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Is there any method to obtain an analytical approximation (probably through a composition of exponential functions) of the solution of a differential equation?? In other words, is is possible to obtain an approximate Green's function of the PDE?

the equation in question:

$$\nabla ^ 2 \phi(\mathbf{x}) = M \frac {\phi(\mathbf{x}) }{\phi(\mathbf{x}) + C}$$

Where M and C are bounded constants of the same order as $\phi$. Additionally $\phi$ is bounded between 0 and 1. The equation is subject to Dirichlet boundary conditions (although that is not really relevant. Could be Neuman or Robin)

Given the non-linear term on the right hand side, the equation does not have an analytical solution (to my knowledge). The obtention of the solution through finite differences (or other numerical methods that do not provide a function as the solution) should be avoided since the goal is to obtain a function of $\mathbf{x}$ that approximates the solution.

My idea is to find a method similar to how the polinomial expansion of Taylor approximates a function around a point. But for the solution of the PDE.

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    $\begingroup$ I suppose that in the large $C$ limit you could try a perturbation expansion $\phi=\phi_0 +{1\over C}\phi_1 \cdots$ $\endgroup$
    – user619894
    Commented Jan 18, 2021 at 12:00
  • $\begingroup$ As pointed out by @user619894, do you have any information on $M$ or $C$? Are they small or large? $\endgroup$
    – Dmoreno
    Commented Jan 18, 2021 at 14:29
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    $\begingroup$ They are both of the same order as $\phi$. In fact $\phi$ represents the concentration of a specific solute, therefore it is bounded between 0 and 1. C will be close to 0.5 and M will be anywhere between 0 and 1. Thank you for pointing that out. $\endgroup$ Commented Jan 18, 2021 at 16:09
  • $\begingroup$ Is the domain of any nice geometry? $\endgroup$
    – Ian
    Commented Jan 18, 2021 at 16:15
  • $\begingroup$ In any case there are numerous numerical methods that provide a function at each point, for example finite element methods. You can also use finite differences and then perform low-order interpolation of the result you get, which is a kind of "poor man's finite element method". $\endgroup$
    – Ian
    Commented Jan 18, 2021 at 16:16

1 Answer 1

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Here is an outline of a numerical approach you can try.

You can search for a power series representation in combination with the convolution theorem of Fourier analysis.

Rewrite like this $$\nabla^2 \phi({\bf x}) \cdot (\phi({\bf x}) + C) = M \phi({\bf x})$$

  1. Right hand side is M times power series representation of $\phi$.

  2. Left hand side left factor is a set of linear combinations of coefficients for this. This can be easily realized as each monomial term of a multidimensional polynomial differentiates to another. Very sparsely.

  3. Left hand side right factor is just +C on the constant term.

Now the multiplication of power series corresponds to a multidimensional convolution of coefficients. Here is where the multidimensional Fourier transform comes into play. We Fourier transform $\nabla^2 \phi({\bf x})$ and $(\phi({\bf x})+C)$ separately and multiply each coefficient. This should equal the fourier transform of $M\phi(\bf {x})$

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