Expressing sums of second partial derivatives

Question

Let $g$ be a twice continuously differentiable function on $\mathbb{R}$. Let $c$ be a positive constant and, for $\mathbf{x}$ in $\mathbb{R}$, let $r = ||\mathbf{x}||$. For $t$ in $\mathbb{R}$, define $$f(\mathbf{x}, t) = f(x_1, x_2, \cdots , x_n, t) = \frac{1}{r} g \left( t - \frac{r}{c} \right) $$ Prove that $$D_{11}f(\mathbf{x}, t)+D_{22}f(\mathbf{x}, t)+\cdots+D_{nn}f(\mathbf{x}, t)=\frac{1}{c^2}D_{n+1, n+1}f(\mathbb{x}, t)$$

My attempt

Using Chain Rule, I was able to go until:

$$D_1f(\mathbf{x}, t) = \frac{x_1}{r^2}\left\lbrace -2g\left( t - \frac{r}{c} \right) + \frac{\partial}{\partial x_1}g\left( t - \frac{r}{c} \right) \right\rbrace$$

But geez, I can't think of differentiating this one more time. Am I on the right path? Or am I missing a clever trick?

Answer 1

Since $r^2 = \|\mathbf x\|^2$, we have $2rD_ir = 2x_i$, or $$D_ir = \frac{x_i}r$$

By the chain rule $D_ih(r) = h'(r)D_i r = \dfrac{x_i}rh'(r)$.

Applying $D_i$ again is not that hard to do, though this time you must also use the product rule to handle the $x_i$.