Let $\Omega\subset R^2$ be a simply connected bounded domain with infinitely differentiable boundary $\partial\Omega$and unit normal vector $v$ directed into the exterior of $\Omega$ $$\Phi{(x,y)}=\dfrac{i}{4}H^{(1)}_{0}(k|x-y|),x\neq y$$ we denote the fundamental solution to the two-dimensional Helmholtz equation in terms of the first kind Hankel function of order zero

where the Helmholtz equation $$\Delta u+k^2u=0, \mbox{in} R^2\overline{\Omega}$$ and $$(T\psi)(x):=\dfrac{\partial}{\partial v(x)}\int_{\partial\Omega}\dfrac{\partial\Phi{(x,y)}}{\partial v(y)}\psi{(y)}ds(y),x\in\partial\Omega.$$

show that: $$(T\psi)(x)=\dfrac{\partial}{\partial s(x)}\int_{\partial\Omega}\Phi{(x,y)}\dfrac{\partial \psi}{\partial s}(y)ds(y)+k^2v(x)\cdot\int_{\partial \Omega}\Phi{(x,y)}v(y)\psi{(y)}ds(y),x\in\partial\Omega $$

This relusut is from this paper:http://num.math.uni-goettingen.de/kress/kress2013.pdf

The author say can in [12], http://link.springer.com/article/10.1007%2FBF02941090#page-1

Now I find this paper,because I don't know french,so I'm not sure this is the proof, if someone understand this proof, can you explain it to me. thank you very much. Thank you

This follow picture is from [12]

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  • 1
    $\begingroup$ This looks like the derivatives are moved around under the integral sign and a "boundary integral" term gets added in. This is always a sign that integration by parts / Green's identities have been used. Are you familiar with these? $\endgroup$ – rajb245 Dec 18 '13 at 22:48
  • $\begingroup$ Yes,I konw Green identities and integration by parts,But this problem I can't have $\Phi{(x,y)}v(y)ds(y)$,can you post you solution? Thank you $\endgroup$ – math110 Dec 19 '13 at 2:55
  • $\begingroup$ I see now that this is more complicated than I thought. I didn't realize that the normal derivatives were replaced by derivatives with respect to the surface area elements. The paper claims that there is a proof in the book "Integral equation methods in scattering theory"; I don't have access to that text but have you checked there? $\endgroup$ – rajb245 Dec 26 '13 at 19:47
  • $\begingroup$ Thank you , I find I have this book,and I can't find this proof in this book,Thank you $\endgroup$ – math110 Dec 27 '13 at 5:34
  • $\begingroup$ The double Layer potential $$\int_{\partial\Omega}\dfrac{\partial\Phi{(x,y)}}{\partial v(y)}\psi{(y)}ds(y),x\in\partial\Omega$$ has a jump relation i.e. discontinuous on $\partial\Omega$. So how is the derivatie to be understood here? $\endgroup$ – Andrew Dec 29 '13 at 11:19

