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I Weak maximum principle

  1. May 11, 2017 #1
    Hello,


    I am reading the link http://math.mit.edu/~jspeck/18.152_Fall2011/Lecture notes/18152 lecture notes - 4.pdf


    Says :
    [tex]w_t-D w_{xx}=f [/tex] with f<0


    w at [tex]\bar{Q}_T[/tex] has its maximum in [tex]\partial_p {Q}_T[/tex]. If w is strictly negative at [tex]\partial_p {Q}_T[/tex] then also is strictly negative in [tex]\bar{Q}_T[/tex]


    (it is OK)


    Says [tex]u=w-\epsilon t [/tex] , [tex]u \leq w [/tex], [tex]w \leq u + \epsilon T [/tex], T is cota,
    then [tex]u_{t}-Du_{xx}=f-\epsilon <0 [/tex] (1)


    (it is OK)


    Says: Claim that the maximum of u in [tex]\bar{Q}_{T-\epsilon}[/tex] is on [tex]\partial_p {Q}_{T-\epsilon}[/tex]. To verify the claim we use [tex](t_0,x_0) \in \bar{Q}_{T-\epsilon}[/tex].


    Says: [tex]t_0 \in (0,T-\epsilon][/tex] since if [tex]t=0[/tex] the claim is true I dont understand this .


    Says [tex]u_t=0[/tex] if [tex]t_0 \in (0,T-\epsilon)[/tex] (it is OK), but says [tex]u_t \geq 0[/tex] if [tex]t_0 =T-\epsilon[/tex] I dont understand this .


    Then using Taylor and claims:


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


    Best regard.
     
  2. jcsd
  3. May 11, 2017 #2
    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
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