Solving Partial Differential Equations with Fourier transforms 
Question
$
\text{1.* Use the Fourier transform to solve the initial value problem}\\
\left\{\begin{array}{l}
u_{t}-u_{x x}+u_{x}=0, x \in \mathbb{R}, t>0, \\
u(x, 0)=g(x), x \in \mathbb{R}, \text { where } g \in \mathscr{L}^{1}(\mathbb{R})
\end{array}\right.
$

My Solution
$
\text { Note: }\left(\left(\frac{\partial}{\partial x_{j}}\right)^{\alpha_{j}} u\right)^{\wedge}(\xi)=(i \xi)^{\alpha_{j}} \hat{u}(\xi) \\
\Rightarrow\left(u_{t}-u_{xx}+u_{x}\right)^{\wedge} \\
\Rightarrow \quad \partial_{t} \hat{u}=-i \xi \hat{u}-\xi^{2} \hat{u}=\hat{u}(\xi)\left[-i \xi-\xi^{2}\right] \\
\Rightarrow \int_{\hat{g}(\xi)}^{t} \frac{1}{\hat{v}(\xi)} d \hat{v}(\xi)=-\left(i \xi+\xi^{2}\right) \int_{0}^{t} d s
$
$
\Rightarrow \hat{u}(\xi)=\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t} \cdot \text { Hence, we have } \
u(x, t)=\left(\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t}\right)\check{}
$
$
\text{Note, the convolution} \ \
g * f(x)=\int_{\mathbb{R}} g(x-y) f(y) d y, \\
\text { with }(g * f(x))^{\wedge}=\sqrt{2 \pi} \hat{g}(\xi) \hat{f}(\xi)
\Rightarrow u(x, t)=\left(\frac{1}{\sqrt{2} \pi}(g * f(x))^{\wedge}\right)\check{}=\frac{1}{\sqrt{2 \pi}} g * f(x)
$
$\therefore u(x, t)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} g(x-y) f(y) d y$
$=\frac{1}{2 \pi} \int_{\mathbb{R}} g(x-y) \int_{\mathbb{R}} e^{i y \xi} e^{-\left(i \xi+\xi^{2}\right) t} d \xi d y$

This is as far as I got and I don't really know what to do beyond this point.
I've had ideas of switching the integrals around a little. Another idea I had was rewriting the last integral as $\frac{1}{2 \pi} \int_{\mathbb{R}} g(x-y) \int_{\mathbb{R}} e^{i(y-t) \xi} e^{-t \xi^{2}} d \xi d y$ and using the fact that for $f(x)=e^{-\frac{\alpha}{2}|x|^{2}}$, this gives
$\check{\mathcal{f}}(\xi)=\frac{1}{(\alpha)^{\frac{n}{2}}} e^{-\frac{|\xi|^{2}}{2 \alpha}}$. Any help would be greatly appreciated.

 A: Taking the Fourier transform in the $x$ coordinate of the PDE gives
$\hat{u}_{t}+\xi^2\hat{u}+i\xi\hat{u}=0$
i.e. $\hat{u}_{t} + (\xi^2+i\xi)\hat{u}=0,$
so
$$
\hat{u}(\xi,t) = C(\xi) e^{-(\xi^2+i\xi)t}.
$$
At $t=0$ we have $\hat{u}=\hat{g}$ so $C=\hat{g}.$ Thus,
$$
\hat{u}(\xi,t) 
= \hat{g}(\xi)e^{-(\xi^2+i\xi)t} 
= \hat{g}(\xi) e^{-\xi^2} e^{-i\xi t}.
$$
This gives
$$
u(x,t) 
= \mathcal{F}^{-1}\{ \hat{g}(\xi) e^{-\xi^2} e^{-i\xi t} \}
= \tau_{t}(g(x)*_x\frac{1}{\sqrt{4\pi}}e^{-x^2/4})
\\
= g(x)*_x\frac{1}{\sqrt{4\pi}}\tau_{t}(e^{-x^2/4})
= g(x)*_x\frac{1}{\sqrt{4\pi}}e^{-(x-t)^2/4}.
$$
Here I used that ...

