Imagine we have the following two transport problems. The first is well known and can be solved using the Fourier transform, but I do not know how to solve the second one.
Problem 1: With Constant Velocity
Let $v \in \mathbb{R}$ be a constant we have:
\begin{align*}
\partial_t u(x,t) = -v\partial_x u(x,t) \ , \quad (x,t) \in \mathbb{R} \times [0,\infty) \ .
\end{align*}
with the the initial condition $u(x,0) = \phi(x)$, where $\phi$ is a Schwartz function.
Solution to problem 1:
We will solve this using the Fourier transform, where $\mathcal{F}_x$ denote the Fourier transform with respect to the $x$ variable, $\mathcal{F}^{-1}_x$ is the inverse and we use the notation $\hat{u}(k,t) = (\mathcal{F}_x u)(k,t)$. We take the Fourier transform on both side:
\begin{align*}
(\mathcal{F}_x \partial_t u)(k,t) &= -v(\mathcal{F}_x \partial_x u)(k,t) \\
\partial_t \hat{u}(k,t) &= -ikv \hat{u}(k,t) \\
\hat{u}(k,t) &= \hat{\phi}(x)\exp{(-ikvt)} \ .
\end{align*}
We now take the inverse Fourier transform:
\begin{align*}
\mathcal{F}^{-1}_x \hat{u}(k,t) &= \mathcal{F}^{-1}_x \hat{\phi}(x)\exp{(-ikvt)}
\end{align*}
where we have $\mathcal{F}^{-1}_x \hat{u}(k,t) = u(x,t)$ for the left hand side and for the right hand side we have
\begin{align*}
\mathcal{F}^{-1}_x \hat{\phi}(x)\exp{(-ikvt)} &= \frac{1}{2 \pi} \int_{-\infty}^\infty \hat{\phi}(x)\exp{(-ikvt)} \exp(ikx) \ dk \\
&= \frac{1}{2 \pi} \int_{-\infty}^\infty \hat{\phi}(x) \exp{(-ik(x-vt))} \ dk \\
&= \phi(x-vt) \ .
\end{align*}
Hence we conclude that the solution to the transport equation is:
\begin{align*}
u(x,t) = \phi(x-vt) \ , \quad (x,t) \in \mathbb{R} \times [0, \infty) \ .
\end{align*}
Problem 2: With Velocity Switch
Let $v \in \mathbb{R}$ be a constant we have:
\begin{align*}
\partial_t u(x,t) = -vH(t)\partial_x u(x,t) \ , \quad (x,t) \in \mathbb{R} \times \mathbb{R} \ .
\end{align*}
with the the initial condition $u(x,t) = \phi(x)$ for all $t \leq 0$, where $\phi$ is a Schwartz function. Here $H(t)$ is the Heaviside step function:
\begin{align*}
H(t) = \begin{cases}
\ 1 \ , \quad t\geq 0 . \\
\ 0 \ , \quad t < 0 .
\end{cases}
\end{align*}
My intuition tells me that these two problems should be equivalent so I would expect the solution to the second problem to be:
\begin{align*}
u(x,t) = \phi(x-H(t)vt) \ , \quad (x,t) \in \mathbb{R} \times \mathbb{R} \ .
\end{align*}
I will like to show this using Tempered Distributions, but alternative approaches are also welcome. Just as a quick disclaimer this is a problem I made up myself, so it may be an ill-posed problem.
Solution Attempt:
As in problem 1 we start by taking the Fourier transform with respect to $x$ then we have
\begin{align*}
(\mathcal{F}_x \partial_t u)(k,t) &= -vH(t)(\mathcal{F}_x \partial_x u)(k,t) \\
\partial_t \hat{u}(k,t) &= -ikvH(t)\hat{u}(k,t) \ .
\end{align*}
We now take take the Fourier transform with respect to time. We denote $g(k,t) = H(t)\hat{u}(k,t)$ and $f(k,t) = \hat{u}(k,t)$. Then we have
\begin{align*}
\mathcal{F}_t\partial_t f(k,t) &= -ikv \mathcal{F}_t g(k,t) \\
i \omega \hat{f}(k,\omega) &= -ikv \hat{g}(k, \omega)
\end{align*}
Let $\phi(\omega)$ be an arbitrary Schwartz function. We multiply the above with $\phi(\omega)$ and integrate
\begin{align*}
\int_{-\infty}^\infty i \omega \hat{f}(k,\omega) \phi(\omega) d\omega &= \int_{-\infty}^\infty -ikv \hat{g}(k, \omega) \phi(\omega) d\omega
\end{align*}
From here I am not really sure how to manipulate it.
Best Answer
Your problems begin when you take $\mathcal{F}_t$. Originally, you have $$\partial_t\hat{u}(k,t)=-ikvH(t)\hat{u}(k,t)$$ This is already solvable as in Problem 1: $$-ikvH(t)=\frac{\partial_t\hat{u}(k,t)}{\hat{u}(k,t)}=\partial_t{\ln{\!(\hat{u}(k,t))}}$$ Now integrate from $0$ to $t$: $$-ikvtH(t)=\ln{\!\left(\frac{\hat{u}(k,t)}{\hat{u}(k,0)}\right)}=\ln{\!\left(\frac{\hat{u}(k,t)}{\hat{\phi}(k)}\right)}$$ Thus $$\hat{u}(k,t)=\hat{\phi}(k)e^{-ikvtH(t)}$$ Taking the inverse transform, $$u(x,t)=\int_{\mathbb{R}}{\hat{\phi(k)}e^{-ikvtH(t)}e^{ikx}\,dk}=\phi(x-vtH(t))$$ as you surmised.