I was working on different forms of the Laplacian of a general product of scalar functions (which I might share in a later post). And I came to a point that I wanted to ask myself when is
$$\nabla{f}\cdot\nabla{f}=\Delta{f}$$
which is similar to asking if
$$\frac{\partial^2 f}{\partial x^2} = \left(\frac{\partial f}{\partial x}\right)^2$$
In fact, the latter is a special case of the former. And if the partials with respect to $x,y,z$ all follow this, then we have a term-wise equality between the dot product of the gradient and the Laplacian. But when does this happen? Well, let's start with a function of 1 variable. So we have the ODE
$$f''(x)- \left(f'(x)\right)^2=0$$
First, substitute $g(x) = f'(x)$.
$$g'-g^2=0$$
Clearly we can write the following
$$\begin{align*}
g&=g\\
g'&=g^2\\
g''=2gg'&=2g^3\\
g^{(3)}=6g^2g'&=6g^4\\
g^{(4)}=24g^3g'&=24g^5\\
&\vdots
\end{align*}$$
And in general is appears that
$$g^{(n)}=n!g^{n+1}$$
We can prove that with induction. The base cases are done. And the induction step is fairly straight forward
$$g^{(n+1)} = \left(g^{(n)}\right)' = \left(n!g^{n+1}\right)' = \left(n+1\right)n!g^ng' = \left(n+1\right)!g^{n+2}$$
Let's try to write the Taylor series for $g$ now letting $g_0=g(x_0)$.
$$g(x) = \sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}\left(x-x_0\right)^n = \sum_{n=0}^\infty\left(g_0\right)^{n+1}\left(x-x_0\right)^n=g_0\sum_{n=0}^\infty\left(g_0\left(x-x_0\right)\right)^n$$
This last part looks an awful lot like the geometric series $(1-x)^{-1}$. But we shift $x$ by $x_0$ scale the value by $g_0$. So really we have
$$g(x) = \frac{g_0}{1-g_0\left(x-x_0\right)}=\frac{-a}{ax+b}$$
where $a=-g_0$ and $b=1+g_0x_0$. Now since we said that $g=f'$ we just integrate once to find $f$
$$f(x) = \int\frac{-a}{ax+b}\,\mathrm dx$$
Simple $u$-substitution for $u=ax+b$ yields
$$f(x) = c - \ln(ax+b)$$
Double checking reveals that this does, in fact, solve our ODE. We could also separate variables and integrate to arrive at this same answer. Also, a buddy of mine shared the intuition almost immediately that $\ln(x)$ is a basic function that gets us close to the solution and putting a negative out front fixes the one sign issue that shows up. So we can just guess that our function looks like
$$f(x) = -\ln(g(x))$$
for some function $g(x)$. Putting this into our original equation yields
$$\frac{-g''(x)g(x)+g'(x)^2}{g(x)^2}=\frac{g'(x)^2}{g(x)^2}$$
So either $g(x) = 0$ or $g''(x)=0$. The former means we'd be dividing by $0$ (plus it's a boring answer), but the latter just implies that our function is linear. So $g(x)=ax+b$. Lastly, we just note that we only restrict the derivative, so we can slap a constant out front.
$$f(x) = c - \ln(ax+b)$$
The same thing we got before! But, how can we come up with a special case where for $x,y,z$ we have $\nabla{f}\cdot\nabla{f}=\Delta{f}$? Well, a really simple solution would be
$$f(x,y,z) = e - \ln(ax+by+cz+d)$$
with $a,b,c,d,e$ constants. The $by+cz+d$ group would be a "constant" when taking the partial derivative with respect to $x$. And the rest of the partials would follow similarly!
Now, I don't know if this is the only solution to our original equation because we worked from the single variable case up. But it's a simple solution.
