1

What is the Green's function of a first order differential operator like $$\vec{a}.\vec{\nabla}+f(x,y)=(a_x\frac{\partial}{\partial x}+a_y\frac{\partial}{\partial y})+f(x,y)$$

where $\vec{a}$ is constant vector.

richard
  • 221
  • The method of characteristics reduces this to more-or-less an ODE, and characteristic lines are straight since $\vec{a}$ is constant. After that you can use integrating factors to reduce the equation further. This will also tell you precisely what the Green's function is. – Willie Wong Dec 19 '13 at 11:22
  • Thank you Willie this seems very helpful – richard Dec 19 '13 at 11:26

1 Answers1

3

You can write $f$ as $\vec a\cdot \nabla F$ for some function $F$ (just integrate $f$ along lines in the direction of $\vec a$). Then the PDE becomes $$\vec a\cdot \nabla (e^F u)=0$$ from where the general solution is clear: $u(x)=e^{-F (x,y)}\,h(P(x,y))$ where $P$ is the projection onto $\{\vec a\}^\perp$ (or onto another hyperplane transverse to $\vec a$).

Green's function is pretty weird, because there is no diffusion in the direction orthogonal to $\vec a$. Any source term only affects the solution on the $\vec a$-line through the location of the source. When the source is $\delta$ placed at some point $\vec p$, we get Green's function. It is a singular distribution supported on the line $\ell=\{\vec p+t\vec a\}$; more precisely, $$G(\vec x,\vec p)=ce^{-F(\vec x)}H((\vec x-\vec p)\cdot \vec a) \, m^1_{\ell}$$ where $m^1_{\ell}$ is the linear measure on $\ell$, and $H$ is the Heaviside function. The constant $c$ is chosen so that you get $\delta$ on the right side of equation; I think it is equal to $e^{F(\vec p)}/|\vec a|$.

Post No Bulls
  • 9,056