Let the complex function $q(x, t)$ satisfy

$$i \frac{\partial q(x, t)}{\partial t}+\frac{\partial^{2} q(x, t)}{\partial x^{2}}=0, \quad 0<x<\infty, 0<t<T$$

where $T$ is a positive constant. The unified transform method implies that the solution of any well-posed problem for the above equation is given by

$$\begin{aligned} q(x, t) &=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i k x-i k^{2} t} \hat{q}_{0}(k) d k \\ &-\frac{1}{2 \pi} \int_{L} e^{i k x-i k^{2} t}\left[k \tilde{g}_{0}\left(i k^{2}, t\right)-i \tilde{g}_{1}\left(i k^{2}, t\right)\right] d k \end{aligned}$$

where $L$ is the union of the rays $(i \infty, 0)$ and $(0, \infty), \hat{q}_{0}(k)$ denotes the Fourier transform of the initial condition $q_{0}(x)$, and $\tilde{g}_{0}, \tilde{g}_{1}$ denote the $t$-transforms of the boundary values $q(0, t), q_{x}(0, t)$ :

$$\begin{gathered} \hat{q}_{0}(k)=\int_{0}^{\infty} e^{-i k x} q_{0}(x) d x, \quad \operatorname{Im} k \leqslant 0, \\ =\int_{0}^{t} e^{k s} q(0, s) d s, \quad \tilde{g}_{1}(k, t)=\int_{0}^{t} e^{k s} q_{x}(0, s) d s, \quad k \in \mathbb{C}, \quad 0<t<T . \end{gathered}$$

Furthermore, $q_{0}(x), q(0, t)$ and $q_{x}(0, t)$ are related via the so-called global relation

$$e^{i k^{2} t} \hat{q}(k, t)=\hat{q}_{0}(k)+k \tilde{g}_{0}\left(i k^{2}, t\right)-i \tilde{g}_{1}\left(i k^{2}, t\right), \quad \operatorname{Im} k \leqslant 0$$

where $\hat{q}(k, t)$ denotes the Fourier transform of $q(x, t)$.

(a) Assuming the validity of (1) and (2), use the global relation to eliminate $\tilde{g}_{1}$ from equation (1).

(b) For the particular case that

$$q_{0}(x)=e^{-a^{2} x}, \quad 0<x<\infty ; \quad q(0, t)=\cos b t, \quad 0<t<T,$$

where $a$ and $b$ are real numbers, use the representation obtained in (a) to express the solution in terms of an integral along the real axis and an integral along $L$ (you should not attempt to evaluate these integrals). Show that it is possible to deform these two integrals to a single integral along a new contour $\tilde{L}$, which you should sketch.

[You may assume the validity of Jordan's lemma.]