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### Separation of Variables

#### Discussion

Consider Laplace's equation on the disk, which we discussed in the last section. $$ \frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r} +\frac{1}{r^2}\frac{\partial^2 u}{\partial\theta^2} = 0 $$ We want to break this partial differential equation in two variables into two ordinary differential equations (each in one variable) so that we can use the techniques we've learned earlier to build a solution. I doubt you will be surprised that the first thing we do is guess the form of the solution, then try to fit that form to solve the equation. In this case, we guess that the function $u(r,\theta)$ is the product of two one-variable functions, $$ u(r,\theta)=R(r)\Theta(\theta) $$ where $\Theta$ is a capital $\theta.$ Then $$ \begin{align} \frac{\partial u}{\partial r} &= \frac{\partial R}{\partial r}\Theta + R\frac{\partial \Theta}{\partial r} \\ &= R'\Theta+0 \\ &=R'\Theta \end{align} $$ since $\Theta$ is a function of $\theta$ alone and doesn't depend on $r.$ Similarly we can compute $$ \frac{\partial^2 u}{\partial r^2} = R''\Theta \qquad\text{and}\qquad \frac{\partial^2 u}{\partial \theta^2}=R\Theta'' $$ Plugging these into Laplace's equation we get $$ \frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r} +\frac{1}{r^2}\frac{\partial^2 u}{\partial\theta^2} =R''\Theta+\frac1rR'\Theta+\frac{1}{r^2}R\Theta''=0 $$ Next we**separate the variables**by putting all the $r\text{'s}$ (and hence all the $R(r)\text{'s}$) on one side and all the $\theta\text{'s}$ (and hence all the $\Theta(\theta)\text{'s}$) on the other side. $$ \begin{align} R''\Theta+\frac1rR'\Theta+\frac{1}{r^2}R\Theta''&=0 \\ (R''+\frac1rR)\Theta&=-\frac{1}{r^2}R\Theta'' \\ \frac{r^2R''+rR'}{R}&=-\frac{\Theta''}{\Theta} \end{align} $$ Now this is a curious equation. The left-hand-side, $\displaystyle \frac{r^2R''+rR'}{R},$ depends only on $r$ and not on $\theta$. The right-hand-side, $\displaystyle\frac{\Theta''}{\Theta},$ depends only on $\theta$ and not on $r$. And yet the two sides are equal to each other. So are $r$ changes, the right-hand-side won't change (since it depends only on $\theta$ and not on $r$) and hence the left-hand-side must not change either. Similarly, as $\theta$ changes, the left-hand-side won't change (since it depends only on $r$ and not on $\theta$) and hence the right-hand-side won't change either. In other words, the two sides must be constant and not depend change with either $r$ or $\theta.$ So there is some constant $\lambda$ such that $$ \frac{r^2R''+rR'}{R}=-\frac{\Theta''}{\Theta}=\lambda. $$ Now we can work we each side separately to get two ordinary differential equations out of the one partial differential equations. $$ \begin{align} r^2R''+rR'-\lambda R &=0 \\ \Theta''+\lambda\Theta&=0 \end{align} $$ The first equation is an Euler equation and the second equation is a constant coefficient equation so we can solve these using the techniques we've learned. Then we multiply these solutions to find a $u(r,\theta)=R(r)\Theta(\theta)$ which solves Laplace's equation in the disk. Actually, we get infinitely many such $u(r,\theta)$, one for every possible choice of constant $\lambda$. Furthermore, since the equation is linear, any linear combination of solutions gives us a new solution. So we can build series not out of $x^n$ but out of our basic solutions to build more complicated solutions to Laplace's equation. We will then use the boundary values to the coefficients of this series to find the unique $u(r,\theta)$ that solves the boundary value problem. This leads to topics like Fourier series which, while beautiful, we don't have time to develop in this class. Math 632, Partial Differential Equations, will go through this in detail. The technique illustrated in this problem is very general and many different partial differential equations are solved using separation of variables.

#### Strings and Drums

As a final comment about applications of variable coefficient problems, we'll consider the tonal qualities of string instruments compared to drums. The wave equation for a string is $$ \frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial^2 u}{\partial x^2} $$ where $t$ is time, $x$ is location along the string, and $c$ is the speed of the wave through the string (which depends on density and other factors). This problem can be solved with separation of variables. When you do this, you get the two equations $$ \begin{align} T''+\lambda T&=0 \\ X''+\lambda X&=0 \end{align} $$ whose solutions are sines and cosines. In two dimensions the wave equation becomes $$ \frac{\partial^2 u}{\partial t^2}= c^2\left(\frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2}\right) $$ and if we are working on a disk (because drums are circular), then we need to rewrite this in polar coordinates as $$ \frac{\partial^2 u}{\partial t^2}=c^2\left( \frac{\partial^2 u}{\partial r^2}+\frac1r\frac{\partial u}{\partial r} +\frac{1}{r^2}\frac{\partial^2 u}{\partial\theta^2}\right) $$ When you apply separation of variables this time you will get three equations (one for $t$, $r,$ and $\theta$). And the equation for $r$ will be Bessel's equation $$ r^2R''+rR'+(r^2-\nu^2)R=0 $$ This is not a solution which has an equation in terms of the functions we already know and love like exponentials, polynomials, sines, or cosines. We will need to do a series solution to find the function that solves this equation. Even more fun, this equation has a singular point at $r=0$ and we can't avoid working at the origin, so we will end up creating a series solution about this singular point (fortunately it is a regular singular point). We call the functions we build here Bessel functions. So for the drum the solution to the wave equations involves Bessel functions while the wave equation on the string involves sines and cosines. Sines and cosines are nice periodic functions and that means the sounds they produce blend together well, which is why you hear the string sections during the romantic scenes in a movie, strings sound pretty. On the other hand, Bessel functions are not periodic so drums don't sound so pretty (but they work well during the action scenes in the movies). Note that this all comes from the geometry of a drum. If you had a nice rectangular drum, then it would sound much prettier. Unfortunately, no one has found a good way to get uniform tension of a membrane stretched over a rectangle, which is why drums are usually circular. On the other hand, if you built a drum based on stiffness rather than tension you could use rectangles or other shapes. And Caribbean steel drums do indeed feature plates that have been shaped to produce specific notes and better tonal qualities. However, stiffness leads to third order equations instead of second order equations so we won't follow this any further. But I did want to give an example of where different geometries led to different functions and different results.

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©2010, 2014 Andrew G. Bennett