Is the function v(r) = (e^(-mr))/r a solution to the Yukawa PDE equation?

[docnet]

Homework Statement

please see below

Relevant Equations

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(1) From "Radial solutions to Laplace's equation", we know that
$$
\Delta u(x) = v(r)''+\frac{n-1}{r}v(r)'
$$
we re-write the PDE
$$
- \Delta u+m^2u=0
$$
in terms of $v(r)$
\begin{equation}
- v(r)''-\frac{n-1}{r}v(r)'+m^2v(r)=0
\end{equation}
to give a linear second order ODE with non-constant coefficients.

(2) To check that the function
$$v(r)=\frac{e^{-mr}}{r}$$
is a solution of our ODE, we compute
$$ v'(r)=\frac{-me^{-mr}r-e^{-mr}}{r^2}$$
and
$$v''(r)=\frac{e^{-mr}(m^2r^2+2(mr+1))}{r^3}$$
and perform pluginology into equation (1)
$$ \frac{-e^{-mr}(m^2r^2+2(mr+1))}{r^3}-\Big(\frac{n-1}{r}\Big)\Big(\frac{-me^{-mr}r-e^{-mr}}{r^2}\Big)+m^2\frac{e^{-mr}}{r}$$
set $n=3$ and simplify
$$\Rightarrow -m^2r^2-2mr-2+2mr+2+m^2r^2=0$$

(3) To define the fundamental solution to the PDE
$$
- \Delta u+m^2u=f
$$
using the function $\Phi:R^n\backslash\{0\} \rightarrow R$ given by
$$
\Phi(x)=\frac{1}{4\pi}\frac{e^{-m||x||}}{||x||}
$$
we modify theorem 3.18Proof of Theorem $3.18$:
We recall
\begin{equation}
\tilde u(x)=\int_{R^n}\Phi(x-y)[f(y)-m^2u(y)]dy
\end{equation}
\begin{equation}=\int_{R^n}\Phi(y)[f(x-y)-m^2u(x-y)]dy\end{equation}
To see $u(x)\in C^2(R)$ we differentiate expression (3) twice
$$\partial_{x_i}\partial_{x_j}u(x)=\int_{R^n}\Phi(y)(\partial_{x_i}\partial_{x_j})[f(x-y)-m^2u(x-y)]$$
to give \begin{equation}\Delta_xu(x)=\sum^n_{i=1}\partial^2_{x_i}u(x)=\int_{R^n }\Phi(y)\Delta_x[f(x-y)-m^2u(x-y)]dy\end{equation}
To show expression (4) equals $f-m^2u$, we deconstruct the integral into two parts
$$\Rightarrow \int_{B_\epsilon(0)} \Phi(y)\Delta_x[f(x-y)-m^2u(x-y)]dy+\int_{R^n\backslash B_\epsilon(0)} \Phi(y)\Delta_x[f(x-y)-m^2u(x-y)]dy$$
We have the following estimate for the first integral
\begin{equation}\Big|\int_{B_\epsilon(0)} \Phi(y)\Delta_x[f(x-y)-m^2u(x-y)]\Big|dy\end{equation} \begin{equation}\leq \int_{B_\epsilon(0)} \Phi(y)\Big|\Delta_x[f(x-y)-m^2u(x-y)]\Big| dy\end{equation} \begin{equation} \leq sup_{y\in B_\epsilon(0)}\Big|\Delta_x[f(x-y)-m^2u(x-y)] \Big|\times \int_{B_\epsilon(0)}\Phi(y)dy\end{equation}
where after a computation in polar coordinates, we observe expression (7) is the size of the largest second derivative times $-\epsilon ^2$.
$$\Rightarrow C(n)||D^2[f(x-y)-m^2u(x-y)]||_{L^\infty(R)}\times -\epsilon^2\rightarrow 0$$
We let $\epsilon\rightarrow 0$ and see that the term goes to $0$.

Next we evaluate the integral in expression (6)
$$\int_{R^n\backslash B_\epsilon(0)} \Phi(y)\Delta_x[f(x-y)-m^2u(x-y)]dy$$$$\Rightarrow \int_{R^n\backslash B_\epsilon(0)} \Phi(y)\Delta_y[f(x-y)-m^2u(x-y)]dy$$
Using integration by parts gives
$$\Rightarrow \int_{\partial\epsilon(0)} \Phi(y) <D_y [f(x-y)-m^2u(x-y)], v(y)> ds(y) $$
\begin{equation} -\int_{R^n\backslash B_\epsilon(0)} <D_y\Phi(y), D_y [f(x-y)-m^2u(x-y)]> dy\end{equation}
We have the following estimate for expression (5)
\begin{equation}\int_{\partial B_\epsilon(0)} \Phi(y) <D_y [f(x-y)-m^2u(x-y)], v(y)> ds(y)\end{equation} \begin{equation}\leq C(n)\Big|\Big|D[f(x-y)-m^2u(x-y)]\Big|\Big|_{L^\infty(R)}\times \epsilon \end{equation}
we send $\epsilon\rightarrow 0$ and see that expression (10) goes to $0$.
Next, we compute expression (9) by integration by parts
\begin{equation}\int_{R^n\backslash B_\epsilon(0)} \Delta \Phi(y) [f(x-y)-m^2u(x-y)]> dy\end{equation} \begin{equation}-\int_{\partial B_\epsilon(0)} <D_y\Phi(y), v(y)> [f(x-y)-m^2u(x-y)] ds(y)\end{equation}
and see expression (11) goes to zero on $R\backslash B_\epsilon(0)$. For expression (12), we compute
$$v(y)=-\frac{y}{\epsilon}$$
$$\Phi(y)=\frac{1}{4\pi}\frac{exp(-m||x||)}{||x||}$$
$$D\Phi(y)=-\frac{y}{4\pi \epsilon^n}$$ to give
$$=-\int_{\partial B_\epsilon(0)} \frac{1}{\epsilon^n}\frac{1}{\epsilon}<y,y> [f(x-y)-m^2u(x-y)] dy$$
$$=\frac{1}{nw_ne^{n-1}}\int_{\partial B_\epsilon(0)}[f(x-y)-m^2u(x-y)] dy$$
where we let $\epsilon\rightarrow 0$ and finish our proof by stating $-\Delta u = f-m^2u$

 

[Nino MMP]

(4) To show the function $$\Phi(x)=\frac{1}{4\pi}\frac{e^{-m||x||}}{||x||}$$is a fundamental solution, we must show that it satisfies the equation\begin{equation}\Delta_x \Phi(x)+m^2\Phi(x)=0\end{equation}We compute$$\Delta_x \Phi(x)=\frac{1}{4\pi}\Delta_x\Big(\frac{e^{-m||x||}}{||x||}\Big)$$$$= \frac{1}{4\pi}\Big[\frac{\Delta_x(e^{-m||x||})}{||x||}-\frac{e^{-m||x||}}{||x||^3}<\nabla ||x||,\nabla ||x||>\Big]$$$$=\frac{1}{4\pi}\Big[\frac{-m^2e^{-m||x||}}{||x||}+\frac{e^{-m||x||}}{||x||^3}n||x||^2\Big]$$$$=\frac{1}{4\pi}\frac{-m^2e^{-m||x||}}{||x||}$$Adding $m^2\Phi(x)$ to expression (13) gives$$\Delta_x\Phi(x)+m^2\Phi(x)=\frac{1}{4\pi}\frac{-m^2e^{-m||x||}}{||x||}+\frac{1}{4\pi}\frac{m^2e^{-m||x||}}{||x||}=0$$Which shows that the function $\Phi$ is indeed a Fundamental Solution to the PDE.