To solve this partial differential equation using the Fourier transform, let’s go through the steps systematically.

Given equation: $u_{tt} + 2u_t + u = u_{xx}$ for $x \in \mathbb{R}$ and $t > 0$, with initial conditions: $u(x, 0) = \text{sinc}(x) \quad \text{and} \quad u_t(x, 0) = -\text{sinc}(x)$ where $\text{sinc}(x) = \frac{\sin x}{x}$ for $x \neq 0$.

Step 1: Apply the Fourier Transform

Let $\hat{u}(k, t)$ be the Fourier transform of $u(x, t)$ with respect to $x$. The Fourier transform properties allow us to handle derivatives as follows:

• The Fourier transform of $u_{xx}$ with respect to $x$ becomes $-k^2 \hat{u}(k, t)$.
• The Fourier transforms of $u_{tt}$ and $u_t$ are taken with respect to $t$ after applying the transform in $x$.

Applying the Fourier transform to both sides of the equation with respect to $x$: $\hat{u}_{tt} + 2 \hat{u}_t + \hat{u} = -k^2 \hat{u}$

This simplifies to: $\hat{u}_{tt} + 2 \hat{u}_t + (1 + k^2) \hat{u} = 0$

Step 2: Solve the Resulting ODE in $t$

This is a second-order linear ordinary differential equation (ODE) in $t$ for $\hat{u}(k, t)$: [ \hat{u}_{tt} + 2 \hat