Question

Let $ u(x,t) $ satisfy the initial boundary value problem $$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}, \quad x \in (0,1), t>0 $$ given that $$ u(x, 0) = \sin(\pi x), \quad x \in [0,1], \quad u(0,t) = u(1,t) = 0, \quad t>0. $$ Then $ u \left( x, \frac{1}{\pi^2} \right) $ for $ x \in (0,1) $ is

• A. \( e \sin(\pi x) \)
• B. \( e^{-1} \sin(\pi x) \)
• C. \( \sin(\pi x) \)
• D. \( \sin(\pi^{-1} x) \)

Hint:

In heat conduction problems, the solution involves exponential decay with respect to time and spatial sine functions due to boundary conditions.

Answer 1

The Correct Option is B

The problem is a standard heat conduction problem where the solution can be obtained using separation of variables. The general solution to the heat equation \( \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} \) with boundary conditions \( u(0,t) = u(1,t) = 0 \) and initial condition \( u(x,0) = \sin(\pi x) \) is: \[ u(x,t) = e^{-\pi^2 t} \sin(\pi x) \] This solution satisfies the given boundary conditions and initial conditions. We are asked to find \( u \left( x, \frac{1}{\pi^2} \right) \), so substitute \( t = \frac{1}{\pi^2} \) into the solution: \[ u \left( x, \frac{1}{\pi^2} \right) = e^{-\pi^2 \cdot \frac{1}{\pi^2}} \sin(\pi x) = e^{-1} \sin(\pi x) \]
Conclusion: Thus, the value of \( u \left( x, \frac{1}{\pi^2} \right) \) is \( e^{-1} \sin(\pi x) \).