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Exercises for March 23rd 2012 entry

March 24, 2012

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}.


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.


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}.


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.


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.


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}).


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

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



From → Analysis, Exercices

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