Physically Interesting Differential Equations 2: The Dampened Harmonic Oscillator

Welcome to the second article in the series: Physically Interesting Differential Equations, where we explore fascinating physical systems that can be modeled with differential equations. This week, we shall look at the Poisson equation. The Poisson equation is a class of partial differential equations that are often useful when doing physics of fields. One such example is the electrostatic field. The Poisson equation for a electrostatic field takes the following form:

\begin{equation}\grad ^2\phi=-4\pi\rho\end{equation}

where $\rho$ is the charge density. 

Upon first glance, this seems like an easy enough differential equation to solve, that is, unless you have taken into the partial nature of the DEs. We can solve this equation using Green’s Function. 

This seemed to be a common example when introducing the idea of solving PDEs with Green’s Functions. This article concerns itself with Fourier transforms of the following forms:

This short document shall provide a concise but nevertheless complete step-by-step account for arriving at the Poisson Equation of the following form:

$$\nabla^2\phi(x,y,z)=-4\pi\rho(x,y,z)$$

This seemed to be a common example when introducing the idea of solving PDEs with Green’s Functions. This document concerns itself with Fourier transforms of the following forms:

$$\mathscr{F}f(x)=\hat{f}(\kappa)=\int_{-\infty}^{\infty}\frac{d\kappa}{2\pi}e^{-i\kappa x}f(x)$$

$$\mathscr{F}^{-1}\hat{f}(\kappa)=f(x)=\int_{-\infty}^{\infty}\frac{d\kappa}{2\pi}e^{i\kappa x}\hat f(x)$$

Please note that depending on your area of study or the reference you consult, your mileage will vary when it comes to the sign/factors of $\pi$.

\section{Workings}

It is helpful to think of multi-dimensional functions, in our case the electrostatic potential at a point $\phi(x,y,z)$ and our charge density at a point $\rho(x,y,z)$, as functions of a single three dimensional vector, i.e.:

$$\vec{\kappa}=\begin{bmatrix} \kappa_x\\ \kappa_y\\ \kappa_z\end{bmatrix} $$

And vector $\vec{x}$ for the function prior to transformation

$$\vec{x}=\begin{bmatrix} x\\ y\\ z \end{bmatrix} $$

We first express our functions in Fourier bases.

$$\phi(x,y,z)= \phi(\vec{x}) =\int_{-\infty}^\infty\frac{d\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\phi}\vec{x}$$

$$\rho(x,y,z)= \rho(\vec{x}) =\int_{-\infty}^\infty\frac{d\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\rho}\vec{x}$$

We may now substitute our newly transformed functions back into our equation. 

$$\nabla^2\int_{-\infty}^\infty\frac{d\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\phi}(\vec{\kappa})=4\pi\int_{-\infty}^\infty\frac{d\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\rho}\vec{x}$$

Since we are working in three dimensions, we should update our integrals to reflect that:

$$\nabla^2\int_{-\infty}^\infty\frac{d^3\kappa}{(2\pi)^3}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\phi}(\vec{\kappa})=4\pi\int_{-\infty}^\infty\frac{d^3\kappa}{(2\pi)^3}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\rho}\vec{x}$$


Let us expand our left hand side of the equation, and rewrite the dot product in the exponential in its full form:


$$\left ( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right )\int_{-\infty}^\infty\frac{d^3\kappa}{2\pi}e^{i(\kappa_xx+\kappa_yy+\kappa_zz)}\hat{\phi}\left (\vec{x}\right)$$

We can now calculate the partial differentials of each variables in space separately:

$$\frac{\partial^2\phi}{\partial x^2}=i^2\kappa_x^2=-\kappa_x^2$$

$$\frac{\partial^2\phi}{\partial y^2}=i^2\kappa_y^2=-\kappa_y^2$$

$$\frac{\partial^2\phi}{\partial z^2}=i^2\kappa_z^2=-\kappa_z^2$$

Putting that back in gives:

$$\int_{-\infty}^\infty\frac{d^3\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\phi}\left (\vec{x}\right)\left(-\kappa_x^2-\kappa_y^2-\kappa_z^2\right)$$

Simplifying gives:

$$\int_{-\infty}^\infty\frac{d^3\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\phi}\left (\vec{x}\right)(-|\vec{\kappa}|^2)$$

We now move the right hand side integral to the left and combine the two:

$$\int_{-\infty}^\infty\frac{d^3\kappa}{2\pi}e^{i\vec{\kappa}\cdot\vec{x}}\hat{\phi}\left (\vec{x}\right)\left[-|\vec{\kappa}|^2+4\pi\hat\rho(\vec{\kappa})\right]=0$$

If you can see it, then hooray! This is exactly the Fourier transform of the following function:

$$\hat\phi(\vec{\kappa})=\frac{4\pi\hat\rho(\vec{\kappa})}{\kappa ^2}$$

We know that the Dirac Delta has the following property:

$$\int f(t)\delta_D(t-t_0)dt=f(t_0)$$

This essentially let us pick specific values from a function, which is very useful for finding the Green’s Function for this PDE. 

$$\nabla^2\mathscr{G}=-4\pi\delta_D$$

$$\mathscr{\hat{G}}(\vec{\kappa})=\frac{4\pi}{\kappa^2}$$

The Green’s Function can then be solved:

$$\mathscr{G}(\vec{x})=\int_{-\infty}^{\infty}\frac{d^3\kappa}{(2\pi)^3}\frac{4\pi e^{i\vec{\kappa}\cdot\vec{x}}}{\kappa^2}$$

We may now work in spherical coordinates for ease of calculations:

$$\frac{4\pi}{(2\pi)^3}\int_0^\infty d\kappa\int_0^\pi d\theta \int_0^{2\pi}d\varphi \kappa^2\frac{e^{i\vec{\kappa}\cdot\vec{x}}}{\kappa^2}$$

And it is excruciatingly obvious that our function does not have any $\varphi$ dependence, whose integral we can immediately demolish, leaving nothing but a factor of $2\pi$.

$$\frac{1}{\pi}\int_0^\infty d\kappa\int_0^\pi d\theta \kappa^2 \sin{\theta}\frac{e^{i\kappa x\cos{\theta}}}{\kappa^2}$$

Integration by substituting $z=\cos\theta$ and $dz=-\sin\theta d\theta$:

$$\frac{1}{\pi}\int_0^\infty d\kappa\int_{-1}^{1} dz -\kappa^2 \frac{e^{i\kappa xz}}{\kappa^2}$$

We cancel the $\kappa^2$ and manipulate the integration boundaries to rid of the minus sign:

$$\frac{1}{\pi}\int_0^\infty d\kappa\int_{1}^{-1} dz e^{i\kappa xz}$$

Author

Leave a Reply

To write mathematics, simply enclose \(\LaTeX\)in \( \) ext{and}\)\).