From Salsa, PDEs in Action, chapter 5.
Let $w_\phi$ be the solution of the Cauchy problem \begin{array}{l} w_{tt}-c^2 \Delta w=0 \quad x\in\mathbb{R}^3,\; t>0 \\ w(x,0) = 0 \quad x\in\mathbb{R}^3 \\ w_t(x,0)=\phi(x) \quad x\in\mathbb{R}^3 \end{array} such that $w\in C^3(\mathbb{R}^3\times[0,+\infty))$.
Then $v\equiv \partial_t w_\phi$ is the solution of the Cauchy problem \begin{array}{l} v_{tt}-c^2 \Delta v=0 \quad x\in\mathbb{R}^3,\; t>0 \\ v(x,0) = \phi(x) \quad x\in\mathbb{R}^3 \\ v_t(x,0)=0 \quad x\in\mathbb{R}^3 \end{array}
This lemma, combined with the superposition principle, simplifies the solution of the Cauchy problem with both Cauchy data.
I cannot figure out how to verify that $ v_t(x,0) = 0$. Here are the steps I have made: \begin{equation*} v_t(x,0) = \partial_{tt} w_\phi(x,0) = \lim_{t\to 0^+} \partial_{tt} w_\phi(x,t) = \lim_{t\to 0^+} c^2\Delta w_\phi(x,t) = c^2\Delta w_\phi(x,0) \end{equation*} where the limits hold due to the regularity of $w_\phi$.
Why is the last term equal to zero? In other words, why is $w_\phi(\cdot,0)$ harmonic?
Thanks in advance.
I have found a reason why $c^2 \Delta w_\phi(x,0)=0$. Since the Laplacian acts only on the space variables $x$ and leaves the time variable $t$ unchanged, we have $$ (\Delta w_\phi)(x,0) = \Delta (w_\phi(x,0)) = 0 \quad\text{since}\quad w_\phi(x,0)=0 $$ In other words, taking the Laplacian and then letting $t=0$ or vice-versa yield the same result.
Incidentally, this is why one must use the wave equation via the limits and cannot simply stop at $\partial_{tt} w_\phi$: actually $$ (\partial_{tt} w_\phi)(x,0) \neq \partial_{tt} (w_\phi(x,0))$$ My comment above on this point was misleading.