2
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thanks for your answer Jacob! That is a simple solution to the problem. $\endgroup$
    – MaxSmask
    Apr 13, 2022 at 19:17
  • $\begingroup$ There is a small typo. The last integral should be with respect to the $k$ variable. $\endgroup$
    – MaxSmask
    Apr 13, 2022 at 19:52
  • $\begingroup$ @MaxSmask: Fixed, thanks! $\endgroup$ Apr 13, 2022 at 19:53
0
$\begingroup$

(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

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – MaxSmask
    Apr 13, 2022 at 17:20
  • $\begingroup$ No. Only the $x$ derivative needs to exist for the first component, and $H(y)$ is constant in $x$. $\endgroup$
    – K.defaoite
    Apr 13, 2022 at 21:28
  • $\begingroup$ arh sure thanks. :) $\endgroup$
    – MaxSmask
    Apr 13, 2022 at 22:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .