Heat equation convection I want to solve the heat equation with convection $F_t = F_{xx} - F_x$ with initial condition $F(x,0) = f(x)$
So far what I've got is that if $F(x,t) = G(x-t,t)$ and G satisfies the heat equation then $∂_tF(x,t) = -G_x(x-t,t) + G_t(x-t,t) = -G_x(x-t,t) + G_{xx}(x-t,t) $
$= -F_x(x,t) + F_{xx}(x,t)$ so $F_t(x,t) = -F_x(x,t) + F_{xx}(x,t)$
Hence

Now I'm stuck. Is this the final answer? Seems a bit short..but if it isn't, how do I get to the final answer?
Thanks in advance to whoever can help me :)
 A: I will provide a solution using Fourier transform. I guess you are familiar with the following notation.
Let's say we have the PDE:
$  u_t(x,t)=u_{xx}(x,t)-u_x(x,t)$, with the initial condition $u(x,0)=f(x)$.
Firstly, we apply Fourier transform ( with respect to the spatial variable $x$). So, we have:
$\begin{array}[t]{l}\hat u_t(\omega,t)=(-i\omega)^2\cdot \hat u(\omega,t)-(-i \omega)\cdot \hat u(\omega,t)\\
\vdots\\
\hat u_t(\omega,t)+(\omega^2-i\omega)\cdot\hat u(\omega,t)=0
\end{array}$
The last equation is an ODE (with respect to $t$), so we have the solution:
$\hat u(\omega,t)=A(\omega)\cdot e^{t\cdot\left(i\omega-\omega^2\right)} $
Moreover, applying Fourier transform to the initial condition we take $\hat u(\omega,0)=\hat f(\omega)$.
$\begin{cases}
\hat u(\omega,0)=A(\omega)\cdot e^{0\cdot(i\omega-\omega^2) }\\\\
\hat u(\omega,0)=\hat f(\omega)
\end{cases}\implies A(\omega)=\hat f(\omega)
$
So, our solution becomes $\hat u(\omega, t) =\hat f(\omega)\cdot e^{t\cdot \left( i\omega - \omega^2\right)}$.
Secondly, we try to find a function $g(x)$ such that $\hat g(\omega)=e^{t\cdot \left( i\omega - \omega^2\right)}$.
Applying inverse Fourier transform to $\hat g(\omega)$, we take:
$g(x)=\dfrac 1 {\sqrt{2\pi}}\cdot \displaystyle \int_{-\infty}^{\infty} \hat g(\omega) \cdot e^{-i\cdot \omega \cdot x}d\omega=\ldots=\dfrac 1 {\sqrt{2t}}\cdot \exp\big[-\dfrac{(t-x)^2}{4t}\big]$.
To define $u(x,t)$, we use the definition and properties of convolution.
We have that $\hat u(\omega, t)=\hat f(\omega)\cdot \hat g(\omega)$, and we also know that $\mathcal F ^{-1}\bigg(\hat f(\omega)\cdot \hat g(\omega)\bigg)(x)=f(x)\ast g(x)$.
So, applying inverse Fourier transform to $\hat u(\omega,t)$, we have:
$
\boxed{u(x,t)=f(x)\ast g(x)\begin{array}[t]{l}=\dfrac 1 {\sqrt{2 \pi}} \cdot \displaystyle \int_\mathbb{R} f(y)\cdot g(x-y)dy=\\=\ldots =\displaystyle \int_\mathbb R\dfrac 1 {\sqrt{4\pi t}}\cdot f(y)\cdot \exp\big[-\dfrac{(t-x+y)^2}{4t}\big]dy
\end{array}} $
