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

In summary, the conversation discusses the application of radial solutions to Laplace's equation in order to rewrite the partial differential equation in terms of a linear second order ordinary differential equation with non-constant coefficients. The function v(r) = e^(-mr)/r is then used to check that the rewritten equation is satisfied, and the fundamental solution to the PDE is defined using the function Φ(x) = (1/4π)(e^(-m||x||))/||x||. The proof of Theorem 3.18 is also discussed, which shows that the function Φ(x) is a fundamental solution to the PDE.
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Homework Statement
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Relevant Equations
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Screen Shot 2021-03-09 at 12.54.02 AM.png

(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##
 
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  • #2
(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.
 

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

1. What is the Yukawa PDE equation?

The Yukawa PDE equation, also known as the Yukawa potential equation, is a partial differential equation that describes the behavior of a scalar field in the presence of a point source. It is commonly used in theoretical physics, particularly in quantum field theory and particle physics.

2. Who discovered the Yukawa PDE equation?

The Yukawa PDE equation was first proposed by Japanese physicist Hideki Yukawa in 1935. He received the Nobel Prize in Physics in 1949 for his work on the theory of nuclear forces, which included the development of the Yukawa potential equation.

3. What is the significance of the Yukawa PDE equation in physics?

The Yukawa PDE equation is significant because it provides a mathematical framework for understanding the forces between subatomic particles. It has been used to study the strong nuclear force, which holds protons and neutrons together in an atomic nucleus, and the weak nuclear force, which is responsible for radioactive decay.

4. How is the Yukawa PDE equation solved?

The Yukawa PDE equation is a second-order linear partial differential equation, and its solutions can be found using various mathematical techniques, such as separation of variables, Fourier transforms, and Green's functions. It can also be solved numerically using computer algorithms.

5. What are some applications of the Yukawa PDE equation?

The Yukawa PDE equation has applications in a wide range of fields, including particle physics, nuclear physics, astrophysics, and condensed matter physics. It has been used to study the behavior of quarks and gluons in quantum chromodynamics, the interactions between nucleons in nuclear physics, and the dynamics of electrons in materials. It also plays a role in the study of black holes and the early universe.

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