Consider the wave equation
$$ \partial^2_tu-\Delta u=F, \text{ on } (0,\infty)\times\mathbb{R}^m\\ u(0,\cdot)=f,\partial_t u(0,\cdot)=g $$ Show that for $f,g\in C^{\infty}_c(\mathbb{R}^m)$ and $F\in C^\infty_c(\mathbb{R}^{m+1}_+)$ the solution can be expressed as $$ u(t)=R_0(t)f+R_1(t)g+\int_0^tR_1(t-s)F(s)ds, $$ where $$ (\mathcal{F}(R_0(t)f))(k)=\cos t|k|(\mathcal{F}f)(k),(\mathcal{F}(R_1(t)g)(\xi)=\frac{\sin(t|k|)}{|k|}(\mathcal{F}g)(k). $$
I tried this two different ways, and got stuck at the same point both times. Here is the first way:
Take the Fourier transform in space to get the ODE $$ \hat{u}''+|k|^2\hat{u}=\hat{F}. $$ Solving the homogeneous equation we get the complementary solution $$ \hat{u}_c=A\cos|k|t+B\sin|k|t. $$ The variation of parameters formula then gives the particular solution $$ \hat{u}_p=-cos|k|t\int\frac{\sin|k|t}{|k|}\hat{F}dt+\sin|k|t\int\frac {\cos|k|t}{|k|}\hat{F} dt. $$
So then $$ \hat{u}=\hat{u}_c+\hat{u}_p=A\cos|k|t+B\sin|k|t-cos|k|t\int\frac{\sin|k|t}{|k|}\hat{F}dt+\sin|k|t\int\frac {\cos|k|t}{|k|} \hat{F}dt. $$
If I apply the initial conditions on just the complementary solution I get $A=\hat{f}$ and $B=\frac{\hat{g}}{|k|}$, which looks like what we want to get. But I feel like I should be able to apply the initial conditions to the general solution and get the same thing. When I try that, however, I get something messy involving integrals I can't integrate.
From the solution we are given it seems like we need to use convolution for the inverse Fourier transform of the last part, but I can't quite seem to get there. I'm also confused by what $R_1(t)$ is. We're told that the Fourier transform of $R_1(t)g$ is $\frac{\sin|k|t}{|k|}(\mathcal{F}g)$ but that doesn't tell us that the Fourier transform of $R_1(t)$ is $\frac{\sin|k|t}{|k|}$, so how can we express it?
The second way I tried to solve the question was, after taking the FT, I let $z=\hat{u}'+i|k|\hat{u}$, showed that this solves the equation $z'-i|k|z=\hat{F}$, found the solution to that, split it into real and imaginary parts, and equated the imaginary part with $|k|\hat{u}$. This gave me
$$ \hat{u}=\frac{1}{|k|}\left[\sin|k|t+\sin|k|t\int\hat{F}\cos|k|tdt-\cos|k|t\int\hat{F}\sin|k|tdt\right] $$
which is similar to the other answers, but without a $\cos$ term and the constant coefficients.
Any help would be very much appreciated!!
Thanks, Tom