May 9th, Thoughts: A Comparison and a Trick

Recall from my previous post: https://menagerieofmathematics.wordpress.com/2025/05/09/may-8th-invariance-of-laplacians-spherically-symmetric-harmonics-in-r2-and-r3/

I have shown that the general form for the spherically symmetric harmonics in $\mathbb{R}^2, \mathbb{R}^3$ are:

$f(x, y) = C_1log(\sqrt{x^2+y^2} + C_2$ and

$f(x, y, z) = \frac{C_1}{\sqrt{x^2 + y^2 + z^2}} + C_2$ , respectively, where $C_1, C_2 \in \mathbb{R}$ .

Now, we will look at the spherically symmetric solutions of $\Delta u = \delta(x)$ , where $\delta(x)$ is defined to be the limit of a sequence of $\delta$ -shaped functions (to be rigorous). But I find the informal definition Arnold gave: the function that equals 0 when $x \neq 0$ and infinity at 0, to be much more applicable. It is this latter definition that we shall use.

So, how do we know such spherically symmetric solutions exist?

Notice that $\delta(x)$ and $\Delta$ are both invariant under rotation (the latter was proven in a previous post). Hence, if $u_0(x)$ is a solution of $\Delta u = \delta(x)$ , it follows that $u_0(R(x))$ is also a solution, where $R \in SO(n)$ . It is also not difficult that a linear combination of such solutions (the coefficients sum to 1) is also a solution. Passing to the limit, we can average a given solution over the group of all rotations to obtain a solution that is invariant under rotations, i.e. spherically symmetric. Thus, such solutions indeed exist.

We now shall find the general form of such solutions in $\mathbb{R}^n$ . Since such solutions depend only on r, it follows that the Laplacian becomes a second-order equation:

$u'' + A(r)u' + B(r)u = \delta(r)$ .

And for $r > 0$ , the above equation becomes:

$u'' + A(r)u' + B(r)u = 0$ (this follows from definition of the $\delta$ -function).

Of course, we can expand the Laplacian to find the coefficients A(r), B(r), but that would be awful. Instead, like solving coefficients using known solutions for algebraic equations, we can do the same things for this differential equation. Isn’t this lovely?

It is obvious that $u(r) = 1$ is a solution (in fact, any constant works). This forces $B(r) = 0$ . Hence, the equation simplifies to:

$u'' + A(r)u' = 0$ .

So now, what do we do?

We observe the properties of the $\delta$ -function, first in $\mathbb{R}$ . Notice that $\delta(2x) = \frac{1}{2}\delta(x)$ (works for any constant). We can argue this as follows: for any $\delta$ -shaped function, say $\delta_{\epsilon}(x)$ , notice that the integral of $\delta_{\epsilon}(2x)$ is a half of the original and so by passing to the limit, the integral of $\delta(2x)$ is also a half of $\delta(x)$ . This means $\delta(2x) = \frac{1}{2}\delta(x)$ .

Ok, so this tells us that $\delta(x)$ is homogeneous of degree -1.

We can kind of see (if you are feeling not in the mood to be rigorous) that the $\delta$ -function in $\mathbb{R}^n$ can be defined as $\delta(x_1, ..., x_n) = \delta(x_1)...\delta(x_n)$ (here, we assume $0 \cdot \infty = 0$ ).

And so, this means that $\delta$ -function in $\mathbb{R}^n$ is homogeneous of degree -n. Thus, the functions that satisfy $\Delta u = \delta(x)$ would need to be homogeneous of degree 2-n. And so, the spherically symmetric solutions we are searching for would be proportional to $r^{2-n}$ .

We now show that $r^{2-n}$ is a solution. The method is very much the same as the one I used to prove the 3-dim. case in: https://menagerieofmathematics.wordpress.com/2025/05/07/may-6th-reading-thoughts-1-r-harmonic-function-relation-between-harmonic-and-holomorphic-functions-space-of-harmonic-polynomials/. I’m currently feeling much too lazy to repeat the proof.

Hence, it follows that the general solution is of the form $\frac{C_1}{r^{n-2}} + C_2$ . Plugging this into the equation, we can also find $A(r)$ .

Now, notice that the spherically symmetric solutions that we obtained for $\Delta u = \delta(x)$ coincide $\Delta u = 0$ for both the 2, and 3-dim. Is this suppose to be obvious?

May 9th, Thoughts: A Comparison and a Trick

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