Consider the partial differential equation for $u(x, t)$,

$$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}+\beta \frac{\partial u}{\partial x}, \quad \beta>0, \quad 0<x<\infty, \quad t>0$$

where $u(x, t)$ is required to vanish rapidly for all $t$ as $x \rightarrow \infty$.

(i) Verify that the PDE $(*)$ can be written in the following form

$$\left(e^{-i k x+\left(k^{2}-i \beta k\right) t} u\right)_{t}=\left(e^{-i k x+\left(k^{2}-i \beta k\right) t}\left[(i k+\beta) u+u_{x}\right]\right)_{x}$$

(ii) Define $\hat{u}(k, t)=\int_{0}^{\infty} e^{-i k x} u(x, t) d x$, which is analytic for $\operatorname{Im} k \leqslant 0$. Determine $\hat{u}(k, t)$ in terms of $\hat{u}(k, 0)$ and also the functions $f_{0}, f_{1}$ defined by

$$f_{0}(\omega, t)=\int_{0}^{t} e^{-\omega\left(t-t^{\prime}\right)} u\left(0, t^{\prime}\right) d t^{\prime}, \quad f_{1}(\omega, t)=\int_{0}^{t} e^{-\omega\left(t-t^{\prime}\right)} u_{x}\left(0, t^{\prime}\right) d t^{\prime}$$

(iii) Show that in the inverse transform expression for $u(x, t)$ the integrals involving $f_{0}, f_{1}$ may be transformed to the contour

$$L=\left\{k \in \mathbb{C}: \operatorname{Re}\left(k^{2}-i \beta k\right)=0, \operatorname{Im} k \geqslant \beta\right\}$$

By considering $\hat{u}\left(k^{\prime}, t\right)$ where $k^{\prime}=-k+i \beta$ and $k \in L$, show that it is possible to obtain an equation which allows $f_{1}$ to be eliminated.

(iv) Obtain an integral expression for the solution of $(*)$ subject to the the initialboundary value conditions of given $u(x, 0), u(0, t)$.

[You need to show that

$$\int_{L} e^{i k x} \hat{u}\left(k^{\prime}, t\right) d k=0, \quad x>0,$$

by an appropriate closure of the contour which should be justified.]