I Weak Maximum Principle Explained

[PeteSampras]

Hello,I am reading the link http://math.mit.edu/~jspeck/18.152_Fall2011/Lecture%20notes/18152%20lecture%20notes%20-%204.pdfSays :
w_t-D w_{xx}=f with f<0w at \bar{Q}_T has its maximum in \partial_p {Q}_T. If w is strictly negative at \partial_p {Q}_T then also is strictly negative in \bar{Q}_T(it is OK)Says u=w-\epsilon t , u \leq w, w \leq u + \epsilon T, T is cota,
then u_{t}-Du_{xx}=f-\epsilon &lt;0 (1)(it is OK)Says: Claim that the maximum of u in \bar{Q}_{T-\epsilon} is on \partial_p {Q}_{T-\epsilon}. To verify the claim we use (t_0,x_0) \in \bar{Q}_{T-\epsilon}.Says: t_0 \in (0,T-\epsilon] since if t=0 the claim is true I don't understand this .Says u_t=0 if t_0 \in (0,T-\epsilon) (it is OK), but says u_t \geq 0 if t_0 =T-\epsilon I don't understand this .Then using Taylor and claims:u_{t}-Du_{xx}&gt;0 (2) and says "which contradicts (1)" I don't understand thisBest regard.

 

[zwierz]

since if

t=0​

t=0 the claim is true I don't understand this .

because $t=0$ corresponds to the boundary (bottom)

but says

ut≥0​

u_t \geq 0 if

t0=T−ϵ​

t_0 =T-\epsilon I don't understand this .

$0\ge u(t,x_0)-u(t_0,x_0)=u_t(t_0,x_0)(t-t_0)+o(t-t_0),\quad t<t_0$

the last question: the left hand side of (1.0.3) is $\le 0$ by assumption and the sign in (1.0.4) is $\ge$