[problem] Let $\Omega = (0, \pi) \subseteq \mathbb{R}$ and $T > 0$. Solve the initial/boundary-value problem \begin{align*} \begin{cases} \partial_t u - \partial_x^2 u = 0 & \text{in } (0, \pi) \times (0, T], \\ u(0, t) = u(\pi, t) = 0 & \text{for } t \in [0, T], \\ u(x, 0) = \sin(x) + 3\sin(2x) & \text{for } x \in (0, \pi). \end{cases} \end{align*} Verify that the solution satisfies the energy identity from the [Energy Dissipation theorem](/theorems/564), and compute the time $t^*$ at which the energy has decayed to half its initial value. [/problem]

[solution] **Step 1: Separation of variables.** We seek solutions of the form $u(x, t) = X(x) T(t)$ satisfying $X T' = X'' T$, i.e., $T'/T = X''/X = -\lambda$ for some constant $\lambda$. The boundary conditions $u(0, t) = u(\pi, t) = 0$ require $X(0) = X(\pi) = 0$. The eigenvalue problem $-X'' = \lambda X$ on $(0, \pi)$ with Dirichlet conditions has eigenfunctions $X_k(x) = \sin(kx)$ with eigenvalues $\lambda_k = k^2$ for $k = 1, 2, 3, \ldots$. The corresponding time factor satisfies $T_k' = -k^2 T_k$, giving $T_k(t) = e^{-k^2 t}$. **Step 2: Superposition.** The general solution is $u(x, t) = \sum_{k=1}^\infty b_k e^{-k^2 t} \sin(kx)$. Matching the initial data $u(x, 0) = \sin(x) + 3\sin(2x)$ identifies $b_1 = 1$, $b_2 = 3$, and $b_k = 0$ for $k \geq 3$. The solution is: \begin{align*} u(x, t) = e^{-t} \sin(x) + 3 e^{-4t} \sin(2x). \end{align*} **Step 3: Energy computation.** Using the orthogonality $\int_0^\pi \sin(kx) \sin(jx) \, dx = \frac{\pi}{2} \delta_{kj}$: \begin{align*} E(t) = \frac{1}{2} \int_0^\pi u^2 \, dx = \frac{1}{2} \left[ e^{-2t} \cdot \frac{\pi}{2} + 9 e^{-8t} \cdot \frac{\pi}{2} \right] = \frac{\pi}{4}\left(e^{-2t} + 9e^{-8t}\right). \end{align*} Differentiating: $E'(t) = \frac{\pi}{4}(-2e^{-2t} - 72e^{-8t})$. The gradient squared integral is: \begin{align*} \int_0^\pi |\partial_x u|^2 \, dx = e^{-2t} \cdot \frac{\pi}{2} + 9 \cdot 4 \cdot e^{-8t} \cdot \frac{\pi}{2} = \frac{\pi}{2}(e^{-2t} + 36 e^{-8t}). \end{align*} Then $-\int |\partial_x u|^2 = -\frac{\pi}{2}(e^{-2t} + 36e^{-8t}) = \frac{\pi}{4}(-2e^{-2t} - 72e^{-8t}) = E'(t)$, confirming the [Energy Dissipation identity](/theorems/564). **Step 4: Half-energy time.** The initial energy is $E(0) = \frac{\pi}{4}(1 + 9) = \frac{10\pi}{4} = \frac{5\pi}{2}$. We seek $t^*$ with $E(t^*) = \frac{5\pi}{4}$, i.e., \begin{align*} e^{-2t^*} + 9e^{-8t^*} = 5. \end{align*} Setting $s = e^{-2t^*}$: $s + 9s^4 = 5$. At $t^* = 0$: $s = 1$, LHS $= 10 > 5$. For small $t^*$, the $9s^4$ term dominates the decay. By monotonicity, a unique solution exists. Newton's method with initial guess $t^* = 0.1$ (giving $s \approx 0.819$, LHS $\approx 4.87$) converges to $t^* \approx 0.094$. The energy halves very quickly because the higher-frequency mode ($k = 2$) decays four times faster than the fundamental. [/solution]