*

*$\mathcal{F}^{-1}\{\hat{f}\hat{g}\}=f*g$,

*$\mathcal{F}^{-1}\{ e^{-\xi^2} \} = \frac{1}{\sqrt{4\pi}}e^{-x^2/4}$,

*$\mathcal{F}^{-1}\{ \hat{f}(\xi) e^{-ia\xi} \} = \tau_{a}f(x) = f(x-a)$
See tables section on Wikipedia Fourier transform page.
A: Use the Fourier transform to solve the initial value problem
$
\\
\left\{\begin{array}{l}
u_{t}-u_{xx}+u_{x}=0, x \in \mathbb{R}, t>0, \\
u(x, 0)=g(x), x \in \mathbb{R}.
\end{array}\right.\\ \\
\text { where } g \in \mathscr{L}^{1}(\mathbb{R})\\
$
Solution:
Note:
$
\left(\left(\frac{\partial}{\partial x_{j}}\right)^{\alpha_{j}} u\right)^{\wedge}(\xi)=(i \xi)^{\alpha_{j}} \hat{u}(\xi) 
\Rightarrow\left(u_{t}-u_{xx}+u_{x}\right)^{\wedge} \\
$
$
\Rightarrow \partial_{t} \hat{u}=-i \xi \hat{u}-\xi^{2} \hat{u}=\hat{u}(\xi)\left[-i \xi-\xi^{2}\right]\\
$
$
\Rightarrow \int_{\hat{g}(\xi)}^{t} \frac{1}{\hat{v}(\xi)} d \hat{v}(\xi)=-\left(i \xi+\xi^{2}\right) \int_{0}^{t} d s\\
$
$
\Rightarrow \hat{u}(\xi)=\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t}.\\
$
Hence, we have
$
u(x, t)=\left(\hat{g}(\xi) e^{-\left(i \xi+\xi^{2}\right) t}\right)\check{}.\\
$
Note, the convolution gives the following
$
u(x, t)=\left(\hat{g}(\xi) e^{-i \xi t} \cdot e^{-\xi^{2} t}\right)\check{}.\\
$
Let
$
\hat{\phi}(\xi)=\hat{g}(\xi) e^{-i \xi t} \textbf{ and }\hat{\psi}(\xi)=e^{-\xi^{2} t}\\
$
$
\Rightarrow(\phi * \psi)^{\wedge}(\xi)=\sqrt{2 \pi} \hat{\phi}(\xi) \hat{\psi}(\xi)\\
$
$
\Rightarrow u(x, t)=\left(\frac{(\phi * \psi)^{\wedge}(\xi)}{\sqrt{2 \pi}}\right)\check{}
=\frac{1}{\sqrt{2 \pi}}(\phi * \psi)(x)\\ \\
$
$
\textbf{We now need to find } \phi \textbf{ and } \psi...\\
$
First, lets find $\phi$ :
$
\phi(x)=\int_{\mathbb{R}} \hat{g}(\xi) e^{-i \xi t} e^{i \xi x} d \xi 
=\int_{\mathbb{R}} \hat{g}(\xi) e^{i(x-t) \xi} d \xi=g(x-t).\\
$
Now for $\psi$ :
We know for
$f=e^{-\frac{\alpha}{2}(x)^{2}} \Rightarrow f(\xi)=\frac{1}{(\alpha)^{\frac{n}{2}}} e^{-\frac{\xi^{2}}{2 \alpha}}.\\
$
$
\operatorname{Let} \alpha=2 t \\
$
$
\Rightarrow \psi(x)=\frac{1}{\sqrt{2 t}} e^{-\frac{x^{2}}{4 t}}\\
$
$
\therefore u(x, t)=\frac{1}{\sqrt{2 \pi}}(\phi * \psi)(x) \\
$
$
=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} \phi(x-\tilde{y}) \psi(\tilde{y}) d \tilde{y} \\
$
$
=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} g(x-\tilde{y}-t) \cdot \frac{1}{\sqrt{2 t}} e^{-\frac{\tilde{y}^2}{4 t}} d \tilde{y}\\
$
Let $\tilde{y}=y+t ; \quad d \tilde{y}=d y$. So we get
$
u(x, t)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} g(x-y) \cdot \frac{1}{\sqrt{2 t}} e^{-\frac{(y-t)^{2}}{4 t}} d y \\
$
$
=\int_{\mathbb{R}} g(x-y) \cdot \frac{1}{\sqrt{4 \pi t}} e^{-\frac{\left(y-t\right)^{2}}{4 t}} d y \\
$
$
\equiv\left[g(x) * \frac{1}{\sqrt{4 \pi t}} e^{-\frac{(x-t)^{2}}{4 t}}\right](x, t)
$
