Solving Partial Differential Equations¶

Method of Separation of Variables¶

Consider the diffusion equation, ([eq:diffusion]), and look for solutions with the form

linear and have no cross-terms. So we now have

So now we have

Separation of Variables in spherical and cylindrical coordinate systems¶

We can decompose Laplace’s equation, equation ([eq:laplace]), into three equations. In spherical coordinates Laplace’s equation becomes

$$\begin{array}{r}\begin{array}{c}{\sin }^2\theta \frac{1}{R}\frac{\partial }{\partial r}\left[r^2\frac{\partial R}{\partial r}\right]+\sin \theta \frac{1}{\mathrm{\Theta }}\frac{\partial }{\partial \theta }[\sin \theta \frac{\partial \mathrm{\Theta }}{\partial \theta }]=-\frac{1}{\mathrm{\Phi }}\frac{\text{difp}2\mathrm{\Phi }}{\text{dif}{\mathrm{\Phi }}^2}\\ Eachsidemustbeconstant,so\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\begin{array}{c}\begin{array}{rl}\frac{1}{\mathrm{\Phi }}\text{dv}[2]\mathrm{\Phi }\varphi & =-m^2\\ {\sin }^2(\theta )\frac{1}{R}\text{dv}r\text{qty}[r^2\text{dv}Rr]+\sin (\theta )\frac{1}{\mathrm{\Theta }}\text{dv}\theta \text{qty}[\sin (\theta )\text{dv}\mathrm{\Theta }\theta ]& =m^2\end{array}\end{array}\\ andthesecanthemselvesbemadeequaltoaconstant,\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\begin{array}{rl}\frac{1}{R}\text{dv}r\text{qty}[r^2\text{dv}Rr]& =-\frac{1}{\sin (\theta )}\frac{1}{\mathrm{\Theta }}\text{dv}\theta \text{qty}[\sin (\theta )\text{dv}\mathrm{\Theta }\theta ]+\frac{m^2}{{\sin }^2(\theta )}=k\end{array}\\ The:math:‘R(r)‘equation:\ast \end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\begin{array}{c}\begin{array}{rl}\text{dv}r\text{qty}[r^2\text{dv}R(r)r]-kR(r)& =0\\ r^2\text{dv}[2]R(r)r+2r\text{dv}R(r)r-kR(r)& =0\end{array}\end{array}\\ Thecoefficientsarepolynomialsof:math:‘r‘,sowetryasolution\end{array}\end{array}$$

of the form $R(r)=r^n$,

$$\begin{array}{r}\begin{array}{c}\begin{array}{c}\begin{array}{rl}{r}^{2}n(n-1)r^{n-2}+2rn{r}^{n-1}-k{r}^n& =0\\ n(n-1)r^n+2n{r}^n-k{r}^n& =0\\ n(n+1)-k& =0\end{array}\end{array}\\ andnowlet:math:‘k=l(l+1)‘,so:math:‘n(n+1)=l(l+1)‘,thus\end{array}\end{array}$$

$n=l$ or :math:`n =

-l-1` making the general solution

$$\begin{array}{r}\begin{array}{c}R(r)=A{r}^l+B{r}^{-l-1}\\ Theequationfor:math:‘\mathrm{\Phi }(\varphi )‘\ast \end{array}\end{array}$$

$$\text{dv}[2]\mathrm{\Phi }(\varphi )\varphi =-m^2\mathrm{\Phi }(\varphi )$$

The solution must have the form $\mathrm{\Phi }(\varphi )=e^{\alpha \varphi }$, so

$$\begin{array}{r}\begin{array}{c}{\alpha }^2\mathrm{\Phi }(\varphi )=-m^2\mathrm{\Phi }(\varphi )\\ so:math:‘\alpha =\pm e^{\alpha \varphi }‘.Thus,thegeneralsolutionis\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\mathrm{\Phi }(\varphi )=A^{\prime }\sin (m\varphi )+B^{\prime }\cos (m\varphi )\\ forconstants:math:‘A^{\prime }‘,andB:math:{‘}^{\prime }‘,and\end{array}\end{array}$$

:math:`m in

mathbb{N}`, since $\mathrm{\Phi }(\varphi +2\pi )=\mathrm{\Phi }(\varphi )$.

$$\sin (\theta )\text{dv}\theta \text{qty}[\sin (\theta )\text{dv}\mathrm{\Theta }(\theta )\theta ]-m^2\mathrm{\Theta }(\theta )+l(l+1){\sin }^2(\theta )\mathrm{\Theta }(\theta )=0$$

which has the form of the Associate Legendre Differential equation, equation ([eq:assoclegendrede]), and the solution is therefore an Associate Legendre polynomial, with a general solution of the form

$$\begin{array}{r}\begin{array}{c}\mathrm{\Theta }(\theta )=A{P}_l^m\text{qty}(\cos (\theta ))\\ Thus,thegeneralsolutionof\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}{\mathrm{\nabla }}^2\psi (r,\theta ,\varphi )=0\\ is\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\psi (r,\theta ,\varphi )=\sum _{l=0}^{\mathrm{\infty }}\sum _{m=0}^l(A_{lm}{r}^l+B_{lm}{r}^{-l-1})P_l^m\text{qty}(\cos \theta )e^{\pm im\varphi }\\ :math:‘A‘,and:math:‘B‘areconstantsdeterminedbytheboundary\end{array}\end{array}$$

conditions of the problem. The functions

$$\begin{array}{r}\begin{array}{c}{Y}_l^m(\theta ,\varphi )=N{e}^{im\varphi }{P}_l^m(\cos (\theta ))\\ with:math:‘l\ge 0‘,and:math:‘|m|\le l‘,are\ast Spherical\end{array}\end{array}$$

Harmonics*.

$$\begin{array}{r}\begin{array}{c}\psi (r,\theta ,\varphi )=\sum _{l=0}^{\mathrm{\infty }}\sum _{m=0}^l{A}_{lm}{r}^l{P}_l^m\text{qty}(\cos \theta )e^{\pm im\varphi }\\ andwehavetheboundaryconditionthat\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\psi (r_0,\theta ,\varphi )={\varphi }_0\cos (\theta )\\ whichhasno:math:‘\varphi ‘dependence,implying:math:‘m=0‘.Then,\end{array}\end{array}$$

$$\begin{array}{r}\begin{array}{c}\psi (r,\theta ,\varphi )=\sum _{l=0}^{\mathrm{\infty }}{A}_l{r}^l{P}_l\text{qty}(\cos \theta )={\varphi }_0\cos (\theta )\\ soonly:math:‘l=1‘cansatisfythisequation,andthusthepotential\end{array}\end{array}$$

is

$$\begin{array}{r}\begin{array}{c}\psi (r,\theta ,\varphi )={\psi }_0\frac{r}{r_0}\cos (\theta )\\ Outsidethesphere\ast \end{array}\end{array}$$

$$R(r)\propto r^l$$

would give an infinity as :math:`r to

infty`. Thus $A_{lm}=0$. The same arguments for angular

depenence also apply, so,

$$\psi (r,\theta ,\varphi )={\psi }_0\text{qty}(\frac{r_0}{r}{)}^2\cos (\theta )$$