I'm looking for a solution of \begin{equation} u_t(x,t)+c\ u_x(x,t)+\gamma\ u(x,t) = 0, \end{equation} for $x,c,\gamma\in\mathbb{R}$ and $t>0$ and $u(x,0)=u_0(x)$. The exercise is to solve it with Fourier transform. My idea was using fourier transform in $x$ so I get \begin{equation} \hat{u}_t(\xi,t)+i \ c\ \xi\ \hat{u}(\xi,t)+\gamma \hat{u}(\xi,t) = 0, \text{or}\\ \hat{u}_t(\xi,t)+ (i \ c\ \xi\ + \gamma) \hat{u}(\xi,t) = 0 \end{equation} Solving the ode gives \begin{equation} \hat{u}(\xi,t)= A\exp{\left(-(i\ c\ \xi + \gamma)t\right)} \end{equation} Now using inverse fourier transform gives \begin{align} u(x,t) &= A \int_{-\infty}^{\infty} \exp{\left(-(i\ c\ \xi + \gamma)t\right)} \exp(i\ x \xi)\ d\xi \\ &= A \exp(-\gamma t) \int_{-\infty}^{\infty} \exp(i \xi (x-c))\ d\xi \end{align} which results in Dirac Delta function?! Is the ansatz wrong or is there a mistake somewhere?

Thank you!

  • 1
    $\begingroup$ Your solution should have some dependence on the initial condition. The $A$ will be a function in $\xi$. That should give you an integral that converges. $\endgroup$ Dec 21, 2022 at 17:31
  • $\begingroup$ Yes, where did $u_0$ go? $\endgroup$ Dec 21, 2022 at 18:22
  • $\begingroup$ Okay, so I got $\hat{u}(\xi,t) = \hat{u_0}(\xi) \exp(...)$. But I dont see how how this helps to calculate the integral if there is $\hat{u_0}(\xi)$ inside. $\endgroup$
    – euklit
    Dec 21, 2022 at 18:28

1 Answer 1


Solving the ODE in Fourier space gives $$\hat u(\xi, t) = \hat u_0(\xi) \exp(-(ic\xi + \gamma)t)$$ where $\hat u_0$ is the Fourier transform of $u(x,0) = u_0(x)$. Then taking the inverse Fourier transform gives $$\begin{align*}u(x,t) &= \int_{-\infty}^\infty \hat u_0(\xi) \exp(-(ic\xi + \gamma)t) \exp(ix\xi) \, d\xi \\ &= \exp(-\gamma t)\int_{-\infty}^\infty \hat u_0(\xi)\exp(i\xi(x-ct)) \, d\xi \\ &= \exp(-\gamma t) u_0(x-ct)\end{align*} $$ where the last line follows from the integral simply being the inverse Fourier transform of $u_0$ evaluated at $x-ct$. You can plug this function into the PDE and see that it is in fact a solution.

  • $\begingroup$ @MarkViola Yeah I wasn't sure what scaling was being used for the Fourier transform, so I used the convention in the question (it is also common to have a factor of $1/\sqrt{2\pi}$). But since we are only using the inverse Fourier transform on $\hat u_0$, the choice of scaling doesn't affect the answer. $\endgroup$ Dec 22, 2022 at 16:32
  • $\begingroup$ Yes, there is no impact on the final result. However, inasmuch as you used the forward transform $\int_{-\infty}^\infty \{\cdot\}(x) e^{-ix\xi}\,dx$, the inverse would have the $\frac1{2\pi}$ scale factor. (+1) for the post $\endgroup$
    – Mark Viola
    Dec 22, 2022 at 17:24

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