I've been asked to solve the problem \begin{equation} \left\{\begin{array}{lc} u_t=u_{xx}+xu_x & \mbox{in }x\in\mathbb{R},t>0,\\ u(x,0)=g(x), & \mbox{on }x\in\mathbb{R}, \end{array}\right. \end{equation} where $g(x)=\delta_{x-x_0}$, using Fourier Transform, i.e. applying the transformation $$\hat u(\xi,t)=\int_\mathbb{R}u(x,t)e^{-2\pi i x\xi}dx.$$ After applying this, the equation above reads \begin{equation} \left\{\begin{array}{lc} \hat u_t=(2\pi i\xi)^2\hat u-(\xi\hat u)_\xi & \mbox{in }\xi\in\mathbb{R},t>0,\\ u(\xi,0)=e^{-2\pi i x_0\xi}, & \mbox{on }\xi\in\mathbb{R}. \end{array}\right. \end{equation} I've tried considering the change $v=\xi e^{2\pi^2\xi^2}\hat u$, which transforms the problem (if I am right) into \begin{equation} \left\{\begin{array}{lc} v_t=-\xi v_\xi & \mbox{in }\xi\in\mathbb{R},t>0,\\ u(\xi,0)=\xi e^{-2\pi i x_0\xi+2\pi^2\xi^2}, & \mbox{on }\xi\in\mathbb{R}. \end{array}\right. \end{equation} Defining $f(s)=v(\xi_0+s\xi,s)$ I found that $f^\prime(s)=0$ and hence $$f(s)=f(0)=v(\xi_0,0)=\xi_0 e^{-2\pi i x_0\xi_0+2\pi^2\xi_0^2},$$ therefore $$v(\xi,t)=\xi(1-t) e^{-2\pi i x_0\xi(1-t)+2\pi^2\xi^2(1-t)^2},$$ since $\xi_0=(1-t)\xi$. Then, one has that the expression for $\hat u$ is $$\hat u(\xi,t)=(1-t) e^{-2\pi i x_0\xi(1-t)+2\pi^2\xi^2(t^2-2t)}.$$
My questions are now: If this is uncorrect, how should I approach this problem? On the other hand, if this is correct, how can I get the expression for the original $u$?
And beyond this, what is the meaning of the term $xu_x$?
Thanks in advance for your help!
What you have done to obtain the expression for $\hat{u}(\xi,t)$ looks alright. If $0<t<2$, then $t^{2}-2t$ is negative. So $e^{2\pi^{2}\xi^{2}(t^{2}-2t)}=e^{-2\pi^{2}\left|t^{2}-2t\right|\xi^{2}}$ is a Gaussian, the Fourier transform of which is also a Gaussian. Observe that
$$\int_{\mathbb{R}}(1-t)e^{-2\pi ix_{0}(1-t)-2\pi^{2}\left|t^{2}-2t\right|\xi^{2}}e^{2\pi i\xi x}d\xi=(1-t)\int_{\mathbb{R}}e^{2\pi i\xi(x-(1-t)x_{0})}e^{-2\pi^{2}\left|t^{2}-2t\right|\xi^{2}}d\xi$$
We make the change of variable $\eta=\sqrt{2\pi\left|t^{2}-2t\right|}\xi$ and use the fact that the Fourier transform of $e^{-\pi\xi^{2}}$ equals $e^{-\pi x^{2}}$ to obtain that the above expression equals
$$\dfrac{1-t}{\sqrt{2\pi\left|t^{2}-2t\right|}}\int_{\mathbb{R}}e^{2\pi i\eta\frac{x-(1-t)x_{0}}{\sqrt{2\pi\left|t^{2}-2t\right|}}}e^{-\pi \eta^{2}}d\eta=\dfrac{1-t}{\sqrt{2\pi\left|t^{2}-2t\right|}}\exp\left[-\pi\left(\dfrac{x-(1-t)x_{0}}{\sqrt{2\pi\left|t^{2}-2t\right|}}\right)^{2}\right]$$
for $(x,t)\in\mathbb{R}\times(0,2)$.