When does a double-layer potential equal to a single-layer one?

1.   Introduction

Let $D$ be a bounded domain in ${{\mathbb R}}^3$ with a closed, smooth, connected boundary $S$, $N = N_s$ be the outer unit normal to $S$ at the point $s\in S$, $k > 0$ be a constant, $g(x, y) = \frac{e^{ik|x-y|}}{4\pi |x-y|}$, $w: = w(x, \mu): = \int_S g_{N}(x, s)\mu(s)ds$ be the double-layer potential, $w^{\pm}$ are the limiting values of $w$ on $S$ from $D$, respectively, from $D'$, $u: = u(x, \sigma): = \int_S g(x, s)\sigma(s)ds$ be the single-layer potential, $u_{N^{\pm}}$ are the limiting values of the normal derivative of $u$ on $S$ from $D$, respectively $D': = {{\mathbb R}}^3\setminus \bar{D}$, $\bar{D}$ is the closure of $D$, $\bar{\sigma}$, denotes the complex conjugate of $\sigma$, $H^0: = L^2(S)$, $H^1: = W^1_2(S)$ is the Sobolev space, $\mu\in H^0$. We write iff for if and only if and use the known formulas for the limiting values of the potentials on $S$, see ^[3], pp. 148,153:

where $A'\mu = \int_Sg_{N_t}(t, s)\mu(s)ds$, $A\sigma = \int_Sg_{N_s}(t, s)\sigma(s)ds$. In this paper it is proved that for every $w$ in $D$ there is a unique $u$, such that $w = u$ in $D$, and vice versa.

Necessary and sufficient conditions are given for $w = u$ in $D'$ and for $u = w$ in $D'$.

In ^[4] the problem for the Laplace's equations was studied in Lipschitz domains. In this case the integral equations, based on potential of double layer, are uniquely solvable for any right-hand side. This is not so, in general, for the corresponding equations for the Helmholtz operator. Our method of proof and the results are new. We assume that the boundary $S$ is smooth and connected. This is done for brevity and simplicity: we do not want to make the presentation more difficult for the reader than it is necessary. Our results and proofs are valid for Lipschitz boundaries. Theorem 1 can be used for looking for the solution of the Dirichlet problem in the form of the single-layer potential.

Let us state these results:

Theorem 1. For every $w$, defined in $D$, there is a unique $u$ such that $w = u$ in $D$, and vice versa.

For every $w$, defined in $D'$, there is a unique $u$ such that $w = u$ in $D'$ iff

For every $u$, defined in $D'$, there is a unique $w$ such that $w = u$ in $D'$ iff

This result is new, although the potential theory has more than 150 years of history.

In Section 2 proofs are given. In Section 3 it is proved that $Q: H^0\to H^1$ is a Fredholm operator.

2.   Proofs

a) Assume that $w = w(x, \mu)$ is given in $D$. Let us prove that $u = u(x, \sigma)$ exists such that $u = w$ in $D$, and $u$ is uniquely defined by $w$. First, let us prove the last claim. Suppose $u_1 = w$ and $u_2 = w$ in $D$. Let $u_1-u_2: = u$, $u = \int_Sg(x, s)\sigma ds$ in $D$, $\sigma = \sigma_1-\sigma_2$. Then $u|_S = 0$, $(\nabla^2+k^2)u = 0$ in $D'$ and $u$ satisfies the radiation condition, so $u = 0$ in $D'$. Thus, $u = 0$ in $D\cup D'$, $u_1 = u_2$, and the claim is proved.

Let us now prove the existence of $u$ such that $w = u$ in $D$. One has

This is an equation for $\sigma$ while $w^{+}$ is given.

Note that $Q = Q_0+Q_1$, where $Q_0\sigma = \int_Sg_0(t, s)\sigma(s)ds$, $g_0(t, s): = \frac 1 {4\pi |t-s|}$. The operator $Q_0$ is an isomorphism of $H^0$ onto $H^1$, see Lemma 1 in Section 3. Therefore, the operator $Q_0^{-1}$ is well defined and maps $H^1$ onto $H^0$.

