Question

Let \( u(x, t) \) be the solution of the following initial-boundary value problem: \[ \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in (0, \pi), \quad t>0, \] with the boundary conditions: \[ u(0, t) = u(\pi, t) = 0, \quad u(x, 0) = \sin 4x \cos 3x. \] Then, for each \( t>0 \), the value of \( u\left( \frac{\pi}{4}, t \right) \) is

• A. \( \frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1) \)
• B. \( \frac{e^{-49t}}{2\sqrt{2}} (1 - e^{48t}) \)
• C. \( \frac{e^{-49t}}{2\sqrt{2}} (1 + e^{48t}) \)
• D. \( \frac{e^{-49t}}{4\sqrt{2}} (1 - e^{48t}) \)

Hint:

For heat equation problems, use separation of variables and apply the initial and boundary conditions to determine the eigenvalues and eigenfunctions. Use trigonometric identities to simplify the initial condition.

Answer 1

The Correct Option is A

Step 1: Separation of Variables
Assume a solution of the form: \[ u(x, t) = X(x) T(t) \] Substitute into the heat equation \( u_t = u_{xx} \): \[ X(x) T'(t) = X''(x) T(t) \] Divide both sides: \[ \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda \] We obtain two ODEs:

• \( X''(x) + \lambda X(x) = 0 \)
• \( T'(t) + \lambda T(t) = 0 \)

Step 2: Solve the Spatial ODE
Boundary conditions: \( X(0) = X(\pi) = 0 \) ⇒ eigenfunctions: \[ X_n(x) = \sin(nx), \quad \lambda_n = n^2,\quad n \in \mathbb{N} \] Step 3: Solve the Time ODE
\[ T_n(t) = e^{-n^2 t} \] Step 4: General Solution
\[ u(x, t) = \sum_{n=1}^{\infty} b_n \sin(nx) e^{-n^2 t} \] Step 5: Apply Initial Condition
Given \( u(x, 0) = \sin(4x) \cos(3x) \), use identity: \[ \sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \] \[ u(x, 0) = \frac{1}{2}[\sin(7x) + \sin(x)] \Rightarrow b_7 = \frac{1}{2}, b_1 = \frac{1}{2} \] Step 6: Construct the Solution
\[ u(x, t) = \frac{1}{2} \sin(7x) e^{-49t} + \frac{1}{2} \sin(x) e^{-t} \] Step 7: Evaluate at \( x = \frac{\pi}{4} \)
\[ u\left( \frac{\pi}{4}, t \right) = \frac{1}{2} \sin\left(\frac{7\pi}{4}\right) e^{-49t} + \frac{1}{2} \sin\left(\frac{\pi}{4}\right) e^{-t} \] \[ = \frac{1}{2} \left(-\frac{1}{\sqrt{2}}\right) e^{-49t} + \frac{1}{2} \left(\frac{1}{\sqrt{2}}\right) e^{-t} = \frac{1}{2\sqrt{2}}(e^{-t} - e^{-49t}) \] \[ = \frac{e^{-49t}}{2\sqrt{2}}(e^{48t} - 1) \]
Final Answer:
\[ \boxed{A \quad \frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1)} \]