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So, my problem is this: I have to solve the heat equation with this boundary condition:

$$ \begin{cases} \dfrac{\partial}{\partial t} u(x, t) - D\ \dfrac{\partial^2}{\partial x^2} u(x, t) = S_0\delta(x)\delta(t) \\\\ u(x, 0) = U_0\delta(x) \end{cases} $$

Where $D, U_0, S_0$ are coefficients.

So the first step was to perform a Fourier transform on the whole equation (on $x \to k$) to get

$$\hat u (k, t) + D k^2 \hat u(k, t) = S_0 \delta(t)$$

And it's ok.

But then in the paper it's written that a Laplace transform (on $t\to s$) is performed, getting:

$$s\ \hat{\tilde u}(k, s) - U_0 + D k^2 \hat{\tilde u} (k, s)= S_0$$

Where of course $s$ (and before: $k$) are the new variables obtained because of the transformations.

My question is: Where does the $U_0$ term comes from, in the Laplace transform?

I cannot find a way to get it, I don't understand how it comes out.

EDIT

Ok, it comes out by LT. But I got stuck in computing it. Shalln't it be also an anti transform? Anybody can help me with the calculation?

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  • $\begingroup$ The constant comes from the Laplace transform of $\hat{u}_{t}$. Integrate by parts and use your initial condition from your PDE transformed into Fourier space. $\endgroup$ Commented Nov 21, 2016 at 14:03
  • $\begingroup$ @Mattos Indeed that is right. The other problem is: how does it come out? I think I got stuck in performing the LT! (Which actually shall be an ANTI transform... right? $\endgroup$
    – Enrico M.
    Commented Nov 21, 2016 at 14:08
  • $\begingroup$ I'll make a post. $\endgroup$ Commented Nov 21, 2016 at 14:08
  • $\begingroup$ @Mattos Wow, thank you super much!! $\endgroup$
    – Enrico M.
    Commented Nov 21, 2016 at 14:09
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    $\begingroup$ @Betty Let $\tilde{\hat{u}}(k,s)=\tilde{v}(k,s)$, then we have $$-U_{0}+s\tilde{v}+Dk^{2}\tilde{v}=S_{0} \implies \tilde{v}=\frac{S_{0}+U_{0}}{s+Dk^{2}}$$ and taking the inverse Laplace transform of both sides gives \begin{align} \mathcal{L}^{-1}\{\tilde{v}(k,s)\}&=v(k,t) \\ &=\mathcal{L}^{-1} \left(\frac{S_{0}+U_{0}}{s+Dk^{2}}\right) \\ &=(S_{0}+U_{0})\mathcal{L}^{-1} \left(\frac{1}{s-(-Dk^{2})}\right) \\ &=(S_{0}+U_{0})\exp\left(-Dk^{2}t\right) \end{align} where in the second to last lines we have $$\mathcal{L}^{-1} \left(\frac{1}{s-a}\right)=\exp(at)$$ $\endgroup$ Commented Apr 11, 2019 at 10:58

1 Answer 1

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Laplace transforming your equation

$$\hat{u}_{t}(k,t) + D k^{2} \hat{u}(k,t) = S_{0} \delta(t)$$

gives

\begin{align} \int_{0}^{\infty} \hat{u}_{t} e^{-st} dt + D k^{2} \underbrace{\int_{0}^{\infty} \hat{u} e^{-st} dt}_{D k^{2} \tilde{\hat{u}}} &= S_{0} \underbrace{\int_{0}^{\infty} \delta(t) e^{-st} dt}_{1} \\ \\ \implies \hat{u} e^{-st} \bigg|_{0}^{\infty} + s \underbrace{\int_{0}^{\infty} \hat{u} e^{-st} dt}_{\tilde{\hat{u}}} + D k^{2} \tilde{\hat{u}} &= S_{0} \end{align}

where we used integration by parts on the first integral with $v = e^{-st}$, $w' = \hat{u}_{t}$. Then note that

\begin{align} \lim_{t \to \infty} \hat{u}(k,t) e^{-st} &\to 0 \\ \hat{u}(k,t) e^{-st} \bigg|_{t = 0} &= \hat{u}(k,0) \\ \end{align}

But $\hat{u}(k,0)$ is just the Fourier transform of our initial condition. The Fourier transform of $\delta(x)$ is $1$ (depending on the convention, I think the transform you are using might have a scaling in front in which case it should actually be $1/\sqrt{2 \pi}$ or $1/2 \pi$). So we get

\begin{align} \hat{u}(k,0) &= U_{0} \\ \implies -U_{0} + s \tilde{\hat{u}} + D k^{2} \tilde{\hat{u}} &= S_{0} \end{align}

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  • $\begingroup$ @AlanTuring All good mate. $\endgroup$ Commented Nov 21, 2016 at 14:36

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