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Let $\delta_0$ be the standard Dirac Delta distribution. I wish to solve the PDE $$u_t+cu_x=\delta_0$$ in the sense of distributions with initial condition $u(x,0)=g(x)$ for some continuous $g$. That is, I wish to find $u(x,t)$ such that $$-\iint_\mathbb{R} u(x,t)(\phi_t+c\phi_x)dA=\phi(0,0)$$ where $\phi$ is any so-called test function.

Can anyone point in me the right direction? I tried to take a Fourier transform but that didn't seem to do much.

Edit:

To respond to a comment, taking the Fourier transform yields: $$\mathcal{F}(u)_t+cik\mathcal{F}(u)=1$$ This is equivalent to the ODE $$f'(t)+cikf(t)=1$$ This ODE is solved by $$\mathcal{F}(u)=f(t)=C e^{-(i kc t)} - i/(kc)$$ I'm unsure of where to go from here, or if this is correct.

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    $\begingroup$ What did you get upon taking the fourier transform? That should lead you the correct way $\endgroup$
    – JMK
    Apr 30, 2020 at 3:32
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    $\begingroup$ I agree that the Fourier transform should lead to the right answer, but there are some alternatives. Have you tried using Duhamel's principle? The delta function is likely easier to deal with as an initial condition since the action of the PDE is just to translate it. $\endgroup$
    – whpowell96
    Apr 30, 2020 at 3:33
  • $\begingroup$ So in taking the fourier transform you've suppressed the $\xi$ variable, which may have created confusion. In particular, your constant, $C$ is actually of the form $e^{k(\xi)}$ for some function of $k(\xi)$ (no $t$ dependence). Use the initial condition of $f(0, \xi) = \hat{g(\xi)}$ to determine $k(\xi)$ in terms of $g(\xi)$. Note: you'll need some stronger requirements than just continuity to make sense of the fourier transform of $g$. $\endgroup$
    – JMK
    Apr 30, 2020 at 3:44
  • $\begingroup$ I find that $C=\mathcal{F}(g)+i/k$, which isn't in the form you suggested. Could you elaboate? $\endgroup$
    – zz20s
    Apr 30, 2020 at 3:47
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    $\begingroup$ @EditPiAf it is, i miscopied the problem (this doesn't really make a difference as far as I'm aware). This is really the 2D dirac distribution $\endgroup$
    – zz20s
    Apr 30, 2020 at 13:37

2 Answers 2

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The r.h.s. of the partially Fourier-transformed equation in OP is incorrect. Indeed, spatial Fourier transformation of the 2D Dirac $\delta_0 =\delta(x)\delta(t)$ gives $\delta(t)$, not $1$. Moreover, the weak form in OP is incorrect too. Integrating by parts, we have \begin{aligned} 0 &= \iint_{\Bbb R\times\Bbb R_+} (u_t + cu_x-\delta_0)\phi\,\text d x\,\text d t \\ &= -\int_{\Bbb R} g\phi|_{t=0}\, \text d x - \iint_{\Bbb R\times\Bbb R_+} u(\phi_t + c\phi_x)\,\text d x\,\text d t - \phi(0,0) \end{aligned} for any test function $\phi$.

The present problem amounts to the computation of the Green's function for the non-homogeneous advection equation $u_t+cu_x=f$. Fourier transformation in space and time of the PDE yields $$ \text i(\omega-kc)\, \mathcal{F}_t\mathcal{F}_x u = 1 $$ where $\mathcal{F}_t = \int\text dt\, e^{-\text i\omega t}$ and $\mathcal{F}_x = \int\text dx\, e^{\text ik x}$. Thus, the solution is represented as \begin{aligned} u(x,t) &= \frac{1}{(2\pi)^2}\iint \frac{e^{\text i(\omega t-kx)}}{\text i (\omega-kc)}\text dk\,\text d\omega \\ &= \frac{\text{sgn}(t)}2 \left( \frac{1}{2\pi} \int e^{-\text i k(x-ct)}\text dk \right) \\ &= \tfrac12 \text{sgn}(t)\, \delta(x-ct) \end{aligned} where the residue theorem was used (singularity at $\omega=kc$ -- see this post). Using the superposition principle, the solution to the initial problem may be expressed as $$ u(x,t) = g(x-ct)+\tfrac12 \text{sgn}(t) \, \delta(x-ct) \, . $$ As pointed out in the comments, an alternative consists in using Duhamel's principle, cf. this article.

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  • $\begingroup$ For the problem of interest, where the solution is to be understood in the sense of distribution, would it be possible to take the Fourier Transform in time only (ie $u\to \mathcal{F}_t u=\hat{u}(x,\omega)$) and then differentiate the resulting Fourier Transform $\hat{u}(x,\omega)$ in $x$? $\endgroup$
    – pluton
    Jun 2, 2020 at 19:11
  • $\begingroup$ @zz20s I might have omitted the causality condition $t>0$ (it's not obvious how it should be included in the above computation), see answer by Qmechanic $\endgroup$
    – EditPiAf
    Jun 3, 2020 at 9:31
  • $\begingroup$ "some care should be taken [...] weak sense": do you know a good reference on this topic in the context of Fourier Transforms? $\endgroup$
    – pluton
    Jun 3, 2020 at 18:32
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    $\begingroup$ @pluton I don't have any reference in mind for PDE and tempered distributions, but this post could help $\endgroup$
    – EditPiAf
    Jun 3, 2020 at 18:53
  • $\begingroup$ yes, the book you recommended in that post says (chapter 10) that "The spatial Fourier transform of a time derivative equals the time derivative of the Fourier transform." However, this is not clear when this statement fails... even in the framework of distributions. $\endgroup$
    – pluton
    Jun 3, 2020 at 21:01
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OP's first-order initial value problem (IVP) is

$$ \frac{\partial u(x,t)}{\partial t}+ c\frac{\partial u(x,t)}{\partial x}~=~\delta(t)\delta(x), \qquad u(x,t\!=\!0)~=~g(x).\tag{1}$$

One idea is to transform the IVP (1) into the form

$$ \frac{\partial v(x^{\prime},t^{\prime})}{\partial t^{\prime}}~=~\delta(t^{\prime})\delta(x^{\prime}), \qquad v(x^{\prime},t^{\prime}\!=\!0)~=~g(x^{\prime}),\tag{2}$$

by make a suitable linear coordinate transformation $(x,t)\mapsto (x^{\prime},t^{\prime})$. A bit of thought using the chain rule reveals that the coordinate transformation $$ x~=~x^{\prime}+ct^{\prime}, \qquad t~=~t^{\prime}, \tag{3}$$ will do the job. The unique solution to the IVP (2) is evidently $$ v(x^{\prime},t^{\prime})~=~\frac{1}{2}{\rm sgn}(t^{\prime})\delta(x^{\prime})+ g(x^{\prime}). \tag{4}$$ Hence the unique solution to the original IVP (1) is $$ u(x,t)~=~\frac{1}{2}{\rm sgn}(t)\delta(x\!-\!ct)+ g(x\!-\!ct). \tag{5}$$

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