The linear differential equation I have been working on is the barotropic Rossby wave equation: $$ L\psi=(\frac{\partial}{\partial t}\nabla^2+\beta\frac{\partial}{\partial x} )\psi=\delta(\vec{x}-\vec{x}_0)\delta(t) $$ where $\psi(x,y,t)$ is the two-dimensional stream function varying with time $t$ and we are going to solve for it, $\beta$ is a real constant, $\nabla^2=\partial^2/\partial x^2+\partial^2/\partial y^2$ is two-dimensional Laplace operator, $\delta$ is the Dirac-delta function representing an impluse located at $\vec{x}_0$ in a 2D-plane $(x,y)$ at the initial moment $t=0$ and it forces the barotropic Rossby wave equation.
Actually, the green function solution for the barotropic Rossby wave equation (in a compact form) has been found by
Kloosterziel, R. C.; Maas, L. R. M., Green’s functions for Rossby waves, J. Fluid Mech. 830, 387-407 (2017). ZBL1421.76052.
Here I put their solution (their eq. 1.3) down: $$\psi(x,y,t)=G_\delta=\frac{H(t)}{4}(J_0(z^+)Y_0(z^-)+J_0(z^-)Y_0(z^+))$$ where $H(t)$ the Heaviside unit step function, $J_0$ and $Y_0$ are the first- and second- kind Bessel functions of zero-order respectively, and $z^{\pm}=\sqrt{\beta t (x\pm yi)}$ with $i^2=-1$. This solution is however only for a infinitely large horizontal plane without considering any boundary conditions, so one may call it the fundamental Green's function solution.
When a solid boundary(impenetrable) is placed at the y axis, $x=0$, the imperetrable boundary condition(no normal flow into/out of the boundary) requests $\psi(x=0,y,t)=0$($\psi$ is the stream function, current flows along it) for sake of simplicity, i.e. our boundary condition is Dirichlet. My question is how to construct the Dirichlet Green function solution using the above fundamental Green-function solution for the positive-x half plane: $\{ (x,y,t)\in \mathbb{R}^3 : x\in[0,\infty),y\in(-\infty,\infty),t\in[0,\infty) \}$, which satisfies the Dirichlet b.c.?**
I am wondering about using the method of mirror image, as inspired by
Haine, T. W. N. & Fuller, A. Boundary β-plumes and their vorticity budgets. Q. J. R. Meteorol. Soc. 142, 2758–2767 (2016).
They solved for a steady beta-plume bounded by a west wall and their governning equation $$\beta\psi+\lambda\nabla^2\psi=-\delta(\vec{x}-\vec{x}_0)$$ contains no time derivative terms, thus is very different case from ours. Here, $\lambda$ is a real constant representing the bottom friction coefficient(Ekman friction).
I guess the Dirichlet Green's function solution can be written as some kind of $G_\delta-G_\delta^{\dagger}$, with $G^{\dagger}_\delta$ denoting the image (adjoint) Green's function solution for $$L^{\dagger}G_\delta^{\dagger}=\delta(\vec{x}-\vec{x}_0)\delta(t)$$ Here $L^\dagger=\frac{\partial}{\partial t}\nabla^2-\beta\frac{\partial}{\partial x}$ is the adjoint operator of $L$. Obviously, $L$ is not self adjoint since $L\ne L^{\dagger}$; the $\beta$-term changes sign in $L^{\dagger}$. Physically, the sign change means that $G_\delta^{\dagger}$ actually describes the Rossby waves radiated from the image source but propagating to the east(positive-x direction, while the original Rossby waves are westward). If the image source/Dirac function locates at $\vec{x}_0=(-x_0,y_0)$ such that it is a mirror reflection to the true souce at $\vec{x}_0=(x_0,y_0)$, the waves radiated from the two point sources will collide at $x=0$ and "compensate" for each other, then the boundary condition $\psi(x=0,y,t)=0$ is satisfied.
However, $$L(G_\delta-G_\delta^\dagger)=\delta(\vec{x}-\vec{x}_0)\delta(t) $$ cannot be satisfied in the right half plane, since $L(G_\delta^\dagger) \ne 0$ there. $G_\delta-G_\delta^\dagger$ cannot be the Dirichlet Green's function solution to our problem.
Then how to crack the problem? any help would be very appreciated!
I think that symmetrization and the method of images may solve the problem.
The delta function located at $(x,y,t)=(a,b,c)$ will be denoted in abbreviated form as $ \delta(a,b,c)= \delta (x-a, y-b, t-c)$.
Consider the change of variables $(x,y,t) \to (-x,y,-t)$. This is a linear transformation of the domain which can be denoted by $T(x,y,t)$. The Rossby operator is defined as $Lf= f_{xxt} + f_{yyt}+ f_x$. By the chain rule one can check easily that the Rossby operator $L$ satisfies the identity $L (f\circ T) = - (Lf)\circ T$. (This is because each partial differential operator term in $L$ undergoes an odd number of sign changes.)
Consider the Green's function $G_{\delta}$ that you cited earlier. If it is not already an odd function under the action of $T$, you can symmetrize it to be so by creating the new function $S= \frac{1}{2}[G_{\delta} - G_{\delta}\circ T]$. Then $LS= \frac{1}{2}[ \delta(0,0,0) + \delta (T(0,0,0))]= \delta(0,0,0)$. This is because $T$ maps the origin to itself.
Next note that if you want a Green's function that produces a delta function at any desired point on one side of the inpenatrable boundary, for example at $P=(x_0,y_0,t_0)= (A,0,0)$ with $A > 0$, you can use the method of images to perform a slightly different type of symmetrization by setting $D(x,y,t)= S(x-A,y,t) +S(A-x,y,-t)$.
Note first that the Dirichlet boundary condition at $x=0$ is satisfied by $D$ because $ D(0,y,t)= S(-A,y,t) + S(A,y,-t) =0$ since $S$ is odd under the map $T$ that sends $(-A,y,t)\to (A,y,-t)$.
Next note that $L(D)= L(S) (x-A,y,t) - L(S)(A-x,y,-t) = \delta(A,0,0)- \delta(-A,0,0)$. In the region where $x>0$, note however that only the first of these delta functions is visible. Thus by restricting our attention to that half-space what we see is a Dirichlet solution to the point source forcing located at $P$.
The delta function solution $D$ associated to point $P=(A,0,0)$ can be translated to any other $t$ or $y$ location since the Rossby operator commutes with translations in those two variables, as does the Dirichlet boundary condition on the boundary $x=0$.