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Consider a PDE

$$F(x,y,\partial_x f, \partial_y f, \partial_x^2 f, \partial_y^2 f,\partial_{xy} f, ...) = 0$$

for $(x,y)\in \mathbb{R}^2$ and $f:\mathbb{R}^2\rightarrow\mathbb{C}$, and some function $g$ that is not a solution to the PDE. If one plugs $g$ into $F$ then one gets

$$F(x,y,\partial_x g, \partial_y g, ...) = G(x,y)$$

with $G\neq 0$ at least for some $(x,y) \in \mathbb{R}^2$.

Starting from this setup, is it possible to somehow use $G$ in order to compute $h=g+g_{correction}$ such that either $h$ is a solution to $F$ or is closer to a solution than the previous? By being closer to a solution I'm thinking of using a condition like

$$\int\int_{\mathbb{R}^2} |H|^2dS<\int\int_{\mathbb{R}^2} |G|^2dS$$

where $H = F(x,y,\partial_x h, \partial_y h, ...)$.

EDIT


Given @hardmath comment I will also mention the particular case that interests me regarding this problem.

I saw some work in which the authors claimed to have solved the following problem. They start from the Helmholtz equation

$$\nabla^2 f + k^2 f = 0$$ with $k\in \mathbb{R}$ a constant and some known function $g$ that is not a solution to the above mention PDE. Then, by calculus of variations they managed to compute another function $h$ that is a solution to the PDE with the condition that it is "similar" to the original function $g$.

Based on their result I first asked myself if the same approach can be used for

$$\partial_z \psi = \frac{i}{2k} \partial_x^2 \psi$$ which from a physics point of view is a particular case of the Helmholtz equation where $f=\exp(ikz)\psi$. I tried to follow their approach and got to the amazing result that if my known function $g$ is already a solution to my PDE, then $g$ is a solution of the PDE.

Then I got stuck. My attempts at considering a known function based on which to find the closest solution to the PDE all the time returned the same thing: If I already have the solution, then guess what, I have the solution.

  • Well if $u$ solves the original PDE then setting $g_{correction} = u-g$ gives $h=u$, so $h$ solves the PDE, but I suspect this isn’t what you had in mind – JackT Jun 07 '21 at 13:43
  • That is true, but I don't know $u$ or $h$. – Victor Palea Jun 07 '21 at 13:49
  • In principle such a program can be carried out under (strong) assumptions about $F$ that makes successive correction terms smaller and smaller, i.e. a contraction mapping. Without a concrete application it is impossible to be more precise. – hardmath Jun 07 '21 at 14:46
  • @hardmath I have added some background to the question. – Victor Palea Jun 07 '21 at 15:27

0 Answers0