Let $\Omega$ an open surface with boundary $\partial \Omega$ with normal vector $\pmb \nu$. For a vector field $\pmb F$, Stokes’s theorem gives $$ \int_\Omega \nu_l\varepsilon_{lmi}\frac{\partial F_i}{\partial x_m}ds(x)=\int_{\partial \Omega} F_i dx_i $$ where $\varepsilon_{ijk}$ is the Levi Civita alternating tensor and $\Omega$ has been oriented in the standard way. We apply Stokes’s theorem to the vector field $$ F_i=\psi \varepsilon_{ijk}\frac{\partial }{\partial x_j}\Phi(P,Q)=-\psi \varepsilon_{ijk}\frac{\partial }{\partial x'_j}\Phi(P,Q) $$ where $Q$ is at $(x_1,x_2,x_3)$, $P$ is at $(x'_1,x'_2,x'_3)$ and $\psi(\pmb{x})$ is a smooth scalar field. We suppose to begin with that $P$ is not on $\Omega$. Hence $$ \begin{align} \int_{\partial \Omega} F_i \operatorname{d}x_i&=\int_\Omega \varepsilon_{lmi}\varepsilon_{ijk}\nu_l\frac{\partial}{\partial x_m} \left(\psi\frac{\partial\Phi}{\partial x_j} \right)\operatorname{d}s(\pmb{x})\\ &=\int_\Omega \left[\nu_j\frac{\partial}{\partial x_k} \left(\psi\frac{\partial\Phi}{\partial x_j} \right)- \nu_k\frac{\partial}{\partial x_j} \left(\psi\frac{\partial\Phi}{\partial x_j} \right) \right]\operatorname{d}s(\pmb{x}) \end{align} $$ using $\varepsilon_{lmi}=\varepsilon_{ilm}$ and $\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$. As $\Phi$ satisfies the Helmholtz equation, we obtain $$ \begin{align} \int_{\partial \Omega} F_i \operatorname{d}x_i&=k^2\int_\Omega \psi\nu_k\Phi \operatorname{d}s(\pmb{x})- \tfrac{\partial}{\partial x'_k}\int_\Omega \psi \tfrac{\partial\Phi}{\partial \nu_q}\operatorname{d}s(\pmb{x})+ \int_\Omega \left(\nu_k\tfrac{\partial\psi}{\partial x_j}- \nu_j\tfrac{\partial\psi}{\partial x_k}\right)\tfrac{\partial\Phi}{\partial x'_j} \operatorname{d}s(\pmb{x}). \end{align} $$ Next, let $P \to p \in\Omega$ to give $$ \lim_{P\to p}\tfrac{\partial}{\partial x'_k}\int_\Omega \psi \tfrac{\partial\Phi}{\partial \nu_q}\operatorname{d}s(x)=\lim_{P\to p}\left\{k^2\int_\Omega \psi\nu_k\Phi \operatorname{d}s(\pmb{x})-\int_{\partial \Omega} F_i dx_i+\tfrac{\partial}{\partial x'_j}\int_\Omega \mu_{jk}(q) \Phi \operatorname{d}s(\pmb{x}) \right\} $$ where $\mu_{jk}=\nu_k\frac{\partial\psi}{\partial x_j}- \nu_j\frac{\partial\psi}{\partial x_k}$. The first two terms on the right-hand side are continuous as $P \to p$. The last term is the gradient of a single-layer potential; its limiting value is $$ \pm\nu_j(p)\mu_{jk}(p)+\int_\Omega \mu_{jk}(q) \frac{\partial \Phi}{\partial x'_j} \operatorname{d}s(\pmb{x}), $$ the sign being $+ (−)$ when $\pmb\nu_p$ points towards (away from) $P$. Hence, multiplying by $\nu_k(p)$, we obtain $$ \begin{align} \frac{\partial }{\partial \nu_p} \int_\Omega \psi(q)\frac{\partial \Phi}{\partial \nu_q}\operatorname{d}s_q&=k^2\int_{\Omega}\nu_k(p)\nu_k(q)\Phi ds_q-\nu_k(p)\int_{\partial\Omega}\psi\varepsilon_{ijk} \frac{\partial \Phi}{\partial x_j} \operatorname{d}x_i+\\ &\quad+\nu_k\int_\Omega \left(\nu_k\frac{\partial\psi}{\partial x_j}- \nu_j\frac{\partial\psi}{\partial x_k}\right)\frac{\partial\Phi}{\partial x'_j} \operatorname{d}s(\pmb{x})\\ &=k^2\int_{\Omega}\pmb\nu(p)\cdot\pmb\nu(q)\Phi (p,q)\psi(q) \operatorname{d}s_q+ \int_{\partial\Omega}\psi(q)\pmb\nu(p)\cdot(\operatorname{d}\pmb r\times\nabla_q\Phi (p,q))+\\ &\quad+\int_{\Omega}(\pmb\nu(q)\times\nabla_q\psi)\cdot(\pmb\nu(p)\times\nabla_p\Phi(p,q))\operatorname{d}s_q \end{align} $$ where we have noted that $\nu_j\nu_k\mu_{jk}= 0$. The line integral around $\partial\Omega$ will vanish if $\Omega$ is a closed surface or if $\psi = 0$ on $\partial\Omega$, and then we obtain Maue’s formula: $$ (T\psi)(p)=\int_{\Omega}(\pmb\nu(q)\times\nabla_q\psi)\cdot(\pmb\nu(p)\times\nabla_p\Phi(p,q))\operatorname{d}s_q+k^2\int_{\Omega}\pmb\nu(p)\cdot\pmb\nu(q)\Phi (p,q)\psi(q) \operatorname{d}s_q $$ where $(T\psi)(p)$ is the normal derivative of a double-layer potential, i.e. $$ (T\psi)(p)=\frac{\partial }{\partial \nu_p} \int_\Omega \psi(q)\frac{\partial \Phi}{\partial \nu_q}\operatorname{d}s_q,\qquad p\in\Omega $$ which is often called a hypersingular operator. The Maue's formula shows that $T\psi$ can be written as an integral involving $\psi$ and its tangential derivatives. It is valid provided that $\Omega$ is a closed surface (or a collection of closed surfaces). It is also valid for open surfaces, but only if $\psi$ vanishes around the boundary of $\Omega$, otherwise there is an additional line-integral contribution from $\partial\Omega$.


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