For a wave equation
$\begin{equation} \begin{cases} u_{tt}- \Delta u =0 \\ u|_{t=0} = f(x) \\ u_{tt}|{t=0} = g(x) \\ \end{cases} \end{equation}$
If we take Fourier transform in time $t$, then we have the Helmholtz equation
$\begin{equation} \begin{cases} -\omega^2v(\omega,x) - \Delta v(\omega,x) =0 \\ \int v(\omega,x)d\omega = f(x) \\ \int i\omega v(\omega,x)d\omega = g(x) \\ \end{cases} \end{equation}$
Now if we have a Helmholtz equation whoes solution $v$ does not satiesfy the conditions $\int v(\omega,x)d\omega = f(x)$ or $\int i\omega v(\omega,x)d\omega = g(x)$, will it still be equivalent to some wave equation?
I read some textbooks. All of them only state that Helmholz equation is the time-harmonic form of wave equation, but do not say anything about the initial conditions.