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[also asked here:http://mathoverflow.net/questions/122191/existence-of-a-function]

All arguments are in $\mathbb{R}^3$.

Suppose $n(x)$ is a smooth function where $\mathbf{supp}(n(x)-1)$ is a compact set $\Omega$. i.e. $n(x) = 1$ when $x$ is outside $\Omega$.

Assume there are some points $x_j\in\Omega$, where $j=1,2,\cdots.m$.

Consider Helmholtz equation

$\Delta u + k^2 n(x) u = 0$

And I want to know if there is a complex-valued function $u:\mathbb{R}^3\rightarrow\mathbb{C}$ satisfies the equation and also vanishes at $x_j$. i.e. $u(x_j) = 0$. [Certainly $u$ $\text{can have more zeros}$ than $x_j$]

Here you may try to give a method to construct $u$, or prove the existence.

And you may take $m=1$ here.

Thanks.


[UPDATE]. Here is a case that this function can exist with some freedom to choose the points.

If we assume the points are on $xy$ plane, i.e. with the corrdinate $(x,y,0)$ in $\mathbb{R}^3$, and $n=1$ as a constant.

Then for $x_j = (a_j,b_j,0)$ in $\mathbb{R}^3$,

we construct function

$\phi(x,y) = \prod_{i=1}^m(x+iy -a_j-ib_j)$

thus $\Delta \phi = 0$, since it is analytic.

My function $u(x,y,z) = \phi(x,y)e^{-ikz}$ satisfies that $u(x_j) = 0$, and satisfies Helmholtz equation.

What I am still thinking about is, can I choose the points randomly, or the points should satisfy some condition?


[UPDATE] For $n(x)$ is not a constant. I have found a way to construct the function for the case that $x_j$ sits on the $xy$ plane.

Take the form of solution as $u(x,y,z) = \phi(x,y)e^{-ik\rho(x,y,z)}$, then

$\rho:\mathbb{R}^3\rightarrow\mathbb{C}$ satisfies that $\Delta \rho + 2\dfrac{\nabla\phi}{\phi}\cdot \nabla\rho + ik(n(x)-\nabla\rho\cdot\nabla\rho) = 0$

Here we cannot say the existence is assured by PDE theories, since $\dfrac{\nabla\phi}{\phi}$ is unbounded, we shall either have $\dfrac{\nabla\phi}{\phi}\cdot \nabla\rho=0$ or $\nabla\rho(x_j) =0$.

I have found a way to construct $\rho$ to saitisfy this, but I still want to know if there is a function such that can have zeros randomly. My example requires the points are on the same plane.


[UPDATE] I have come up with an easier proof. We can prove the existence by induction. The only thing we have to ensure is the $\nabla u\neq\mathbf{0}$ in $\Omega$.

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  • $\begingroup$ Crossposted to MathOverflow. $\endgroup$ Feb 18, 2013 at 16:27
  • $\begingroup$ I assume you mean other than $u=0$. $\endgroup$
    – Ray Yang
    Feb 20, 2013 at 20:09
  • $\begingroup$ sure, non-zeros solution. $\endgroup$
    – Yimin
    Feb 20, 2013 at 20:23
  • $\begingroup$ @RayYang Even more so, $u$ must vanish only at $x_j$, if I read the question right. (Would be easier otherwise). $\endgroup$
    – user53153
    Feb 20, 2013 at 21:04
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    $\begingroup$ I believe I can prove it is possible. However, I'm just buried in other projects now, some of which are long overdue. Do you have any urgency with this question (e.g., you need it as a lemma for something else, etc.)? If you do, I'll try to check and post my proof soon, but if it was asked of pure curiosity, I'll prefer to finish other things first :) $\endgroup$
    – fedja
    Mar 3, 2013 at 14:02

1 Answer 1

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I have given a proof here. In the existence chapter.

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