Question

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

Hint:

For boundary value problems involving the heat equation, the solution can often be expressed as a Fourier series, and as \( t \to \infty \), the higher-frequency terms decay, leaving the steady-state solution.

Answer 1

We are given the initial boundary value problem and need to find the limit of \( u \left( \frac{3\pi}{4}, t \right) \) as \( t \to \infty \). The solution to the heat equation can be represented using Fourier series expansion. We can write \( u(x, t) \) as: \[ u(x, t) = \sum_{n=1}^{\infty} B_n \sin(nx) e^{-n^2 t}. \] The initial condition \( u(x, 0) = 2 \sin \left( \frac{3x}{2} \right) \cos \left( \frac{x}{2} \right) \) suggests we use a Fourier expansion to match the initial profile. First, we apply the product-to-sum identity for the cosine term: \[ u(x, 0) = 2 \sin \left( \frac{3x}{2} \right) \cos \left( \frac{x}{2} \right) = \sin \left( \frac{3x}{2} \right) + \sin \left( \frac{5x}{2} \right). \] Thus, we have: \[ u(x, t) = \sum_{n=1}^{\infty} B_n \sin(nx) e^{-n^2 t}, \] where \( B_n \) are the Fourier coefficients. For large \( t \), the exponential term \( e^{-n^2 t} \) decays, and only the lowest-frequency terms (those with the smallest values of \( n \)) contribute significantly to the value of \( u(x, t) \). At \( t \to \infty \), all terms with non-zero frequency decay, and we are left with the constant term corresponding to the lowest-frequency mode. Therefore, evaluating at \( x = \frac{3\pi}{4} \), we get: \[ u\left( \frac{3\pi}{4}, t \right) \to 0.71 { as } t \to \infty. \] Thus, the value of \( \lim_{t \to \infty} u \left( \frac{3\pi}{4}, t \right) \) is approximately 0.71.