Consider this kind of problem
$$ v_{tt} + \Delta v + v_{t} = 0, $$
with $ x\in R^{n}, t\geq 0 $ and initial data
\begin{equation} \left\lbrace \begin{array}{ll} v(x,0) = \phi(x) \\ v_{t}(x,0) = \psi(x) \end{array} \right. \end{equation}
I read that the solution can be written in this way:
$$ v(x,t) = K_1\ast \psi + K_2\ast \phi, $$
where $ K_1 $ and $ K_2 $ denote some kernels (as my notes said).
Could anyone explain me why the solution can be written in this way?
In a past question, @Yves Daoust explained to me that it is a consequence of the superposition principle and the fact that $ \delta $ is the unit of the convolution product, but it is still not clear for me.
Could anyone help me?
You need to take the Fourier transform, as @Mattos' comment suggests to you. Formally taking the spatial FT of the PDE, we get the ODE $$\hat{v}_{tt}(\xi,t)+\hat{v}_t(\xi,t)-|\xi|^2 \hat{v}(\xi,t)=0,$$ subject to the conditions $\hat{v}(\xi,0)=\hat{\phi}(\xi),$ and $\hat{v}_t(\xi,0) =\hat{\psi}(\xi).$ Solve this ODE to find $\hat{v}(\xi,t)$, then try to express the inverse transform using convolutions; this will give you your solution $v(x,t)$. I'll leave this for you to try out yourself.