# Solving Transport Equation with Velocity Switch using Tempered Distributions

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.

## 2 Answers

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.

• Thanks for your answer Jacob! That is a simple solution to the problem. Apr 13, 2022 at 19:17
• There is a small typo. The last integral should be with respect to the $k$ variable. Apr 13, 2022 at 19:52
• @MaxSmask: Fixed, thanks! Apr 13, 2022 at 19:53

(Because we are allowing $$t$$ to take negative values, which is nonstandard, I will rename it to $$y$$.)

The problem as it stands is ill-posed, because the derivatives break down at $$y=0$$. But we can look for a weak solution..

Let $$\varphi:\Bbb R^2 \to \mathbb R$$ be a smooth function with compact support. Multiply the entire equation by $$\varphi$$: $$\varphi(x,y) H(y) c\partial_x u(x,y)+\varphi(x,y)\partial_yu(x,y)=0$$ Now integrate over $$\Bbb R^2$$: $$c\int_\Bbb{R}\int_\Bbb{R}\varphi(x,y)H(y)\partial_x u(x,y)\mathrm dx\mathrm dy+\int_\Bbb{R}\int_\Bbb{R}\varphi(x,y)\partial_y u(x,y)\mathrm dx\mathrm dy=0$$ Now we use a little integration by parts. This can be rewritten as (exercise) $$c\int_\Bbb{R}\int_\Bbb{R}u(x,y)H(y)\partial_x\varphi(x,y)\mathrm dx\mathrm dy+\int_\Bbb{R}\int_\Bbb{R}u(x,y)\partial_y\varphi(x,y)\mathrm dx\mathrm dy$$ Or as $$\int_\Bbb R\int_{\Bbb R}(u\mathbf v\cdot\nabla\varphi)(x,y)\mathrm dx\mathrm dy=0$$ Where $$\mathbf v(x,y)=\big(cH(y),1\big)$$. Now recall the product rule for divergence: $$\nabla\cdot (f \mathbf u)=f\nabla\cdot\mathbf u+\mathbf u\cdot \nabla f$$ But clearly $$\nabla\cdot \mathbf v=0$$ hence our integral becomes $$\int_\Bbb R\int_\Bbb R \big(u\nabla\cdot (\varphi \mathbf v)\big)(x,y)\mathrm dx\mathrm dy=0$$ This is the weak form of our PDE. Let's see if your trial solution of $$u(x,y)=u_0(x-H(y)cy)$$ satisfies this weak form, and assume that $$u_0$$ is a smooth Schwartz function. $$\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\big(u\nabla\cdot(\varphi \mathbf v)\big)(x,y)\mathrm dx\mathrm dy = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty u_0(x-H(y)cy)~\nabla\cdot(\varphi \mathbf v)(x,y)\mathrm dy\mathrm dx \\ =\int\limits_{-\infty}^\infty\left[\int_{-\infty}^0u_0(x)~\nabla\cdot (\varphi\mathbf v)(x,y)\mathrm dy+\int_0^\infty u_0(x-cy)~\nabla\cdot(\varphi\mathbf v)\mathrm dy\right]\mathrm dx$$ Recall $$\nabla\cdot(\varphi \mathbf v)(x,y)=cH(y)\partial_x\varphi(x,y)+\partial_y\varphi(x,y)$$ So we can simplify this integral more to $$=\int\limits_{-\infty}^\infty\left[\int_{-\infty}^0u_0(x)~\partial_y\varphi(x,y)\mathrm dy+\int_0^\infty u_0(x-cy)~\big(c\partial_x\varphi(x,y)+\partial_y\varphi(x,y)\big)\mathrm dy\right]\mathrm dx$$

Unfortunately I am now busy so I will have to come back to this at a later stage. Hopefully this is helpful in some way

• Thanks for you reply! You statement "But clearly $\nabla \cdot \mathbf{v} = 0$" does it not require that the derivative of $H(t)$ exists in the classical sense? Apr 13, 2022 at 17:20
• No. Only the $x$ derivative needs to exist for the first component, and $H(y)$ is constant in $x$. Apr 13, 2022 at 21:28
• arh sure thanks. :) Apr 13, 2022 at 22:43