I have a question regarding the form of the general solution to a PDE in terms of its Green's function. For example, consider the heat equation: \begin{equation} \frac{\partial u}{\partial t}-\Delta u=0 \quad x \in \mathbb{R}^n, t>0 \\ u(0,x)=u_{0}(x) \end{equation} The solution is given by $u(t,x)= (p_{t} \ast u_{0})(x)$, where $\{p_{t}\}_{t \geq 0}$ is the heat semigroup (or the Green's function in free space).
Now consider the wave equation: \begin{equation} \frac{\partial^2 u}{\partial^2 t}-\Delta u=0 \quad x \in \mathbb{R}^n, t>0 \\ u(0,x)=u_{0}(x) \\ \frac{\partial u}{\partial t}(0,x)=v_{0}(x) \end{equation} The solution is apparently given by: \begin{equation} u(t,x)=(G_{t} \ast u_{0})(x)+ \left(\frac{d G_{t}}{dt} \ast v_{0}\right)(x), \end{equation} where $G_{t}$ is again, the Green's function. I don't see why this is the case. Since $G_{t}(x)$ is the solution to the PDE with initial value as the delta distribution, one would obtain $u(0,x)=u_{0}(x)+v_{0}(x)$, which is clearly not true. What am I missing here?
You have $G_0(x) = \delta(x),$ but $G'_0(x) = 0.$ Therefore, $u(0,x)=u_0(x).$