Question

ut =9uxx

u(0,t)=u(2π,t) 0, t > 0

u(x,0) = {x , 0 < x < π

{- 3x, π < x < 2π

Answer 1

$Given -$

$u_{t}=9u_{xx}$

so we can write this equation as =

$=\dfrac{\partial u}{\partial t}=9\dfrac{\partial^{2} u}{\partial u^ {2}}.....................................1)$

$=$ $Boundry \ conditions$

$u(0,t)=u(2{\pi},t)=0,t>0$

Since $u(0,$ $t)$ and $u(2{\pi},t)$ is given , so we will apply fourier sine transfrom with

$=\ Initial\ condition$

$u(x,0)=\ {x},\ 0<x<{\pi}$

$\ -{3x}$ $,{-\pi}<x<2{\pi}$

$=$ $\int \dfrac{\partial u}{\partial t}sin\dfrac{2{\pi}x}{2{\pi}}dx=\int 9\dfrac{\partial^{2} u}{\partial u^ {2}}sin\dfrac{2{\pi}x}{2{\pi}}dx$

$=\dfrac{d}{dt}\int_{0}^{2{\pi}}usinxdx=$ $9\int_{0}^{2{\pi}}\dfrac{\partial^{2} u}{\partial u^ {2}}sin{x}dx$

$=\dfrac{d\hat{u_s}(s,t)}{dt}=-9s[u(x,t)cossx|_{0}^{2{\pi}}+9s\int_{0}^{2{\pi}}usinsxdx]$

$=$ $\dfrac{d\hat{u_s}(s,t)}{dt}=9s[u(2{\pi},t)coss2{\pi}-u_{(0,t)}]-9s^{2}\int_{0}^{2{\pi}}u(x,t)sinsxdx$

$\because$ $u({2{\pi},t})=0=u_{(0,t)}$

$=$ $\dfrac{d\hat{u_s}(s,t)}{dt}=-9s^{2}\hat{u_s}(s,t).............................................4)$

where $\hat{u_s}(s,t)=\int_{0}^{2\pi}u(x,t)sinsxdx$

Now solving equation 4) we get ,

$\hat{u_s}(s,t)=9Ce^{-s^{2}t}..........................................5) , where \ C\ is\ an\ arbitrary\ constant$

Put t=0 in equation 5) , we get ,

$c=\hat{u_s}(s,0)=\int_{0}^{2{\pi}}u(x,0)sinsxdx$

As interval 0 \ to\ 2{\pi}\ is given in interval , so we have to also break the above integeral , so the integeral can be break like this .

$=$ $\hat{u_s}(s,0)=\int_{0}^{2{\pi}}u(x,0)sinsxdx=\int_{0}^{{\pi}}xsinsxdx-3\int_{\pi}^{2{\pi}}xsinsxdx$

$=\dfrac{\pi(-1)^{s+1}}{s}-3\dfrac{[{\pi}s(-1)^{s}-2{\pi}s]}{s^{2}}...................6)$

From 5) and 6) , we have -

$\hat{u_s}(s,t)=$$[\dfrac{\pi(-1)^{s+1}}{s}-3\dfrac{[{\pi}s(-1)^{s}-2{\pi}s]}{s^{2}}]e^{-s^{2}t}$

Now taking inverse fourier sin trasnform , we get -

$u(x,t)=\dfrac{18}{\pi}\Sigma_{s=0}^{1}[\dfrac{\pi(-1)^{s+1}}{s}-3\dfrac{[{\pi}s(-1)^{s}-2{\pi}s]}{s^{2}}]e^{-s^{2}t}sinsx$