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I know a few solutions of the equation $\Delta \phi + \phi =0$ on $\mathbb{R}^2$. They can be described as the Fourier transform of any finite measure $\mu$ on $S^1$. In particular take $\mu$ a finite, positive measure on $S^1$ and consider $G$ defined on $\mathbb{R}^2$ by $$G(x,y) = \int_{0}^{2\pi} e^{i\langle(x,y),(\cos\theta,\sin \theta)\rangle} \mathrm d \mu(\theta)$$ It is easy to check that $$\Delta G+G=0$$ on $\mathbb{R}^2$ and also $G$ is bounded on $\mathbb{R}^2$. I have the following two questions:

  1. Does this equation have any unbounded solutions? Is it possible to describe all solutions of this equation?

  2. Suppose $H$ is a bounded solution of this equation, then can we conclude that it necessarily has to be the Fourier transform of a finite measure on $S^1$.

Any help or reference will be very helpful.

Thanks!

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3 Answers 3

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Every solution of $\Delta\phi+\phi=0$ on $\mathbf R^2$ can be written as the “Poisson integral” (where for short $(u,v)=w$, $(x,y)=z$) $$ \phi(z)= \left\langle T, e^{i\langle z,\cdot\rangle}\right\rangle=\int_{\mathrm S^1}e^{i\langle z,w\rangle}dT(w) $$ for a unique entire functional $\,T$ on the circle $\mathrm S^1$; these include Radon measures, Schwartz distributions, hyperfunctions, and more. Details and proofs see Hashizume et al. (1972, p. 543), Helgason (1974, p. 348; 1984, p. 5), or Agmon (1999). For example, if I am not mistaken,

  • $T_1=$ Dirac measure at $(-i,\sqrt2)\in (\mathrm S^1)^{\mathbf C}$ gives $\phi_1(z)=e^{x+\sqrt2iy}$ (found by Maple);
  • $T_2=$ derivative of $T_1$ in the direction $(\sqrt2,i)$ gives $\phi_2(z)=(\sqrt2x+iy)e^{x+\sqrt2iy}$ (new).
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Regarding the other question in your post: not quite.

If $\phi$ is a bounded function such that $\Delta \phi + \phi = 0$, then $\phi$ is a tempered distribution such that the Fourier transform of $\phi$ (which is another tempered distribution) satisfies $(-|\xi|^2 + 1) \hat\phi = 0$. This easily implies that $\hat\phi$ is supported in the unit sphere $\partial B = \{\xi : |\xi| = 1\}$.

However, $\hat\phi$ need not be a finite measure. If I am not mistaken, the distribution $$ \langle \hat\phi, u \rangle = \lim_{\varepsilon \to 0^+} \int_{\{|\xi_1| > \varepsilon\} \cap \partial B} \frac{u(\xi)}{i \xi_1} \, \sigma(d\xi) $$ (where $\sigma$ is the usual surface measure on $\partial B$) is the Fourier transform a bounded function, the indefinite integral (with respect to $x_1$) of the Bessel function $J_0(|x|)$. (Just as in dimension one the distribution $1/x$ is the Fourier transform of a bounded function.)

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The answer to your first question is affirmative. The command of Maple 2019.1

pdsolve(VectorCalculus:-Laplacian(phi(x, y), [x, y]) = -phi(x, y), phi(x, y), explicit);

produces

$$\phi \left( x,y \right) ={\it \_C1}\,{{\rm e}^{\sqrt {{\it \_c}_{{1}}} x}}{\it \_C3}\,\sin \left( \sqrt {{\it \_c}_{{1}}+1}y \right) +{\it \_C1}\,{{\rm e}^{\sqrt {{\it \_c}_{{1}}}x}}{\it \_C4}\,\cos \left( \sqrt {{\it \_c}_{{1}}+1}y \right) +{\frac {{\it \_C2}\,{\it \_C3}\, \sin \left( \sqrt {{\it \_c}_{{1}}+1}y \right) }{{{\rm e}^{\sqrt {{ \it \_c}_{{1}}}x}}}}+{\frac {{\it \_C2}\,{\it \_C4}\,\cos \left( \sqrt {{\it \_c}_{{1}}+1}y \right) }{{{\rm e}^{\sqrt {{\it \_c}_{{1}}} x}}}}. $$

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    $\begingroup$ Great! This answers first part of my first question. Just a small clarification - what really is the Maple command supposed to produce? Is it a class of functions which solves the equation? I am just wondering how/why it chooses one class of examples over some other class (for instance the class of functions $G$ I mentioned in the question)? $\endgroup$
    – April
    Oct 6, 2019 at 9:35
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    $\begingroup$ From the help: "The strategy pdsolve uses is to look for the most general solution to the given PDE or, in the worst case, to look for a complete separation of variables. Thus, when successful, the command returns one of the following: - A general solution, - A quasi-general solution (a solution containing arbitrary functions, but not in sufficient number or not having enough variables to constitute a general solution), or - A set of uncoupled ODEs with all the variables separated, or a complete solution obtained after integrating this set (when the option INTEGRATE is indicated) ". $\endgroup$
    – user64494
    Oct 6, 2019 at 10:30
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    $\begingroup$ This set is however not the full set of solutions, though. The second example given in Francois' answer is not in this form. This looks like the set of stuff you get from separation of variables (which incidentally, immediately tells you that $e^{\pm x}$ and $e^{\pm y}$ are unbounded solutions to the original PDE). $\endgroup$ Oct 7, 2019 at 14:19

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