Part(a)

\begin{array}{l}

\mathop u\nolimits_t = - 2x,\\

\mathop u\nolimits_x = - 2t - 2x\\

\mathop u\nolimits_{xx} = - 2

\end{array}

plugging in the equation, so âˆ’2x = âˆ’2x. u(x,t) is the solution which satisfies the equation

part(a):

By principle of maximum theorem. If a maximum is at an interior point, then ux=ut=0,then x=t=0, which lies on the bottom boundary, for u = 0 âˆ€x âˆˆ (âˆ’2, 2). That is there is no maximum in the interior. Now consider case of the two side edges, firstly on the left side, where {(x, t)|x = âˆ’2, 0 â‰¤ t â‰¤ 1}, u(âˆ’2, t) = 4(t âˆ’ 1) â‰¤ 0 , for any 0 â‰¤ t â‰¤ 1. Next consider the right side, where {(x, t)|x = 2, 0 â‰¤ t â‰¤ 1}, u(2, t) = âˆ’4(t + 1) < 0, for all 0 â‰¤ t â‰¤ 1.Therefore we can conclude that the maximum value of u on the bottom and sides is 0. Consider the case of the top, where {(x, t)| âˆ’ 2 < x < 2, t = 1},

u(x, 1) = âˆ’2x âˆ’ x^2. If the maximum exists, then x âˆˆ (âˆ’2, 2) and u_x (x,1) = 0.

From part(a), u_x (x,1) = âˆ’2 âˆ’ 2x. Taking u_x=0, then x=-1, hence u_max=u(-1,1)=1

part(b):

consider these situations below:

u_t = âˆ’2x, u_t(1,1)=2>0

u_xx = âˆ’2, u_xx(-1,1)=-2<0

but u(x,t) still satisfies the PDE equation.