Finding all radial solutions for biharmonic equation $\Delta^2u=0$ in $n$ dimensions

Question

I need help finishing the proof:

Let $u(x) = v(|x|)$ be a solution of the problem $\Delta^2u = 0$. Where $\Delta^2u = \Delta(\Delta(u))$ is the Bilaplacian operator.

We have by standart calculus that $$\frac{\partial}{\partial x_i}v(|x|) = v'(|x|)\frac{x_i}{|x|} \implies \nabla v(|x|) = v'(|x|)\frac{x}{|x|}.$$ $$\frac{\partial^2}{\partial x_i^2}v(|x|) = v''(|x|)\frac{x_i^2}{|x|^2}+v'(|x|)\left[\frac{1}{|x|}+\frac{x_i^2}{|x|^3}\right] \implies \Delta v(|x|) = v''(|x|)+v'(|x|)\frac{(n-1)}{|x|}.$$

and repeating the same calculations with $\Delta v$, using $r$ in place of $|x|$ (to shorten the notation) will result in:

$$\Delta^2 v(r) = v^{(4)}(r)+2(n-1)\frac{v'''(r)}{r}+(n-3)(n-1)\frac{v''(r)}{r^2}-(n-3)(n-1)\frac{v'(r)}{r^3}$$

As $v(r)$ is solution of the biharmonic equation, multiplying the relation above by $r^{n-1}$ and rearranging the therms will give

$$0 = r^{n-1}\Delta^2 v(r) = \left(v^{(4)}(r)r^{n-1}+(n-1)v'''(r)r^{n-2}\right)+(n-1)\left(v'''(r)r^{n-2}+(n-2)v''(r)r^{n-3}\right)-(n-1)\left(v''(r)r^{n-3}+(n-3)v'(r)r^{n-4}\right)$$ $$ = \frac{d}{dr}\left(v'''(r)r^{n-1}\right)+(n-1)\frac{d}{dr}\left(v''(r)r^{n-2}\right)-(n-1)\frac{d}{dr}\left(v'(r)r^{n-3}\right)$$ $$=\frac{d}{dr}\left(v'''(r)r^{n-1}+(n-1)v''(r)r^{n-2}-(n-1)v'(r)r^{n-3}\right) = 0$$

Therefore there's a constant $c_1$ such that

$$v'''(r)r^{n-1}+(n-1)v''(r)r^{n-2}-(n-1)v'(r)r^{n-3} = c_1$$ $$\frac{d}{dr}\left(v''(r)r^{n-1}\right)-(n-1)v'(r)r^{n-3} = c_1$$

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How do I continue from here? The calculations I'm doing are the same used to find the radial solution of the laplace equation. On this point with just the laplacian we would be able to solve the resulting ODE easily. But in this case it seems more complicated. Is there a product I'm missing that could help solving the remaining equation?

Thanks in advance.

Answer 1

We can define $\Delta$ in a bit different way and it will help you to find a general solution. If the system is spherically symmetric we have to deal only with one variable and $f(\vec{r})=f(r)$.

$\Delta{f(r)}=div(\vec\nabla{f(r)})$; gradient has only one component: $\vec\nabla{f(r)}=\vec{e_r}\frac{\partial}{\partial{r}}f(r)$

We can get the expression for $\Delta{f(r)}$ in $n$ dimensions in different ways; for instance, from the Stokes theorem for an arbitrary radial vector $\vec{A(r)}$

$div({A(r)}\vec{e_r})\sim\lim_{\delta{r}\to0}\frac{1}{r^{n-1}\delta{r}}\left((r+\delta{r})^{n-1}A(r+\delta{r})-r^{n-1}A(r)\right)=\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}A(r)\right)$

If we take $\vec{A(r)}=\vec\nabla{f(r)}$ we get (accurate to constant factors) $$\Delta{f(r)}\sim\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)$$

(you can check this expression, for instance, for $n=3$)

And $$\Delta\Delta{f(r)}\sim\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}\left(\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)\right)\right)=0$$

Integrating step by step we get

$$r^{n-1}\frac{\partial}{\partial{r}}\left(\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)\right)=C_1$$

$$\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)=\frac{C_1}{2-n}r^{2-n}+C_2$$ and so on (you can easily do it yourself).

Finally, I got (please, check it!) $$f(r)=\frac{C_1}{2(2-n)(4-n)}r^{4-n}+\frac{C_2}{3n}r^{2}+\frac{C_3}{2-n}r^{2-n}+C_4$$

The uncertainties at points $n=2$ and $n=4$ could be disclosed and give additional terms $\sim\log(r)$