Here are the solutions to the exercises of March 23rd entry, A Motivational Problem for the Study of Fourier Series.

Exercise 1: Show that in polar coordinates, the Laplacian is given by

$\bigtriangleup u = \frac {\partial ^2u}{\partial x^2} + \frac {1}{r} \frac {\partial u}{\partial r} + \frac {1}{r^2} \frac {\partial ^2u}{\partial \theta ^2}$.

Solution:

In terms of $x$ and $y$, $r$ and $\theta$ are functions given by $r = \sqrt {x^2 + y^2}$ and $\theta = arctan(\frac{y}{x})$. We thus have :

• $\frac{\partial r}{\partial x} = \frac{x}{\sqrt {x^2 + y^2}} = \cos \theta$
• $\frac{\partial r}{\partial y} = \frac{y}{\sqrt {x^2 + y^2}} = \sin \theta$
• $\frac{\partial \theta}{\partial x} = \frac {1}{1 + (y/x)^2} \frac {-y}{x^2} = \frac {-y}{x^2 + y^2} = - \frac{\sin \theta}{r}$
• $\frac{\partial \theta}{\partial y} = \frac {1}{1+(y/x)^2} \frac {1}{x} = \frac {x}{x^2 + y^2} = \frac {\cos \theta}{r}$

Therefore, by the chain rule,

1. $\frac {\partial }{\partial x} = \frac {\partial }{\partial r} \frac {\partial r}{\partial x} + \frac {\partial }{\partial \theta} \frac {\partial \theta}{\partial x} = \cos\theta\frac{\partial}{\partial r} - \frac {\sin \theta}{r}\frac {\partial }{\partial \theta}$
2. $\frac {\partial }{\partial y} = \frac {\partial }{\partial r} \frac {\partial r}{\partial y} + \frac {\partial }{\partial \theta} \frac {\partial \theta}{\partial y} = \sin\theta\frac{\partial}{\partial r} + \frac {\cos \theta}{r}\frac {\partial }{\partial \theta}$.

It is important at this point to realize that 1. and 2. are differential operators while their evaluation at $u$, for example $\frac {\partial u}{\partial x}$, are functions. So that $\frac {\partial ^2u}{\partial x^2} = \frac {\partial}{\partial x} \left(\frac{\partial u}{\partial x}\right)$ is the evaluation of 1. at the function $\frac {\partial u}{\partial x}$. With this clarified, we can write explicitly what we have to calculate :

$\frac {\partial ^2u}{\partial x^2} = \frac{\partial}{\partial x} \left(\cos\theta\frac{\partial u}{\partial r}-\frac {\sin \theta}{r}\frac {\partial u}{\partial \theta}\right)\\ \\= \cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial u}{\partial r}-\frac {\sin \theta}{r}\frac {\partial u}{\partial \theta}\right)-\frac {\sin \theta}{r}\frac {\partial}{\partial \theta}\left(\cos\theta\frac{\partial u}{\partial r}-\frac {\sin \theta}{r}\frac {\partial u}{\partial \theta}\right)$

Applying distributivity of the differential operators on sums of functions and then the Leibniz rule (product rule) when the operator acts on a product of functions, we get, after simplifying,

$\frac {\partial ^2u}{\partial x^2} = \cos ^2 \theta \frac {\partial ^2u}{\partial r^2} - \frac {2\cos \theta \sin \theta}{r} \frac {\partial ^2}{\partial r \partial \theta} + \frac {2\cos \theta \sin \theta}{r^2} \frac {\partial u}{\partial \theta} + \frac {\sin ^2 \theta}{r^2} \frac {\partial ^2u}{\partial \theta ^2} + \frac {\sin ^2 \theta}{r} \frac {\partial u}{\partial r}$

In a similar way, we obtain (using the commutativity of $\frac {\partial}{\partial r}$ and $\frac {\partial}{\partial \theta}$)

$\frac {\partial ^2u}{\partial y^2} = \sin ^2 \theta \frac {\partial ^2u}{\partial r^2} + \frac {2\cos \theta \sin \theta}{r} \frac {\partial ^2}{\partial r \partial \theta} - \frac {2\cos \theta \sin\theta}{r^2} \frac {\partial u}{\partial \theta} + \frac {\cos ^2 \theta}{r^2} \frac {\partial ^2u}{\partial \theta ^2} + \frac {\cos ^2 \theta}{r} \frac {\partial u}{\partial r}$.

Thus, summing these two expressions and using the identity $\cos ^2 \theta + \sin ^2 \theta = 1$, we get

$\bigtriangleup u = \frac {\partial ^2u}{\partial x^2} + \frac {\partial ^2u}{\partial y^2} = \frac {\partial ^2u}{\partial r^2} + \frac {1}{r^2} \frac {\partial ^2u}{\partial \theta ^2} + \frac {1}{r} \frac {\partial u}{\partial r}$

as wanted.

QED

Exercise 2: Show that if $f : \textbf{R} \rightarrow \textbf{R}$ is a twice differentiable function satisfying $f''(\theta ) + m^2 f (\theta ) = 0$ for some $m\in \textbf{Z}$, then there exists $a, b \in \textbf{C}$ such that $f (\theta) = a.e^{im\theta} + b.e^{-im\theta}$.

