Euler-Cauchy equation

While trying to solve for the Laplacian in polar coordinates, one encounters the famous Euler-Cauchy differential equation.

x^{2}{\frac  {d^{2}y}{dx^{2}}}+2x{\frac  {dy}{dx}}-l(l+1)y=0

or

{\frac {d^{2}y}{dx^{2}}}+ \frac{2}{x}{\frac {dy}{dx}}- \frac{l(l+1)}{x^2}y=0

How does one find a solution to this differential equation ? Well, most places that I have read simply dictate that the solutions to this differential equation as y = x^l, x^{-(l+1)} without explaining why only these two are the only solutions. The purpose of this post is to explain why!

We can clearly see that x = 0 is a singular point. Therefore a simple power series solution won’t work. Hence we use the frobenius method to get rid of the singularity i.e

y= \sum\limits_{n=0}^{\infty} a_{n}x^{n + \lambda}

Let’s now compute its derivatives wrt x.

\frac{dy}{dx}= \sum\limits_{n=0}^{\infty} (n+\lambda)a_{n}x^{n +\lambda -1}

\frac{d^{2}y}{dx^{2}}= \sum\limits_{n=0}^{\infty} (n + \lambda)(n + \lambda -1)a_{n}x^{n + \lambda -2}

Putting the values for y,\frac{dy}{dx}  and \frac{d^{2}y}{dx^{2}}  back into the differential equation, we get the following form.

x^{2}{\sum\limits_{n=0}^{\infty} (n + \lambda)(n + \lambda -1)a_{n}x^{n-2}}

+2x{\sum\limits_{n=0}^{\infty}(n+\lambda)a_{n}x^{n-1}}

- l(l+1){\sum\limits_{n=0}^{\infty} a_{n}x^{n}}=0

Bringing the x,x^{2}  terms inside the summation, we obtain the following form:

\sum\limits_{n=0}^{\infty}x^n \left({(n+\lambda)(n+\lambda -1)a_n + 2(n+\lambda)a_n -l(l+1)a_n}\right)=0

\sum\limits_{n=0}^{\infty}a_n x^n \left({(n+\lambda)(n+\lambda +1) -l(l+1)}\right) = 0

Obviously, the coefficients of the summation cannot be zero and x=0 is a trivial solution. Therefore, we get the indicial equation.

(n+\lambda)(n+\lambda+1) - l(l+1)= 0  or

(n+\lambda)(n+\lambda+1) = l(l+1)

(i) When n = 0, the indicial equation becomes

\lambda(\lambda+1) = l(l+1)

The values of the \lambda that solve for this equation are l,-(l+1)

\lambda = l , -(l+1)

(ii) When n = 1, the indicial equation becomes

(\lambda + 1)(\lambda+2) = l(l+1)

The values of the \lambda that solve for this equation are (l-1),-(l+2)

\lambda = (l-1) , -(l+2)

\vdots

And in general:

\lambda_n = (l-n), -(l+n+1)

\lambda_{n_1} = (l-n)  , \lambda_{n_2} = -(l+n+1)

Here is the crux of it all.
Lets now write down the solution for this differential equation in its glory:

y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{n + \lambda_{n_1}} + b_n x^{n+ \lambda_{n_{2}}}\right)

y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{n + (l-n)} + b_n x^{n-(l+n+1)}\right)

y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{l} + b_n x^{-(l+1)}\right)

y = \left(a_0 + a_1 + a_2 + \hdots \right) x^{l} + \left(b_0 + b_1 + b_2 + \hdots \right) x^{-(l+1)}

y = A x^{l} + B x^{-(l+1)}

where A and B are constants.

This is the general solution to the Euler-Cauchy differential equation. 😀

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