Problem:
$u_{tt} = u_{xxxx}, t>0$
$u(0,x) = p(x)$
$u_t(0,x) = 0$
Attempt:
$F(u_{tt}) = F(u_{xxxx})$
$F(u_{tt}) = F(u)_{tt}$
$F(u_{xxxx}) = -k^4F(u)$ so $F(u)_{tt} + k^4F(u) = 0$
$e^{k^4t}F(u)_{tt} + k^4e^{k^4t}F(u) = 0$
$(e^{k^4t}F(u)_{tt} = 0$ so $e^{k^4t}F(u) = a(k)t + b(k)$
Hence $F(u) = (a(k)t + b(k))e^{-k4t}$
$u(0,x) = p(x)$
$F(u(t,k)) = (c(k)t + d(k))e^{-k^4t}$ so $F(u(0,k)) = a(k)t + b(k)$
$F(u(0,k)) = F(p(x))$ and $a(k)t + b(k) = F(p(x))$
$F(u(t,k)) = F(p(x))e^{-k^4t} := F(p(x))F(q(t,k))$
$g(t,x) = ∫_Re^{-k^4t-ikx}dk = ie^{-tk^4-ikx}/k$
$F(u) = F(pq)$ so $u(t,x) = p(x)ie^{-tk^4-ikx}/k$
I don't think the answer is correct. What am I doing wrong?
HINT :
$$F(u)_{tt}+k^4F(u)=0 \quad\to\quad F(u)=a(k)e^{ik^2t}+b(k)e^{-ik^2t}$$