# I Weak maximum principle

1. May 11, 2017

### PeteSampras

Hello,

Says :
$$w_t-D w_{xx}=f$$ with f<0

w 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 $$u_{t}-Du_{xx}=f-\epsilon <0$$ (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 dont 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 dont understand this .

Then using Taylor and claims:

$$u_{t}-Du_{xx}>0$$ (2) and says "which contradicts (1)" I dont understand this

Best regard.

2. May 11, 2017

### zwierz

because $t=0$ corresponds to the boundary (bottom)
$0\ge u(t,x_0)-u(t_0,x_0)=u_t(t_0,x_0)(t-t_0)+o(t-t_0),\quad t<t_0$

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

Last edited: May 11, 2017