Consequently, equation (2.1) is equivalent to

The operator $Q_1Q_0^{-1}$ is compact in $H^0$, see Section 3. Therefore a necessary and sufficient condition for the solvability of equation (2.2), and the equivalent equation (2.1), is:

where $N(B)$ is the null space of the operator $B$ and $B^\star$ is the adjoint operator to $B$ in $H^0$, $B = I+Q_1Q_0^{-1}$, $B$ is of Fredholm type in $H^0$. The kernel $g(t, s)$ of $Q$, the function $g(t, s) = \frac{e^{ik|t-s|}}{4\pi |t-s|}$, is symmetric: $g(t, s) = g(s, t)$. Therefore, the kernel of $Q^\star$ is $\bar{g}(t, s)$. Clearly, $(I+Q_1Q_0^{-1})^{\star} \eta = 0$ iff $(I+{Q}^{-1}_0\bar{Q}_1) \eta = 0$, or, taking the complex conjugate,

where we have used the real-valuedness of the kernel of $Q_0$. Applying the operator $Q_0$ to the last equation, one gets an equivalent equation

since $Q_0$ is an isomorphism. Let $u = u(x, \bar{\eta})$. Then $u|_S = 0$ according to equation (2.5). Since $(\nabla^2+k^2)u = 0$ in $D'$ and $u$ satisfies the radiation condition, it follows that $u = 0$ in $D'$ and $\bar{\eta} = u^{+}_N-u^{-}_N = u^{+}_N$. Therefore, using the Green's formula, one obtains:

and, since $u = 0$ on $S$, it follows that condition (2.3) is always satisfied.

Thus, the necessary and sufficient condition (2.3) for the solvability of equation (2.1) is always satisfied. Therefore, $u(x, \eta) = w(x, \mu)$ in $D$, $Q_0^{-1}\eta = \sigma$.

b) Asume now that $u$ is given in $D$ and let us prove the existence of a unique $w$ such that $u = w$ in $D$. First, we prove that $w$ is uniquely determined by $u$ if $u = w$ in $D$.

Indeed, assume that there are two $w_j$, $j = 1, 2,$ such that $u = w_j$ in $D$. Then $w: = w_1-w_2 = 0$ in $D$. Therefore $w_{N^+} = 0$ on $S$. It is known (see ^[3], p. 154) that $w_{N^+} = w_{N^-}$, so $w_{N^-} = 0$. Therefore, $(\nabla^2+k^2)w = 0$ in $D'$, $w_{N^-} = 0$ on $S$, and $w$ satisfies the radiation condition at infinity. This implies $w = 0$ in $D'$, so $w = 0$ in $D\cup D'$. Therefore $\mu = w^–w^+ = 0$, so $w_1 = w_2$ if $u = w_j$, $j = 1, 2,$ in $D$. We have proved that $w$ is uniquely determined by $u$ if $u = w$ in $D$.

Let us now prove the existence of the solution $\mu$ to equation (2.1) and the relation $u = w$ in $D$.

The operator $A'-I$ is Fredholm in $H^0$, so a necessary and sufficient condition for the equation (2.1) to be solvable is:

One has $(A'-I)^{\star} h = (\bar{A}-I)h = 0$ iff $(A-I)\bar{h} = 0$.

If $(A-I)\bar{h} = 0$, then $u_N^{-}(s, \bar{h}) = 0$, so $u(s, \bar{h}) = 0$ in $D'$. Note that $u(s, \bar{h}) = Q(\bar{h})$. Therefore, $Q(\bar{h}) = 0$ in $D'$. Since $Q$ is a symmetric operator in $H^0$, one has:

Consequently, condition (2.6) is always satisfied, the solution $\mu$ to equation (2.1) exists and $u = w$ in $D$.

c) Assume that $w$ is given in $D'$. Let us prove that $u$ exists such that $u = w$ in $D'$ iff

Consider the equation for $\sigma$:

