Let's solve the given PDE step by step. --- ### **Step 1: Write Down the Problem** The equation is the **heat equation** or diffusion equation: \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, \quad -\infty < x < \infty,\ t > 0 \] with the **initial condition**: \[ u(x, 0) = e^{-|x|} \] --- ### **Step 2: Use the Fourier Transform** Let’s take the Fourier transform of \(u(x, t)\) with respect to \(x\): Define: \[ \hat{u}(k, t) = \int_{-\infty}^{\infty} u(x, t) e^{-ikx} dx \] Taking the Fourier transform of both sides of the PDE: \[ \mathcal{F}\left[\frac{\partial^2 u}{\partial x^2}\right] = -k^2 \hat{u}(k, t) \] \[ \mathcal{F}\left[\frac{\partial u}{\partial t}\right] = \frac{\partial \hat{u}}{\partial t} \] So the transformed equation becomes: \[ \frac{\partial \hat{u}}{\partial t} = -k^2 \hat{u} \] --- ### **Step 3: Solve the ODE in \(t\)** This is a simple ODE: \[ \frac{\partial \hat{u}}{\partial t} + k^2 \hat{u} = 0 \] \[ \implies \hat{u}(k, t) = A(k) e^{-k^2 t} \] Where \(A(k)\) is determined from the initial condition. --- ### **Step 4: Apply Initial Condition** At \(t = 0\): \[ u(x, 0) = e^{-|x|} \implies \hat{u}(k, 0) = \int_{-\infty}^{\infty} e^{-|x|} e^{-ikx} dx \] Let's compute this Fourier transform: \[ \hat{u}(k, 0) = \int_{-\infty}^{\infty} e^{-|x|} e^{-ikx} dx \] Split into two parts: \[ = \int_{-\infty}^0 e^{x} e^{-ikx} dx + \int_0^{\infty} e^{-x} e^{-ikx} dx \] \[ = \int_{-\infty}^0 e^{(1-ik)x} dx + \int_0^{\infty} e^{-(1+ik)x} dx \] Calculate both integrals: - For \(x < 0\): \[ \int_{-\infty}^0 e^{(1-ik)x} dx = \left[ \frac{e^{(1-ik)x}}{1-ik} \right]_{x=-\infty}^0 = \frac{1}{1-ik} \] (as \(x \to -\infty\), \(e^{(1-ik)x} \to 0\) since \(\text{Re}(1-ik) = 1 > 0\)) - For \(x > 0\): \[ \int_0^{\infty} e^{-(1+ik)x} dx = \left[ \frac{-e^{-(1+ik)x}}{1+ik} \right]_{x=0}^{x=\infty} = \frac{1}{1+ik} \] Sum up: \[ \hat{u}(k, 0) = \frac{1}{1-ik} + \frac{1}{1+ik} \] \[ = \frac{1+ik + 1-ik}{(1-ik)(1+ik)} = \frac{2}{1 + k^2} \] So, \(A(k) = \frac{2}{1 + k^2}\). --- ### **Step 5: Write Full Solution in \(k\)-space** \[ \hat{u}(k, t) = \frac{2}{1 + k^2} e^{-k^2 t} \] --- ### **Step 6: Inverse Fourier Transform** Now invert to get \(u(x, t)\): \[ u(x, t) = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{u}(k, t) e^{ikx} dk \] \[ = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{2}{1 + k^2} e^{-k^2 t} e^{ikx} dk \] \[ = \frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{-k^2 t} e^{ikx}}{1 + k^2} dk \] --- ### **Final Answer** \[ \boxed{ u(x, t) = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{-k^2 t} e^{ikx}}{1 + k^2} dk } \] This is the explicit solution. Alternatively, the integral can be left as is, or evaluated further using tables or residues for special cases, but this form is a complete answer for arbitrary \(x\) and \(t > 0\). --- **Summary of Steps:** 1. Apply Fourier transform to convert PDE to ODE. 2. Solve ODE in \(k\)-space. 3. Use initial condition to find \(A(k)\). 4. Invert Fourier transform to get solution.

