Function Whose Second Derivative is the Square of it's First Dervative

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.

Alternative Definition for a Skew-Symmetric Matrix

I was working on some advanced solid mechanics homework and a problem asked to prove a particular identity for $2^{\text{nd}}$ order tensors. I really enjoyed the solution that I came up with. It's the type of proof that goes

Suppose some statement $A$ is true for all $x$. We want to show that statement $B$ is true. Consider the very special case where $x=(\text{randomnesspulled}$ $\text{outofthinairbutmakestheproblemverysimple})$... and the proof goes on very smoothly because of that particular choice of $x$.

Well, I finally managed to come up with one of those proofs! I believe that book only intended for the problem to be solved for the 3-dimensional case where writing out the explicit possibilities is very feasible and most likely the expected solution. I will share my (arguably prettier) solution. But I will translate it into terms of linear algebra, matrices, and vectors instead of tensors. But first some background.

Definition 1. A skew-symmetric matrix is a matrix $\mathbf{A}$ such that $\mathbf{A}=-\mathbf{A}^T$.

Where $^T$ denotes the transpose operator (i.e. flip along the main diagonal). We will use the notation $A_{ij}$ to denote the element of $\mathbf{A}$ in the $i^{\text{th}}$ row and $j^{\text{th}}$ column. Therefore, this definition can be written as $$A_{ij}=-A_{ji}$$ Here are a few examples of skew symmetric matrices: $$\left[\begin{matrix}0&1&1\\ -1&0&1\\ -1&-1&0\end{matrix}\right] \quad\left[\begin{matrix}0&1&-2&3\\ -1&0&-4&5\\ 2&4&0&6\\ -3&-5&-6&0\end{matrix}\right] \quad\left[\begin{matrix}0&a\\ -a&0\end{matrix}\right]$$ Note that all have $0$'s down the main diagonal because $0$ is the only number that solves $x=-x$. Now, the alternative definition to this type of matrix is:

Definition 2. For all vectors $\mathbf{u}$, $\mathbf{A}$ is skew-symmetric if $\mathbf{u}^T\mathbf{Au}=0$.

And to show that this definition is the same as the previous one, all we need to do it show the following theorem. Which is the problem in the book but in terms of matrices and vectors and not tensors (maybe another day!).

Theorem 1. For all vectors $\mathbf{u}$, $\mathbf{u}^T\mathbf{Au}=0$ if and only if $A_{ij}=-A_{ji}$.

Proof.
$(\Leftarrow)$ Assume that $A_{ij}=-A_{ji}$. We want to show that for all $\mathbf{u}$, $\mathbf{u}^T\mathbf{Au}=0$. Consider $\mathbf{u}^T\mathbf{Au}$. It can be shown that $$\mathbf{u}^T\mathbf{Au}=\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j$$ where the vectors are of length $n$ and the matrix is of size $n\times n$. And since $A_{ij}=-A_{ji}$, we know that $$\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j=\sum_{i=1}^n\sum_{j=1}^n-A_{ji}u_iu_j$$ But $i$ and $j$ are just dummy variable names and we can swap them, as long as we swap all of them. So: $$\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j=\sum_{j=1}^n\sum_{i=1}^n-A_{ij}u_ju_i$$ But summation order doesn't matter, multiplication is commutative, and we can pull the negative out. $$\left(\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j\right)=-\left(\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j\right)$$ And, as stated above, the only solution to $x=-x$ is $0$. Thus $\mathbf{u}^T\mathbf{Au}=0$. That was the easy direction. Now for the fun one!
$(\Rightarrow)$ Assume $\mathbf{u}^T\mathbf{Au}=0$ for all $\mathbf{u}$. We want to show that $A_{ij}=-A_{ji}$. Going back to the summations we have $$\mathbf{u}^T\mathbf{Au}=\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j=0$$ Now, consider the vectors $\mathbf{u}^{(r,s)}$ such that $u_i=\delta_{ri}+\delta_{si}$ where $\delta_{ij}$ is the Kronecker delta function. $$\delta_{ij}=\begin{cases}1&i=j\\ 0&i\ne j\end{cases}$$ Some examples of these vectors are $$\begin{align*} \mathbf{u}^{(1,1)}=\left[2,0,0\right]\\ \mathbf{u}^{(2,1)}=\left[1,1,0\right]\\ \mathbf{u}^{(3,1)}=\left[1,0,1\right] \end{align*}$$ Those were all examples in three dimensions, however, these can be in any number of dimensions. For instance this ten dimensional version $$\mathbf{u}^{(4,7)}=\left[0,0,0,1,0,0,1,0,0,0\right]$$ Basically, if $r=s$, then the vector has all zero components with the exception of the $r^{\text{th}}$ component which is $1+1=2$. Or if $r\ne s$ then the vector still has all zero components with the exception of the $r^{\text{th}}$ component which is $1+0=1$ and the $s^{\text{th}}$ component which is $0+1=1$.

