing with non-periodic functions and partial differential equations.

Heimdall
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Hi,

I'm looking for an efficient algorithm to solve this kind of equation :

[tex]S = (1-\nabla^2)B[/tex]

where both S(x,y) and B(x,y) can both be non-periodic functions. We know S and want to find out what is B.

I was wondering if there was a 'well known' method to solve this kind or problem in the case where both S and B are non-periodic functions...

I've started to write something...

Let [tex]S=S_0+S_*[/tex] where S0 and S* are non periodic and periodic functions respectively. I take S0 such that I have [tex]S_0 = S[/tex] on the boundaries of my domain, so S* is null there.

you have : [tex]S_* = (1-\nabla^2)(B+S_1)[/tex]

with [tex]-(1-\nabla^2)S_1 = S_0[/tex]

You can Fourier transform and obtain :

[tex]\mathcal{F}\left(S_*\right)= \mathcal{F}\left((1-\nabla^2)(B+S_1)\right) = (1+k^2) \mathcal{F}(B+S_1)[/tex]

so that you can find :

[tex]B = \mathcal{F}^{-1}\left(\frac{\tilde{S_*}}{1+k^2}\right) - S_1[/tex]

you can then have a solution of the problem by finding the analytical easy-to-integrate function S0.

In 1D it seems ok, but in 2D S0 must have the correct values on all borders which seems a bit complicated...
 
Last edited:
Physics news on Phys.org
Looks like ##B=S+c_1\exp(B) +c_2\exp(-B)##.
 

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