# Reduce PDE to ODE

Maybe you don't want to check all the details, but could look at a few equations here. Would you mind leaving a comment that you at least some part looks okay?- This way, I know that at least somebody checked it, too (although maybe not every step).

I want to reduce the PDE

$$\sum_{i=1}^{n} \partial_i \left( \frac{\partial_i u(x)}{\sqrt{1+ \sum_{j=1}^{n} ( \partial_j u(x)^2 ) }} \right)=0$$

to an ODE by looking at solutions of the form $u(x):=f(||x||).$

We have $\partial_i u(x) = f'(||x||) \frac{x_i}{||x||}$ and $\partial_{i,j}u(x) = f''(||x||) \frac{x_i x_j}{||x||^2} + f'(||x||) \frac{\delta_{i,j}||x||^2 - x_j x_i }{||x||^3}.$

Applying the product rule to the PDE and multiplying by the square root in the demoninator, I get

$$\left(\sum_{i=1}^{n} \partial_i^2 u(x) \right) \left(1+ \sum_{j=1}^{n} ( \partial_j u(x)^2) \right) - \left(\sum_{i,k=1}^n \partial_i u(x) \partial_k u(x) \partial_{i,k} u(x) \right) =0$$

Rewriting gives

$$\left(\sum_{i=1}^{n} \partial_i^2 u(x) \right) = f''(||x||)+f'(||x||) \frac{(n-1)}{||x||}$$

$$\left(1+ \sum_{j=1}^{n} ( \partial_j u(x)^2) \right) = 1+f'(||x||)^2$$

and finally

$$\left(\sum_{i,k=1}^n \partial_i u(x) \partial_k u(x) \partial_{i,k} u(x) \right) = f'(||x||)^2 f''(||x||).$$

Is this all correct?

Yes. What you have computed all appear to be correct.

Just for sanity check you can simplify to

$$f'' + \frac{n-1}{\|x\|} [1 + (f')^2] f' = 0$$

When $n = 2$ a solution should be $f = \cosh^{-1}(\|x\|)$. (This is because what you wrote down is the minimal surface equation and we know that one of them is given by the catenoid.) We know that for this $f$

$$f' = \frac{1}{\sqrt{r^2 - 1}}$$

and

$$f'' = -\frac{ r}{(r^2 - 1)^\frac32}$$

so you have confirmation that your formula works at least when $n = 2$.