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

### Problem Summary
We have the PDE:
\[
u_{tt} + 2u_t + u = u_{xx}
\]
where \( u = u(x, t) \) is a function of \( x \) and \( t \), and it holds for \( x \in \mathbb{R} \) and \( t > 0 \). The initial conditions are:
\[
u(x, 0) = \text{sinc}(x) \quad \text{and} \quad u_t(x, 0) = -\text{sinc}(x),
\]
where \( \text{sinc}(x) = \frac{\sin x}{x} \) for \( x \neq 0 \).

### Solution Using Fourier Transform
1. **Take the Fourier Transform in \( x \):**
   Let \( \hat{u}(k, t) \) be the Fourier transform of \( u(x, t) \) with respect to \( x \):
   \[
   \hat{u}(k, t) = \int_{-\infty}^{\infty} u(x, t) e^{-ikx} \, dx.
   \]
   Applying the Fourier transform to both sides of the PDE, we use the properties of the Fourier transform:
   - \( \mathcal{F}(u_{tt}) = \hat{u}_{tt}(k, t) \),
   - \( \mathcal{F}(u_t) = \hat{u}_t(k, t) \),
   - \( \mathcal{F}(u_{xx}) = -k^2 \hat{u}(k, t) \).

   Thus, the Fourier-transformed equation becomes:
   \[
   \hat{u}_{tt}(k, t) + 2 \hat{u}_t(k, t) + \hat{u}(k, t) = -k^2 \hat{u}(k, t).
   \]

2. **Rewrite the Equation:**
   Combine terms involving \( \hat{u} \):
   \[
   \hat{u}_{tt}(k, t) + 2 \hat{u}_t(k, t) + (1 + k^2) \hat{u}(k, t) = 0.
   \]
   This is now a second-order ordinary differential equation in \( t \) for \( \hat{u}(k, t) \).

3. **Solve the Characteristic Equation:**
   The characteristic equation associated with this ODE is:
   \[
   r^2 + 2r + (1 + k^2) = 0.
   \]
   Solving for \( r \) gives:
   \[
   r = -1 \pm i \sqrt{k^2}.
   \]
   Thus, the solution to the ODE for \( \hat{u}(k, t) \) is of the form:
   \[
   \hat{u}(k, t) = \left( A(k) e^{(-1 + i |k|) t} + B(k) e^{(-1 - i |k|) t} \right).
   \]

4. **Apply Initial Conditions in Fourier Space:**
   We need to determine \( A(k) \) and \( B(k) \) using the initial conditions:
   - **Initial Condition for \( u(x, 0) \):** Taking the Fourier transform, we get:
     \[
     \hat{u}(k, 0) = \mathcal{F}(\text{sinc}(x)) = \pi \chi_{[-1, 1]}(k),
     \]
     where \( \chi_{[-1, 1]}(k) \) is the indicator function over the interval \( [-1, 1] \).
   - **Initial Condition for \( u_t(x, 0) \):** Taking the Fourier transform, we get:
     \[
     \hat{u}_t(k, 0) = -\pi \chi_{[-1, 1]}(k).
     \]

5. **Substitute Initial Conditions and Solve for \( A(k) \) and \( B(k) \):**
   Use these Fourier-transformed initial conditions to solve for \( A(k) \) and \( B(k) \) explicitly.

6. **Inverse Fourier Transform to Find \( u(x, t) \):**
   Once \( \hat{u}(k, t) \) is fully determined, apply the inverse Fourier transform:
   \[
   u(x, t) = \int_{-\infty}^{\infty} \hat{u}(k, t) e^{ikx} \, dk.
   \]

This outlines the general approach. Would you like to see the explicit calculations for finding \( A(k) \) and \( B(k) \), or do you have questions about any particular step?

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### Follow-Up Questions
1. What is the characteristic equation used to solve ordinary differential equations in Fourier space?
2. How do initial conditions transform when applying the Fourier transform to partial differential equations?
3. How can the sinc function's Fourier transform properties be applied to solve wave-like equations?
4. What is the physical interpretation of solving this PDE in the Fourier domain?
5. How would the solution change if we altered the damping term in the PDE?

### Tip
Always check that the initial and boundary conditions match when transforming PDEs to the Fourier domain, as they guide the form of the solution significantly.

Use the Fourier transform to solve the equation: u_tt + 2u_t + u = u_xx for x in R and t > 0, with initial conditions u(x, 0) = sinc(x) and u_t(x, 0) = -sinc(x).

Learn how to use the Fourier transform to solve the PDE u_tt + 2u_t + u = u_xx with initial conditions u(x, 0) = sinc(x) and u_t(x, 0) = -sinc(x). This solution involves solving a transformed ODE using characteristic equations and applying inverse Fourier transform techniques.

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