Heat equation with Gaussian boundary condition 
Let $$ S(x,t)= \frac{1}{2\sqrt{\pi t}}e^\frac{-x^2}{4t},\quad
-\infty < x < \infty,\quad t>0$$
a. Find the solution to the equation 
  $$u_t = u_{xx},\quad -\infty < x < \infty,\quad 0<t<\infty$$
$$u(x,0) = S(x,1)$$
b. Find a solution $v$ satisfying $v_t + v_{xx} = 0$ in 
  $\lbrace (x,t) : −∞ < x < ∞,\; −∞ < t < 1\rbrace$
  and $v(x, 1) = S(x, 1)$.

a. my approach is to apply $t=1$ to $S$, which will get us $u(x,0)$. Then, by using separation of variables of $u$ (since it is homogeneous) $u = X(x)T(t)$ so $u(x,0)=X(x)T(0)$, so we can conclude $$X(x) = \frac{1}{2\sqrt\pi}e^\frac{-x^2}{4}$$
but I'm not sure whether the result of  integration his $X(x)$ by $x$ twice, and then $t$ would get me the right solution. 
If it is not valid, could you explain the proper approach for this problem, possibly relating it to the question b also?  
 A: This problem deals actually with change of variable, more specifically with time-shifting and time-reversal. Consider the change of variable $\sigma(x,\tau) = S(x, t)$ where $\tau = \theta\pm t$ and $\theta$ is an arbitrary constant -- conversely, $t = \pm(\tau - \theta)$. According to the chain rule, we have
$$
S_t = \frac{x^2 - 2 t}{8 t^2\sqrt{πt}} e^{-x^2/(4 t)} = \pm\sigma_\tau ,\qquad
S_{xx} = \frac{x^2 - 2 t}{8 t^2\sqrt{πt}} e^{-x^2/(4 t)} = \sigma_{xx} \, .
$$
In particular, we find that $S(x,t)$ solves the heat equation for $t>0$ with the boundary condition 
$S(x,1) = \tfrac12 {e^{-x^2/4}}/{\sqrt{\pi}}$.
Thus, we also find that $\sigma(x,\tau)$ solves the PDE $\sigma_\tau = \pm\sigma_{xx}$ for $\pm\tau > \pm\theta$ with the boundary condition $\sigma(x, \theta\pm 1) = S(x,1)$.
Hints:
[a.] To conclude, use $\tau = t-1$, which corresponds to the '+' sign and $\theta = -1$.
[b.] Use $\tau = 2- t$, i.e. the '-' sign with $\theta = 2$.

Note: $S$ is the Green's function of the initial value problem for the 1D heat equation over $\Bbb R$. Indeed, we have
$$
\lim_\limits{t\to 0^+} S(x,t) = \lim_\limits{t\to 0^+} \delta_{\sqrt{4t}}(x) = \delta(x)
$$
where $\delta$ denotes the Dirac distribution, represented as the limit of a sequence of Gaussian functions $\delta_a$ (normal distributions).
