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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.