Question

Let $u(x,t)$ be the solution of the initial boundary value problem: \[ \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0, 0<x<\pi, \; t>0, \] \[ u(x,0) = 2 \sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{x}{2}\right), 0<x<\pi, \] \[ u(0,t) = u(\pi,t) = 0, t>0. \] Then the value of $\lim_{t\to\infty u\!\left(\tfrac{3\pi}{4},t\right)$ is equal to (rounded off to two decimal places).}

Hint:

In heat equations with homogeneous boundary conditions, the long-term behavior is dominated by the lowest eigenmode (here $\sin(x)$). Higher modes vanish exponentially faster.

Answer 1

Correct Answer: 0.68

Step 1: Recognize PDE type.
The given PDE is: \[ u_t = u_{xx}, \] which is the classical one-dimensional heat/diffusion equation on $(0,\pi)$ with homogeneous Dirichlet boundary conditions.

Step 2: Eigenfunction expansion.
The general solution is expanded in Fourier sine series: \[ u(x,t) = \sum_{n=1}^{\infty} A_n e^{-n^2 t} \sin(nx). \]

Step 3: Simplify initial condition.
Given initial function: \[ u(x,0) = 2 \sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{x}{2}\right). \] Use identity $2\sin A \cos B = \sin(A+B)+\sin(A-B)$: \[ u(x,0) = \sin\!\left(\tfrac{3x}{2}+\tfrac{x}{2}\right) + \sin\!\left(\tfrac{3x}{2}-\tfrac{x}{2}\right). \] \[ = \sin(2x) + \sin(x). \] So the Fourier sine expansion has only terms $\sin(x)$ and $\sin(2x)$.

Step 4: Construct solution.
Thus, \[ u(x,t) = e^{-1^2 t}\sin(x) + e^{-2^2 t}\sin(2x). \]



Step 5: Long-time behavior.
As $t \to \infty$, the higher-order exponential terms vanish faster. The smallest eigenvalue ($n=1$) dominates. Hence, \[ \lim_{t \to \infty} u(x,t) = \sin(x). \]

Step 6: Evaluate at $x=\tfrac{3\pi{4}$.}
\[ \sin\!\left(\tfrac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.71. \] But note: check sign carefully — since $u(x,0)=\sin(2x)+\sin(x)$, both coefficients were $+1$. So indeed final steady dominant term is $+\sin(x)$. Therefore, \[ \lim_{t\to\infty} u\!\left(\tfrac{3\pi}{4},t\right) = \sin\!\left(\tfrac{3\pi}{4}\right) = +0.71. \] Final Answer:
\[ \boxed{0.71} \]