Let $ u = u(x, t) $ be the solution of $$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}, \quad 00, $$ with boundary conditions $ u(0, t) = u(1, t) = 0 $ and initial condition $ u(x, 0) = \sin(\pi x) $. Define $$ g(t) = \int_0^1 u^2(x, t) \, dx. $$ Which one of the following is correct?

Question

Let $ u = u(x, t) $ be the solution of $$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}, \quad 0<x<1, \, t>0, $$ with boundary conditions $ u(0, t) = u(1, t) = 0 $ and initial condition $ u(x, 0) = \sin(\pi x) $. Define $$ g(t) = \int_0^1 u^2(x, t) \, dx. $$ Which one of the following is correct?

cdquestions Admin · Accepted Answer

Step 1: Analyzing the heat equation. The solution $ u(x, t) = e^{-\pi^2 t} \sin(\pi x) $. Substituting this into $ g(t) $, we get: $$ g(t) = \int_0^1 \left( e^{-\pi^2 t} \sin(\pi x) \right)^2 \, dx = e^{-2\pi^2 t} \int_0^1 \sin^2(\pi x) \, dx. $$ Step 2: Simplifying $ g(t) $. $$ \int_0^1 \sin^2(\pi x) \, dx = \frac{1}{2}, $$ so $$ g(t) = \frac{1}{2} e^{-2\pi^2 t}. $$ Step 3: Behavior of $ g(t) $. - $ g(t) $ is decreasing as $ e^{-2\pi^2 t} $ decreases over $ t>0 $. - As $ t \to \infty $, $ g(t) \to 0 $. Step 4: Conclusion. The correct statement is $ {(1)} $.

Answer

$ g $ is decreasing on $ (0, \infty) $ and $ \lim_{t \to \infty} g(t) = 0 $

Answer

$ g $ is decreasing on $ (0, \infty) $ and $ \lim_{t \to \infty} g(t) = \frac{1}{4} $

Answer

$ g $ is increasing on $ (0, \infty) $ and $ \lim_{t \to \infty} g(t) $ does not exist

Answer

$ g $ is increasing on $ (0, \infty) $ and $ \lim_{t \to \infty} g(t) = 3 $

Show Hint

The Correct Option is A

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