Consider first, the case where $r=s$. We know that $u_i^{(r,r)}=2\delta_{ri}$ and thus $$0=\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j=\sum_{i=1}^n\sum_{j=1}^nA_{ij}(2\delta_{ri})(2\delta_{rj})$$ Since $\delta_{ri}$ and $\delta_{rj}$ are zero unless $i=r$ and $j=r$, we can drop the summation and just use $i=j=r$ and the deltas become multiplications by $1$. $$0=A_{rr}(2)(1)(2)(1)=4A_{rr}$$ Thus $A_{ii}=0$ and we have zeros down the main diagonal. Step in the right direction!

Now consider the case where $r\ne s$. We know that $u_i^{(r,s)}=\delta_{ri}+\delta_{si}$. Thus $$0=\sum_{i=1}^n\sum_{j=1}^nA_{ij}u_iu_j=\sum_{i=1}^n\sum_{j=1}^nA_{ij}(\delta_{ri}+\delta_{si})(\delta_{rj}+\delta_{sj})$$ We can foil and distribute and get $$0=\sum_{i=1}^n\sum_{j=1}^n\left(A_{ij}\delta_{ri}\delta_{rj}+A_{ij}\delta_{ri}\delta_{sj}+A_{ij}\delta_{si}\delta_{rj}+A_{ij}\delta_{si}\delta_{sj}\right)$$ The summand is only non-zero for four cases: $(r=i=j)$, $(r=i,s=j)$, $(s=i,r=j)$, $(s=i=j)$. If one of these isn't true, then the summand is $0+0+0+0=0$. Thus, we can take just the four terms with those $i$ and $j$ values and drop the summation. $$\begin{align*} 0=&\left(A_{rr}\delta_{rr}\delta_{rr}+A_{rr}\delta_{rr}\delta_{sr}+A_{rr}\delta_{sr}\delta_{rr}+A_{rr}\delta_{sr}\delta_{sr}\right)+\\ &\left(A_{rs}\delta_{rr}\delta_{rs}+A_{rs}\delta_{rr}\delta_{ss}+A_{rs}\delta_{sr}\delta_{rs}+A_{rs}\delta_{sr}\delta_{ss}\right)+\\ &\left(A_{sr}\delta_{rs}\delta_{rr}+A_{sr}\delta_{rs}\delta_{sr}+A_{sr}\delta_{ss}\delta_{rr}+A_{sr}\delta_{ss}\delta_{sr}\right)+\\ &\left(A_{ss}\delta_{rs}\delta_{rs}+A_{ss}\delta_{rs}\delta_{ss}+A_{ss}\delta_{ss}\delta_{rs}+A_{ss}\delta_{ss}\delta_{ss}\right) \end{align*}$$ Now remember that $A_{ii}=0$ so the first and last lines are all $0$. And we can cross out any term with a $\delta_{rs}$ or $\delta_{sr}$. And we can replace any $\delta_{rr}$ or $\delta_{ss}$ with a multiplication by $1$. This leaves us with $$0=A_{rs}+A_{sr}$$ And this implies that $A_{ij}=-A_{ji}$.

I will say that this was both simpler and much prettier in Einstein tensor notation. But with the choice of $\mathbf{u}$ we had all but $4$ of $n^2$ terms cancel out in one step and the $14$ of $16$ terms cancel out in the second step. The remainder of the proof was easy!