Let's carefully solve the given PDE step by step with detailed calculations. --- ### **Given PDE and initial condition:** \[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, \quad -\infty < x < \infty, \ t > 0 \] with \[ u(x, 0) = e^{-|x|} \] --- ### **Step 1: Recognize the PDE** This is the classical heat equation: \[ u_t = u_{xx} \] --- ### **Step 2: Apply Fourier Transform in \(x\)** Define the Fourier transform: \[ \hat{u}(k, t) = \int_{-\infty}^\infty u(x, t) e^{-ikx} dx \] Transform the PDE: - Fourier transform of \(u_{xx}\): \[ \mathcal{F}[u_{xx}] = -k^2 \hat{u}(k,t) \] - Fourier transform of \(u_t\): \[ \mathcal{F}[u_t] = \frac{\partial}{\partial t} \hat{u}(k, t) \] So, the PDE becomes: \[ \frac{\partial}{\partial t} \hat{u}(k, t) = -k^2 \hat{u}(k, t) \] --- ### **Step 3: Solve the ODE in Fourier space** This is a first-order linear ODE: \[ \frac{d}{dt} \hat{u}(k,t) = -k^2 \hat{u}(k,t) \] with solution: \[ \hat{u}(k, t) = A(k) e^{-k^2 t} \] Determine \(A(k)\) from initial condition: \[ \hat{u}(k, 0) = \int_{-\infty}^\infty u(x, 0) e^{-ikx} dx = \int_{-\infty}^\infty e^{-|x|} e^{-ikx} dx \] --- ### **Step 4: Compute the initial Fourier transform** Break into two integrals: \[ \hat{u}(k, 0) = \int_{-\infty}^0 e^{x} e^{-ikx} dx + \int_0^{\infty} e^{-x} e^{-ikx} dx \] Calculate each: 1. For \(x<0\): \[ I_1 = \int_{-\infty}^0 e^{x} e^{-ikx} dx = \int_{-\infty}^0 e^{(1 - ik)x} dx \] Since \(\text{Re}(1 - ik) = 1 > 0\), the integral converges: \[ I_1 = \left[ \frac{e^{(1 - ik)x}}{1 - ik} \right]_{x=-\infty}^0 = \frac{1}{1 - ik} \] 2. For \(x > 0\): \[ I_2 = \int_0^{\infty} e^{-x} e^{-ikx} dx = \int_0^{\infty} e^{-(1 + ik)x} dx \] \[ I_2 = \left[ \frac{-e^{-(1 + ik)x}}{1 + ik} \right]_0^{\infty} = \frac{1}{1 + ik} \] (since the exponential decays to zero as \(x \to \infty\)) Adding both: \[ \hat{u}(k, 0) = I_1 + I_2 = \frac{1}{1 - ik} + \frac{1}{1 + ik} \] Combine: \[ \hat{u}(k, 0) = \frac{(1 + ik) + (1 - ik)}{(1 - ik)(1 + ik)} = \frac{2}{1 + k^2} \] --- ### **Step 5: Write the solution in Fourier space** \[ \hat{u}(k, t) = \frac{2}{1 + k^2} e^{-k^2 t} \] --- ### **Step 6: Inverse Fourier Transform to find \(u(x, t)\)** \[ u(x, t) = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{u}(k, t) e^{ikx} dk \] Substitute: \[ u(x, t) = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{2}{1 + k^2} e^{-k^2 t} e^{ikx} dk \] Simplify: \[ u(x, t) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{-k^2 t} e^{ikx}}{1 + k^2} dk \] --- ### **### **Final expression for the solution:** \[ \boxed{ \boxed{ u(x, t) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{-k^2 t} e^{ikx}}{1 + k^2} \, dk } } \] This integral expresses the solution explicitly. For particular values of \(x\) and \(t\), it can be evaluated numerically or approximated further if needed. --- ### **Summary of key steps:** - Transformed PDE to ODE in Fourier space. - Solved the ODE using initial Fourier data. - Computed the Fourier transform of the initial condition. - Inverted the Fourier transform to get the solution in integral form. This completes the detailed solution.