Uniqueness of constructed solutions to the Helmholtz equation

Question

My question is regarding the inhomogeneous Helmholtz equation on $\mathbb{R}^3$ with real wavenumber $k$ and outgoing radiation condition

\begin{equation} \Delta u + k^2 u = - f \quad \text{and} \quad \lim_{r\to \infty} r(\partial_r - ik)u = 0 \quad \text{on } S^2. \end{equation} It is well known that for, say, $f \in L^2(\mathbb{R}^3)$ there is a unique solution $u \in H^2_{loc}(\mathbb{R}^3)$ and that the solution is given by a convolution with the fundamental solution, i.e.,

$$ u(x) = \int_{\mathbb{R}^3} \frac{e^{ik|x-y|}}{4\pi |x-y|} f(y) dy. $$

My confusion stems from the following considerations (which again stem from trying to manufacture a solution for testing the convergence of a numerical method):

For a ball $B_r$ centered at $0$ and radius $r$, take $v \in C^\infty_c(\mathbb{R}^3)$ and assume $v$ is real valued and such that $\text{supp}(v) \subset B_{r}$. Then set $\tilde{f} = - (\Delta + k^2 )v$.

By construction, $v$ satisfies the Helmholtz equation with right-hand side $-\tilde{f}$, and $v = 0$ on $\mathbb{R}^3\setminus B_r$, and so it satisfies the radiation condition. Therefore, it should be the unique, real-valued solution to the problem.

However, if we write

\begin{align} V(x) &= v_1(x) + iv_2(x) = \int_{\mathbb{R}^3} \frac{e^{ik|x-y|}}{4\pi |x-y|} \tilde{f}(y) dy\\ &= \int_{\mathbb{R}^3} \frac{\cos(k|x-y|)}{4\pi |x-y|} \tilde{f}(y) dy + i\int_{\mathbb{R}^3} \frac{\sin(k|x-y|)}{4\pi |x-y|} \tilde{f}(y)dy, \end{align}

there is no reason for the imaginary part $v_2(x)$ to vanish everywhere.

But we should have $$(\Delta + k^2)V = (\Delta + k^2)v_1 + (\Delta + k^2)iv_2 = -\tilde{f}$$ and as $\tilde{f}$ is real-valued, we must have $(\Delta + k^2)v_2 = 0$, and by uniqueness $v_2 = 0$.

My question is therefore what to make of the imaginary part $v_2$ of the solution $V$ given by the convolution with the fundamental solution? And is there something wrong with my initial constructed solution $v$ that disqualifies it from being a solution to the Helmholtz equation with Sommerfeld radiation condition?

All help is much appreciated!

Answer 1

The old Sherlock Holmes adage

When you have eliminated the impossible, whatever remains, however improbable, must be the truth.

applies here. Since nothing else you did was wrong, it must be your idea that

there is no reason for the imaginary part $v_2$ to vanish everywhere

that is faulty. So let me give you the reason.

First consider the function $\frac{\sin(kr)}{r}$ on $\mathbb{R}^3$. Note that the sinc function is real analytic and even, so $\frac{\sin(kr)}{r}$ is a smooth function on $\mathbb{R}^3$.

Now you can compute, in polar coordinates for convenience, $$ \Delta \frac{\sin(kr)}{r} = \frac{1}{r} \partial^2_{rr} \sin(kr) = - k^2\frac{\sin(kr)}{r} $$ which shows the key fact that

$$(\Delta + k^2) \frac{\sin(kr)}{r} = 0.$$

Given any function $w$, smooth with compact support, you can integrate by parts

$$ \int \frac{\sin(k|x|)}{|x|} (\Delta + k^2) w(x) ~dx = \int w(x) (\Delta + k^2) \frac{\sin(k|x|)}{|x|} ~dx = 0. $$

Now for any given $y\in \mathbb{R}^3$, set $w(x) = -v(y-x)$ for the $v$ in your question, and you check therefore the imaginary part of $v_{2}(y) = 0$.

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Note, however, the final integration by parts may fail to be valid when $w$ is not decaying sufficiently fast. Taking care of this "boundary condition at infinity" is why you impose the radiation condition.