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For almost one year, I am searching for the Green's function for wave equation in R² or R³ with some boundary conditions. As far as I know, when the boundaries permit the method of images, we can get the Green's function. But this requirement is too strong.

What we would like to have is that: the wave is confined in a convex, sufficiently smooth domain. On the boundary, either Dirichlet or Neumann's conditions can be put. To impose these conditions is just to avoid the diffraction problem, which can be too much complicated for us.

During this searching, I encountered books by Prof. Melrose, Prof. Michael E. Taylor and also the formidable three volumns by Prof. Hormander. I still feel hopeless in finding that.

Thanks in advance for any comments!


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What do you mean by getting the Green's function ? If you mean in closed form, then this is hopeless for most domains.

Otherwise, the proper way to express the solution of $$u_{tt}=\Delta u,\qquad u(0)=u_0,\qquad u_t(0)=u_1$$ with homogeneous boundary conditions BC (say Dirichlet or Neumann) is to use the Laplace transform $\hat u$ of $u$ as an auxiliary function: $$\hat u(z):=\int_0^{+\infty}\exp(-sz)u(s)ds.$$ For each $z$ of positive real part, $\hat u(z)$ solves the elliptic problem $$(-\Delta+z^2)w=u_1+zu_0.$$ The above problem, with BC, is well-posed, for every $z$ away from the imaginary axis, and the map $z\mapsto w$ is holomorphic. One recovers $u$ through a Cauchy integral along an appropriate contour in the complex plane. This amounts to express the Green's function of the wave equation as a Cauchy integral in terms of the Green's functions of the elliptic problems parametrized by $z$. This expression may be used to analyze the singularities of the Green's function.

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Thanks Prof. Serre. I am wondering whether the geometric optics would work under our assumptions on the boundaries (smooth concave closed domain). From geometric optics, people use ray-tracing method in computer graphics and acoustics as well. It is an approximation to the real physics phenomenon. But under our assumptions, we hope that there was no diffraction and hence the approximation was indeed accurate. –  Anand Sep 15 '10 at 13:23
Do you mean smooth convex close domain ? –  Denis Serre Sep 15 '10 at 13:33
Yes. Like interior of a ball. :-) –  Anand Sep 15 '10 at 14:15
Well, geometric optics is a vast topic. It is impossible to give even a flavour of it in a few lines, because it involves the theory of pseudo-differential operators, even for a domain without boundary (${\mathbb R}^n$, compact manifolds). In presence of a boundary, it can be a nightmare when rays reach the boundary tangentially. Fortunately, this does not happen if the domain is convex, and thus the theory is essentially ray-tracing plus ordinary reflexion. –  Denis Serre Sep 16 '10 at 8:24
Dear Prof. Serre, that's true. Even when the domain has a conner such as edges of a box, there will be very complicated diffractions. That's why we impose the strong conditions that the domain is convex (to avoid tangent incident rays) plus sufficiently continuous (to avoid diffractions by conner). You said that it is "essentially"...., what do you mean by "essentially"? I am wondering under these conditions, whether we can prove strictly that ray-tracing plus ordinary reflexion can work. Thank you very much for your useful comments! :-) –  Anand Sep 16 '10 at 9:28

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