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For concreteness, consider the Poisson equation: $$\Delta \phi = f$$ in $\mathbb{R}^3$. $\phi$ is subject to appropriate fall-off conditions at infinity. Suppose that I know an approximate solution to this equation in some bounded domain D, say a ball of radius $R$. By approximate, I would mean here that one is given a power series solution, truncated to some order, to this equation, centered around some point $x_0 \in D$.

Question: Is there a way to extend this known solution to a larger, bounded domain $\tilde{D} \subset D$? This extension is ideally smooth, but it should be at least $C^2$, so that this extension is still an approximate solution to the PDE in a strong sense. This extension should be ideally as accurate as the known solution in $D$. (in the sense that the (norm of the) error is of the same order in $D$ as it is on $\tilde{D} \setminus D$).

A naive thing to do would be to simply consider the analytical extension of the known solution in $D$. That is, take the power series solution to be valid in $\tilde{D}$. (We can assume here that the power series would indeed converge in this larger domain. So, such an extension makes at least mathematical sense). However, this analytical continuation, due to its power series representation, becomes less and less accurate as we go further way from $x_0$. It would therefore be generally less accurate in $\tilde{D} \setminus D$ than in $D$.

For this to work, I would assume one has to invoke the underlying PDE again, but I am not sure how. Of course, one would want to avoid actually solving the full PDE again, and instead somehow make use of the approximate solution we do already have. Has there been some work done that tackle this type of question?

Addressing @Sal remark concerning the boundary data on $D$ and $\tilde{D}$: the power series solution in $D$ is the general solution of the above PDE in $D$. For a unique solution, boundary data on $D$ needs to be supplied. This boundary data then fixes some of the coefficient of the power series which were not determined by the PDE alone. I would hope that one can do the same for the extended solution $\tilde{D}$. If this is not possible, one may then assume that boundary data is given on $\tilde{D}$, specifically, the value of $\phi$ and its normal derivative on $\tilde{D}$.

Patrick.B
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  • Is the boundary value problem in $\tilde{D}$ well posed? You do not specify any boundary conditions? – Sal Mar 03 '22 at 12:45
  • @Sal: Thank you. I edited my question to address this. – Patrick.B Mar 03 '22 at 12:57
  • I'm not sure if I understand correctly. Would an example (in $\mathbb{R}^2)$ be: $D$ the interior of the circle $x^2+y^2=1$ and $\tilde{D}$ the interior of the circle $x^2+y^2=2$? – Sal Mar 03 '22 at 14:24
  • For example, yes. Although I was hoping for $\tilde{D}$ to be a less "trivial" extension of $D$. For example, where $D=(x-1)^2+(y-1)^2=1$ and $\tilde{D}$ is the torus centered at the origin which circumscribes $D$. – Patrick.B Mar 03 '22 at 15:36
  • In this case, you have two separate problems: the one in $D$ with boundary data on $\partial D$, and the one in $ \tilde{D}/D$ with boundary data on $\partial D \ \cup \partial \tilde{D}$. In general it will not be possible for both $\phi$ and $\partial \phi /\partial n$ (the normal derivative) to be continuous across $\partial D$. This is because the Laplacian BVP becomes over-determined if you specify both on $\partial D$ – Sal Mar 03 '22 at 16:33
  • In $D$ the general solution is known, as a function of the boundary data $\partial D$, whatever those may be. You then want to know the (unique) solution in $\tilde{D}$, given specific, known, boundary data on $\partial \tilde{D}$. Instead of just solving the full problem in $\tilde{D}$, you want to make use of a known power-series solution in $D$. – Patrick.B Mar 03 '22 at 17:06
  • It seems to me that since one of $\phi$ or $\partial \phi/\partial n$ is discontinuous across $\partial D$, analytic continuation is not useful. Or do I misunderstand something? – Sal Mar 03 '22 at 17:10
  • Why would they need to be discontinuous? – Patrick.B Mar 03 '22 at 17:23
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    Suppose $\phi$ is specified on $\partial D$, then the solution in the region $D$ is fully determined. This means that $\partial \phi/\partial n$ is determined on $\partial D$. But for the solution in the region $\tilde{D}/D$, the boundary data on $\partial D$ cannot specify both $\phi$ and $\partial \phi/\partial n$, so in general one must be discontinuous. Perhaps for very special choices of the boundary data on $\partial \tilde{D}$ they both happen to be continuous, but that is a sort of inverse problem which would involve determining that boundary data on $\partial \tilde{D}$ – Sal Mar 03 '22 at 17:40

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