Decay estimate for the heat equation: $sup_t>0int_mathbbR t^alpha |u_x|^2 dx$

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Let $u$ be a solution of the heat equation $$u_t - u_xx = 0, quad t>0, x in mathbbR$$
with initial data $u(0,cdot) = u_0$.
Fix $alpha >0$. How can I estimate (without using explicitly the heat kernel)
$$sup_t>0int_mathbbR t^alpha |u_x|^2 dx,$$
in terms of the initial data? Could you point out a reference where such an estimate is obtained?



Is it fair to call what we obtain a decay estimate?




calculus real-analysis differential-equations reference-request pde




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    Let $u$ be a solution of the heat equation $$u_t - u_xx = 0, quad t>0, x in mathbbR$$
    with initial data $u(0,cdot) = u_0$.
    Fix $alpha >0$. How can I estimate (without using explicitly the heat kernel)
    $$sup_t>0int_mathbbR t^alpha |u_x|^2 dx,$$
    in terms of the initial data? Could you point out a reference where such an estimate is obtained?



    Is it fair to call what we obtain a decay estimate?




    calculus real-analysis differential-equations reference-request pde




    share|cite|improve this question





















      up vote
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      up vote
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      down vote

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      Let $u$ be a solution of the heat equation $$u_t - u_xx = 0, quad t>0, x in mathbbR$$
      with initial data $u(0,cdot) = u_0$.
      Fix $alpha >0$. How can I estimate (without using explicitly the heat kernel)
      $$sup_t>0int_mathbbR t^alpha |u_x|^2 dx,$$
      in terms of the initial data? Could you point out a reference where such an estimate is obtained?



      Is it fair to call what we obtain a decay estimate?




      calculus real-analysis differential-equations reference-request pde




      share|cite|improve this question











      Let $u$ be a solution of the heat equation $$u_t - u_xx = 0, quad t>0, x in mathbbR$$
      with initial data $u(0,cdot) = u_0$.
      Fix $alpha >0$. How can I estimate (without using explicitly the heat kernel)
      $$sup_t>0int_mathbbR t^alpha |u_x|^2 dx,$$
      in terms of the initial data? Could you point out a reference where such an estimate is obtained?



      Is it fair to call what we obtain a decay estimate?





      calculus real-analysis differential-equations reference-request pde





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      asked Jul 25 at 12:44









      Riku

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          The initial condition has been badly stated since the answer would then be$$u(x,t)=u_0quad,quad forall x,t$$this makes sense since when all the bar has the same temperature, there is no cause to change it anywhere and anytime, but if a only a limited part of bar has a different temperature and the rest have different ones, heat then can be propagated (from high temperature parts to low temperature parts).






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            The initial condition has been badly stated since the answer would then be$$u(x,t)=u_0quad,quad forall x,t$$this makes sense since when all the bar has the same temperature, there is no cause to change it anywhere and anytime, but if a only a limited part of bar has a different temperature and the rest have different ones, heat then can be propagated (from high temperature parts to low temperature parts).






            share|cite|improve this answer

























              up vote
              0
              down vote













              The initial condition has been badly stated since the answer would then be$$u(x,t)=u_0quad,quad forall x,t$$this makes sense since when all the bar has the same temperature, there is no cause to change it anywhere and anytime, but if a only a limited part of bar has a different temperature and the rest have different ones, heat then can be propagated (from high temperature parts to low temperature parts).






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                The initial condition has been badly stated since the answer would then be$$u(x,t)=u_0quad,quad forall x,t$$this makes sense since when all the bar has the same temperature, there is no cause to change it anywhere and anytime, but if a only a limited part of bar has a different temperature and the rest have different ones, heat then can be propagated (from high temperature parts to low temperature parts).






                share|cite|improve this answer













                The initial condition has been badly stated since the answer would then be$$u(x,t)=u_0quad,quad forall x,t$$this makes sense since when all the bar has the same temperature, there is no cause to change it anywhere and anytime, but if a only a limited part of bar has a different temperature and the rest have different ones, heat then can be propagated (from high temperature parts to low temperature parts).







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Aug 2 at 9:10









                Mostafa Ayaz

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