I want to solve the following: $$ u_t = ku_{xx} \quad x>0 \quad t>0 \\ u(x, 0) = 0 \quad x>0 \\ u_x(0, t) = g(t) $$ plus, $u$ and $u_x$ approach 0 when $ x $ approaches $ \infty$
I want to derive the solution is given by:
$$ \Large u(x, t) = -\sqrt{\frac{k}{\pi}} \int_0^t \frac{g(\tau)}{\sqrt{t - \tau}} e^{- \frac{x^2}{4k(t-\tau)}}$$
I went about it with fourier transforming both sides of the PDE:
$$ \frac{d(\hat u(\omega)) }{dt} = -\omega^2k \ \hat u (\omega)$$
$$\implies \hat u(\omega) = Ae^{-\omega^2\ k \ t} $$ but from the initial conditions: $$ A = \hat u (\omega, 0) = 0$$
and from here I don't know how to proceed, Help would be appreciated!
Hint.
Using the Laplace transform we have
$$ sU(x,s)-kU_{xx}(x,s) = 0, \ \ U_x(0,s) = G(s),\ \ \lim_{x\to \infty}U(x,s)=0 $$
Solving for $x$ we have
$$ U(x,s) = \psi_1(s)e^{\sqrt{\frac sk}x}+\psi_2(s)e^{-\sqrt{\frac sk}x} $$
but as $\lim_{x\to \infty}U(x,s)=0$ we follow with
$$ U(x,s) = \psi_2(s)e^{-\sqrt{\frac sk}x}, \ \ U_x(0,s) = G(s) $$
so
$$ \phi_2(s) = -\sqrt{\frac ks}G(s) $$
and
$$ U(x,s) = -\sqrt{\frac ks}G(s)e^{-\sqrt{\frac sk}x} $$
and finally we can use the convolution theorem
$$ u(x,t) = \mathcal{L}^{-1}\left(-\sqrt{\frac ks}e^{-\sqrt{\frac sk}x}\right)\circledast g(t) $$
NOTE
$$ \mathcal{L}^{-1}\left(-\sqrt{\frac ks}e^{-\sqrt{\frac sk}x}\right) = -\sqrt{\frac {k}{\pi}}\frac{e^{-\frac{x^2}{4k t}}}{\sqrt{t}} $$