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
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
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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.
Updated On: Feb 2, 2026
- \( \frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1) \)
- \( \frac{e^{-49t}}{2\sqrt{2}} (1 - e^{48t}) \)
- \( \frac{e^{-49t}}{2\sqrt{2}} (1 + e^{48t}) \)
- \( \frac{e^{-49t}}{4\sqrt{2}} (1 - e^{48t}) \)
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The Correct Option is A
Solution and Explanation
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:
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:
\[ \boxed{A \quad \frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1)} \]
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:
- \( X''(x) + \lambda X(x) = 0 \)
- \( T'(t) + \lambda T(t) = 0 \)
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:
\[ \boxed{A \quad \frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1)} \]
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