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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

Use the Fourier transform to solve the following equation:

u_tt + 2u_t + u = u_xx for x ∈ R and t > 0

With initial conditions:
u(x, 0) = sinc(x)
u_t(x, 0) = -sinc(x)

where sinc(x) = sin(x) / x for x ≠ 0.

Learn to solve the partial differential equation u_tt + 2u_t + u = u_xx using Fourier transforms. Includes initial conditions u(x, 0) = sinc(x) and u_t(x, 0) = -sinc(x). This guide covers Fourier transform methods, handling initial value problems, and properties of the sinc function.

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