given
$u_t=u_{xx}+e^t\sin(x)$
$u(x,0)=x^2$
$u(0,t)=u(1,t)=t^2$
and
$v_t=v_{xx}+e^t\sin(x)$
$v(x,0)=x$
$v(0,t)=v(1,t)=t$
Does $u(1/2,1/2)>v(1/2,1/2)$?
My attempt:
Define $w=u-v$ and the system of equations become
$w_t=w_{xx}$
$w(x,0)=x^2-x$
$w(0,t)=w(1,t)=t^2-t$ which is the same as $w(1/2,1/2)>0$
Setting $W=X(x)T(t)$
solving first for $X(x)$ we get
$X''(x)+\lambda X(x)=0\to X_n(x)=A\cos(\omega x)+B\sin(\omega x),\lambda=\omega^2$ using the boundary conditions we get
$t^2-t=X(0)=A\cos(\omega 0)+B\sin(\omega 0)=A$
$t^2-t=X(1)=A\cos(\omega 1)+B\sin(\omega 1)\to B=\frac{A(1-\cos(\omega))}{\sin(\omega)}$ afterwards solving for T(t) we get that $T(t)=c_ne^{-t\lambda}$
$w(x,t)=\sum^{\infty}(A\cos(\omega x)+B\sin(\omega x))c_ne^{-t\lambda}$
and when i tried to use $w(x,0)=x^2-x$ i got stuck any help please :)
Are you sure you aren't suppose to show $u(1/2,1/2)<v(1/2,1/2)$?
You are on the right track with defining $w$, but instead of trying to solve for $w$ you should use the maximum principle. Indeed, $w_t=w_{xx}$ in $(0,1)\times (0,1)$. When $x=0$ or $x=1$ then $w(x,t)=t^2-t\leqslant 0$ for all $0\leqslant t <1$. Moreover, when $t=0$, $w(x,0)=x^2-x\leqslant 0$ for all $x\in [0,1]$. Hence, $w\leqslant 0$ on the parabolic boundary of $(0,1) \times (0,1)$, so the weak maximum principle implies $w\leqslant 0$ in $[0,1]\times [0,1]$, that is, $u\leqslant v$ in $[0,1]\times [0,1]$.
Finally, if $u(1/2,1/2)=v(1/2,1/2)$ then $w(1/2)=0$, so by strong maximum principle, $w$ is constant in $[0,1]\times [ 0,1/2]$, but this contradicts your boundary conditions. Thus, $u(1/2,1/2)<v(1/2,1/2)$.