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]