0
$\begingroup$

Let $u(x,t)$ be the solution to the problem $u_{t} = 2u_{xx}-2u_{x}-u$, $0

I need to find the best estimates for the bounds $a$ and $b$.

I actually solved a very similar problem here What I'm doing then, is I'm taking the general form of the solution to the linked question,

$\displaystyle u(x,t) = e^{ct}\left( \delta e^{\lambda_{1}x} + \gamma e^{\lambda_{2} x}\right)$,

where $\lambda_{1} = \frac{\sqrt{2c-1}}{2}$ and $\lambda_{2} = -\frac{\sqrt{2c-1}}{2}$, and somehow I need to apply the max/min principle here to find the maximum or minimum that $u(x,t)$ can be in this problem with these boundary condition and initial conditions.

The idea that I had was that $u(x,t)$ is increasing in $x$, and so $u(x,t)$ should increase as $x$ increases. If this were the case, then, $u(0,t) \leq u(x,t) \leq u(1,t)$. I'm not sure if these are the absolute best bounds, though, and if there's anything more that needs to be done.

Could somebody please help me figure this out? I've been coming back to this problem over and over again all day and I still am not sure what to do, although I'm certainly closer than I've ever been before. Thank you.

  • 0
    Please refer to [this](https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_20.pdf) document and let me know if it helps2017-01-12
  • 0
    @caverac there's nothing in there about finding bounds on a solution2017-01-12
  • 0
    @JessyunBourne, do you have condition for $u_x(0,t)$? You state that $u(x,t)$ is increasing in $x$? I can't see it.2017-01-12
  • 0
    @xpaul actually I thought about it a little more last night and I don't think it is either. The opposite in fact - I thought maybe it was increasing in t and decreasing in x2017-01-12

0 Answers 0