Question

Let $u(x,t)$ be the solution of the initial–boundary value problem \[ \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=0, x\in(0,2),\ t>0, \] \[ u(x,0)=\sin(\pi x), x\in(0,2),\qquad u(0,t)=u(2,t)=0. \] Then the value of $e^{\pi^2}\!\left(u\!\left(\tfrac{1}{2},1\right)-u\!\left(\tfrac{3}{2},1\right)\right)$ is ________________ (in integer).

Hint:

On $(0,L)$ with Dirichlet BCs, the heat equation expands in $\sin\!\left(\frac{n\pi x}{L}\right)$ with decay $e^{-(n\pi/L)^2 t}$. If the initial data is itself one eigenfunction, the solution is just that single mode.

Answer 1

Correct Answer: 2

Step 1: Eigenfunctions for the heat equation on $(0,2)$.
With Dirichlet boundaries, separated solutions are \[ u_n(x,t)=\sin\!\left(\frac{n\pi x}{2}\right)e^{-\left(\frac{n\pi}{2}\right)^2 t}, n=1,2,\ldots \]
Step 2: Match the initial condition.
$u(x,0)=\sin(\pi x)=\sin\!\left(\frac{2\pi x}{2}\right)$ corresponds to the single mode $n=2$. Hence \[ u(x,t)=\sin(\pi x)\,e^{-\pi^2 t}. \]
Step 3: Evaluate the requested expression.
At $t=1$, \[ u\!\left(\tfrac{1}{2},1\right)=\sin\!\left(\frac{\pi}{2}\right)e^{-\pi^2}=1 . e^{-\pi^2}, u\!\left(\tfrac{3}{2},1\right)=\sin\!\left(\frac{3\pi}{2}\right)e^{-\pi^2}=-1 . e^{-\pi^2}. \] Therefore \[ e^{\pi^2}\!\left(u\!\left(\tfrac{1}{2},1\right)-u\!\left(\tfrac{3}{2},1\right)\right) =e^{\pi^2}\!\left(e^{-\pi^2}-(-e^{-\pi^2})\right)=1-(-1)=2. \] Final Answer: \[ \boxed{2} \]