Solution:

We proceed in two steps. First, we show that $f$ can be written as $a' \cos m\theta + b' \sin m\theta$ for some $a', b' \in \textbf{R}$. We’ll then show how this gives us the desired result.

Let $f$ be a twice differentiable function satisfying $f''(\theta) + m^2f(\theta) = 0$. But then we have $f'' = -m^2f$ so $f''$ is also twice differentiable. It’s first and second differentials are given respectively by $f^{(3)} = -m^2f'$ and $f^{(4)} = -m^2f'' = m^4f.$ By induction, $f$ is then infinitely differentiable and we have $f^{(k)} = m^kf$ for even values of $k$ and $f^{(k)} = m^{k-1}f'$ for odd values of $k$. The Taylor series expansion of $f$ at $0$ is thus given by

$f(\theta) = f(0) + f'(0)\theta - \frac{m^2f(0)\theta ^2}{2!} - \frac{m^2f'(0) \theta ^3}{3!} + \frac{m^4f(0) \theta ^4}{4!} + \frac{m^4f'(0) \theta ^5}{5!} - \cdots$

But, summing the Taylor series representation of $\cos$ and $\sin$, we get

$\cos (\alpha x) + sin (\alpha x) = (\alpha x)^0 + (\alpha x) - \frac{(\alpha x)^2}{2!} - \frac{(\alpha x)^3}{3!} + \frac{(\alpha x)^4}{4!} + \frac {(\alpha x)^5}{5!} - \cdots$

Taking $a' = f(0)$ and $b' = f'(0)/m$, we get

$f(\theta) = a'\cos(m\theta ) + b'\sin(m\theta )$

as wanted.

To conclude, let $a = \frac{a' - ib'}{2}$ and $b = \frac {a' + ib'}{2}$. We get, with Euler’s formula,

$a.e^{im\theta } + b.e^{-im\theta } =a(\cos(m\theta) + i\sin(m\theta))+b(\cos(-m\theta) + i\sin(-m\theta)).$

But since $\cos$ is an even function and $\sin$ an odd function, this can be rewritten as

$(a+b)\cos(m\theta) + i(a-b)\sin(m\theta)$

We also have

$(a+b) = (\frac{a'-ib'}{2} + \frac{a' + ib'}{2}) = a'$ and $(a-b) = (\frac{a'-ib'}{2} - \frac{a'+ib'}{2}) = -ib'.$

So putting all this together, we get

$a.e^{im\theta } + b.e^{-im\theta } = a'\cos{m\theta} + b'\sin{m\theta} = f(\theta)$

as desired.

QED

Exercise 3: Show that if $n \in \textbf{Z}$ the only solutions of the differential equation

$r^2F''(r) + rF'(r) - n^2F(r) = 0,$

which are twice differentiable when $r > 0$, are given by linear combinations of $r^n$ and $r^{-n}$ when $n \neq 0$, and 1 and $\log r$ when $n=0$.

Solution:

If $n = 0$, then we have $r^2F''(r) + rF'(r) = 0$ so that $F''(r) = - \frac{F'(r)}{r}.$ This means that $\frac{d}{dr}F'(r) = \frac{-1}{r}F'(r)$ so $F'(r) = \frac{1}{r}$. We conclude that $F(r) = \log r + C$ for some $C \in \textbf{R}$.

In case $n \neq 0$, we make the substitution $r = e^t$. We then have

$\frac{dF}{dt} = \frac{dF}{dr}\frac{dr}{dt} = \frac{dF}{dr}e^t = r\frac{dF}{dr}.$

So we get $\frac{d^2F}{dt^2} = \frac{d}{dt}(r\frac{dF}{dr})$. By the chain rule, this can be rewritten as $\frac{d}{dr}(r\frac{dF}{dr})\frac{dr}{dt}$ which, by the product rule, is equal to $(\frac{dF}{dr} + r\frac{d^2F}{dr^2})\frac{dr}{dt}.$ But since $\frac{dr}{dt} = r$, this gives us $\frac{d^2F}{dt^2} = r^2\frac{d^2F}{dr^2} + r\frac{dF}{dr} = r^2F'' + rF'$ which is equal to $n^2F$ by our original hypothesis.

The problem thus reduces to solving the resulting differential equation $F''(t) = n^2F(t)$. But this is very similar to what we have done in exercise 2! In fact, it can be shown that there exists $a, b \in \textbf{R}$ such that $F(t) = a.e^{nt}+b.e^{-nt}$ (the argument is the same as in exercise 2 but we consider $\cosh$ and $\sinh$ instead of $\cos$ and $\sin$) so, since $r = e^t$, we have

$F(r) = a\exp(n\log r) + b\exp(-n\log r)$

which is

$a\exp(\log r^n) + b\exp(\log r^{-n}).$

Hence

$F(r) = ar^n + br^{-n}$

for some $a,b \in \textbf{R}$ as wanted.

QED