-1

$\text{I was asked to solve the equation}\, \frac{\partial u}{\partial t} = \frac{\partial ^2 u}{\partial x^2} -au, $

$\text{for } 0<x<\pi ,\, t>0 ,\ \text{with boundary values} $

$u(0,t)=u(\pi,t)=0,\ when,\ t\ge0,\ and,\ u(x,0)=(sinx)^{3}.$

$\text{I solved it using separation of variables and got the solution to be equal to}$ $$ u(x,t)=\frac{1}{4}e^{-(1+a)t}sinx-\frac{3}{4}e^{-(9+a)t}sin(3x)$$

$\text{But now I'm asked to find for which values of} \ a \ \text{does}$

$$ \ \lim_{t \to \infty} u(x,t)=0.$$

$\text{I'm given the hint: set} \ u(x,t)=\sum_{1}^\infty b_{n}(t)\sin nx$

I have a formula for the $b_{n}$s in my book:

$b_{n}(t)=\frac{2}{\pi}\int_{0}^{\pi} u(x,t)\sin nx dx$

Anyone have any ideas?

Jean Marie
  • 81,803
fejz1234
  • 542

1 Answers1

3

I guess that hint was for the first part. You got the answer without it anyway. For the second part, you want both (1+a) and (9+a) to be positive. Otherwise, one or both of the terms blow(s) up.

Thus, a>-1 is the required answer.

Saurish Chakrabarty
  • 31
  • So there is no way to deduce formally that a should be bigger than -1, other then inspecting when (1+a) and (9+a) are positive? – fejz1234 Feb 26 '17 at 14:16