The operator $Q:H^0\to H^1$ is Fredholm-type. Thus, a necessary and sufficient condition for the solvability of the above equation is equation (2.7). If $\sigma$ solves (2.8), then $u(x, \sigma) = w$ in $D'$ because the value of $u$ on $S$ determines uniquely $u$ in $D'$.

d) Assume now that $u$ is given in $D'$. Let us prove that $w$ exists such that $u = w$ in $D'$ iff

Note that $p \in N(A+I)$ iff $\bar{p}\in N(\bar{A}+I)$. The equation for $\mu$, given $u$ in $D'$, is:

Since the operator $A'+I$ is Fredholm in $H^0$, a necessary and sufficient condition for the solvability of (2.10) for $\mu$ is:

If $p\in N((A'+I)^{\star})$, then $p\in N(\bar{A}+I)$, so condition (2.9) is the same as (2.11). As in section c), the relation $u = w$ in $D'$ is a consequence of the fact that $u = w^{-}$ on $S$.

Theorem 1 is proved.

Remark 1. Our proofs remain valid if $k = 0$, that is, for the potentials corresponding to the Laplace equation, rather than the Helmholtz equation.

3.   Auxiliary lemma

Recall that $H^1$ is the Sobolev space on $S$, $H^0 = L^2(S)$.

Lemma 1. The operator $Q = Q_0+Q_1: H^0\to H^1$ is of Fredholm-type, where $Q_0$ is the operator with the kernel $\frac 1 {4\pi |t-s|}$ and $Q_1$ has the kernel $\frac {e^{ik|t-s|}-1}{4\pi |s-t|}$. The operator $Q_0$ is an isomorhism of $H^0$ onto $H^1$, which has a continuous inverse. The operator $Q_1Q_0^{-1}$ is compact in $H^0$.

Proof. Let us check that $Q_0: H^0 \to H^1$ is an isomorphism. The Fourier transform of the kernel $\frac 1 {4\pi |x-y|}$ is positive: $\int_{{{\mathbb R}}^3}\frac {e^{i\xi \cdot x}}{4\pi |x|}dx = \frac 1 {|\xi|^2}$. So, $Q_0: H^0 \to H^1$ is injective. The kernel of the operator $Q_1$ is smooth enough for $Q_1: H^0 \to H^1$ to be compact. Let us check that $Q_0: H^0 \to H^1$ is surjective. Let $f\in H^1$ and $Q_0\sigma = f$. Then $u: = u(x, \sigma) = \int_Sg(x, s)\sigma ds$ solves the problem: $\nabla^2 u = 0$ in $D$, $u|_S = f$. By the known elliptic estimates (see, e.g., ^[1]) one has $\|u\|_{H^{3/2}(D)}\le \|u\|_{H^1(S)}$. Therefore, $\nabla u\in H^{1/2}(D)$ and, by the trace theorem, $u|_S\in H^0(S)$. This proves surjectivity of $Q_0: H^0 \to H^1$. Thus, $Q_0$ is an isomorphism of $H^0$ onto $H^1$ which has a continuous inverse. The sum of an isomorphism $Q_0$ and a compact operator $Q_1$ is a Fredholm operator, see, e.g., ^[2]. The operator $Q_1Q_0^{-1}$ is compact in $H^0$ because the kernel of $Q_1$ is sufficiently smooth. Although the operator $Q_1Q_0^{-1}$ is defined on a dense subset $H^1$ of $H^0$, but since this operator is bounded in $H^0$ its closure is a bounded operator in $H^0$. Since the kernel of $Q_1$ is $O(|s-t|)$, the kernel of $Q_1Q_0^{-1}$ is a continuous function of $|s-t|$ and the surface $S$ is a compact set. Therefore, the operator $Q_1Q_0^{-1}$ is compact in $H^0$.

Lemma 1 is proved.

4.   Conclusions

It is proved that every double layer potential $w$ in a bounded domain is equal to a single layer potential $u$ in a bounded domain $D$ with a smooth closed connected boundary. Necessary and sufficient conditions are given for $w = u$ in the exterior domain $D'$.

Conflict of interest

The authors declare that there are no conflicts of interest.