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Given $u$ solves $u_t - u_{xx} = 16u$ on $(0,\pi)$, with Neumann conditions $u_x (0, t) = u_x(\pi, t) = 0$. Demonstrate the initial value $u_0 = u(x,0)$ such that $u(x,t)$ is bounded as $t\rightarrow \infty$.

My attempt: As $t\rightarrow \infty$, $u_t\rightarrow 0$, and we have: -$u_{xx} = 16u$ in $(0,\pi)$ and $u_x(0, t) = u_x(\pi, t) = 0$ on the boundary. Now, we see that when $u(x,t) = \cos(4x)$, $u$ satisfies the equations, as well as the original equation: $u_t - u_{xx} = 16u$. Since this is a Neumann problem in a bounded domain, the solution $u(x,t)$ has the form $\cos(4x) + f(t)$

If $f(t) = 0$, as $t\rightarrow \infty$, we see that for every values of $u_0$, $u(x,t)$ is always bounded as $|\cos(4x)|\leq 1$.

If $f(t) \neq 0$, then we plug back $u(x,t)= \cos(4x) + f(t)$ into the original equation, and we would need $f'(t) = 16\ f(t)$. This implies $f(t) = e^{16t}$, but then this means $u(x,t)\rightarrow \infty$ as $t\rightarrow \infty$ for any initial data $u_0(x)$. Thus, by choosing $u_0(x)\neq 1 + \cos(4x)$, we can eliminate $u(x,t)= \cos(4x) + f(t)$ as a solution to our original PDE.

My question: From these weird results that I have achieved, I'm not sure why $u_0$ really matters in the behavior of $u(x,t)$ as $t\rightarrow \infty$. Even when I use Variation of Constant formula to get: $u(x,t) = e^{\ t\triangle}u_0 + \int_{0}^{t} 16e^{(t-s)\triangle}u$ (since $f = 16u$ in this case). When $t\rightarrow \infty$, $u_0 = 0$ is the only possible choice, but then the integral might still approach infinity in that case, so why $u_0$ matters?Can someone please help me solve this problem?

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As $t\to\infty$, $u_t\rightarrow 0$

This is incorrect. For example, $u(x,t)=e^{16t}$ is a solution of this boundary value problem with initial data $u(x,0)=1$, and it is unbounded with unbounded time derivative.

Thinking in physical terms: you have a process where the substance does not escape from the interval (Neumann boundary) and is generated inside of it at the rate proportional to the quantity ($16u$ term). The only way this can stay bounded is if the net amount is zero.

Justification: let $m(t) = \int_0^\pi u(x,t)\,dx$, the total amount at time $t$. Then (using the boundary condition) $$m'(t) = \int_0^\pi (u_{xx}+16u)\,dx = \int_0^\pi (16u)\,dx = m(t)$$ Hence $m(t)=e^{16t}m(0)$, making the solution unbounded as $t\to\infty$ unless $m(0)=0$.

So, a necessary condition for boundedness is $\int_0^\pi u_0(x)\,dx=0$. This condition is also sufficient: indeed, $u(x,t)\to 0$ since the mean value remains at $0$ and the diffusion equation evens out the rest. To demonstrate this rigorously, use the Fourier series of $u_0$: the terms with nonzero frequency are killed by diffusion.

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  • $\begingroup$ Thank you very much for your help! I will have to think more about your argument, but it seems to be correct. I solved this problem a while ago by using Fourier series, and figure out the convergence conditions for an infinite series with $\sin$ and $\cos$. Can you please help with another problem here: math.stackexchange.com/questions/1602925/… $\endgroup$ – user177196 Jan 10 '16 at 4:01
  • $\begingroup$ I'm not so sure how your $u(x,t)$ satisfies $u(x,0) = 0$? $\endgroup$ – user177196 Jan 10 '16 at 4:02
  • $\begingroup$ Typo; should have been $1$. $\endgroup$ – user147263 Jan 10 '16 at 4:11
  • $\begingroup$ thanks for your great help. I think your proof is correct! Great one btw:) Can you please help me with another problem - the one in the link? $\endgroup$ – user177196 Jan 10 '16 at 7:13

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