# Solve PDE using fourier transform

I'm looking for a solution of $$$$u_t(x,t)+c\ u_x(x,t)+\gamma\ u(x,t) = 0,$$$$ for $$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$$ so I get $$$$\hat{u}_t(\xi,t)+i \ c\ \xi\ \hat{u}(\xi,t)+\gamma \hat{u}(\xi,t) = 0, \text{or}\\ \hat{u}_t(\xi,t)+ (i \ c\ \xi\ + \gamma) \hat{u}(\xi,t) = 0$$$$ Solving the ode gives $$$$\hat{u}(\xi,t)= A\exp{\left(-(i\ c\ \xi + \gamma)t\right)}$$$$ 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!

• 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. Dec 21, 2022 at 17:31
• Yes, where did $u_0$ go? Dec 21, 2022 at 18:22
• 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. Dec 21, 2022 at 18:28

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.
• @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. Dec 22, 2022 at 16:32
• 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 Dec 22, 2022 at 17:24