Existance on a basic differential equation

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I have a few questions about a basic differetial equation. Below, there are some of my considerations. I would be very greateful for your help.



Let $fcolon[0,T]timesmathbbRto mathbbR$ be continuous and Lipschitz with respect to the second argument. Consider the following Cauchy problem:
$$u'(t)=f(t,x(t)),qquad u(0)=u_0.$$
Then, Pickard's theorem gives unique solution with the following regularity $C([0,T])$. I don't want to look at this problem this way anymore. Hence, I introduce the framework of evolution triples $Vsubset Hsubset V^*$. In what follows, I assume that $fcolon [0,T]times Vto V$ and for every $vin V$ $f(cdot,,v)in L^2(0,T,V)$. Consider the following problem:



Find $uin W^1,2(0,T;V,H)$ such that
$$u'(t)=f(t,x(t))quadtexta.e.quad tin(0,T),qquad u(0)=u_0.qquad (*)$$



Suppose that $u$ is a solution to $(*)$. Since $f$ is integrable, we have that
$$int_0^tu'(s),ds=int_0^tf(s, u(s)),ds,$$
hence
$$u(t)=u(0)+int_0^tf(s, u(s)),ds.qquad (**)$$
Now, I want to apply the following proposition:



Proposition. If $f,gin L^1([0,T],X)$, then the following conditions are equivalent:



a) $f(t)=v+int_0^tg(s),ds$, for almost all $tin [0,T]$,



b) $int_0^tf(s)varphi'(s),ds=-int_0^tg(s)varphi(s),ds$,



c) for every $v^*in V^*$,
$$fracddtlangle v^*, f(cdot)rangle_V=langle v^*, g(cdot)rangle_V$$



in the distributional sense on $(0,T)$.



This gives that $u$ is absolutely continuous ($uin AC^1,2([0,T],V)$). Since the spaces $W^1,2(0,T;V,H)$ and $AC^1,2([0,T],V)$ can be indentified, then, every solution to (*) is of regularity $W^1,2(0,T;V,H)$. Also, a solution of $(*)$ is also a solution to $(**)$. The problem is and it's not clear to me, that if $uin W(0,T;V,H)$, then $u'in L^2(0,T;V^*)$. So, now $u'(t)in V^*$ while from $(*)$ $u'(t)in V$. I think, using the framework of evolution triples, we may understand it two ways - it depends on what we want ???
What is the regularity of $tto int_0^tf(s,u(s)),ds$?



Conversly, suppose that $u$ is a solution to $(**)$. Then, $u$ is absolutely continuous and hence differentiable almost everywhere. So every solution of $(**)$ solves also $(*)$. Once again, the regularity of the solution is $W^1,2(0,T;V,H)$.



Having this, do you think I can repeat fixed point argumentation from Pickard's theorem and prove the existance of $(*)$ adding some hypothesis on $f$?







share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I have a few questions about a basic differetial equation. Below, there are some of my considerations. I would be very greateful for your help.



    Let $fcolon[0,T]timesmathbbRto mathbbR$ be continuous and Lipschitz with respect to the second argument. Consider the following Cauchy problem:
    $$u'(t)=f(t,x(t)),qquad u(0)=u_0.$$
    Then, Pickard's theorem gives unique solution with the following regularity $C([0,T])$. I don't want to look at this problem this way anymore. Hence, I introduce the framework of evolution triples $Vsubset Hsubset V^*$. In what follows, I assume that $fcolon [0,T]times Vto V$ and for every $vin V$ $f(cdot,,v)in L^2(0,T,V)$. Consider the following problem:



    Find $uin W^1,2(0,T;V,H)$ such that
    $$u'(t)=f(t,x(t))quadtexta.e.quad tin(0,T),qquad u(0)=u_0.qquad (*)$$



    Suppose that $u$ is a solution to $(*)$. Since $f$ is integrable, we have that
    $$int_0^tu'(s),ds=int_0^tf(s, u(s)),ds,$$
    hence
    $$u(t)=u(0)+int_0^tf(s, u(s)),ds.qquad (**)$$
    Now, I want to apply the following proposition:



    Proposition. If $f,gin L^1([0,T],X)$, then the following conditions are equivalent:



    a) $f(t)=v+int_0^tg(s),ds$, for almost all $tin [0,T]$,



    b) $int_0^tf(s)varphi'(s),ds=-int_0^tg(s)varphi(s),ds$,



    c) for every $v^*in V^*$,
    $$fracddtlangle v^*, f(cdot)rangle_V=langle v^*, g(cdot)rangle_V$$



    in the distributional sense on $(0,T)$.



    This gives that $u$ is absolutely continuous ($uin AC^1,2([0,T],V)$). Since the spaces $W^1,2(0,T;V,H)$ and $AC^1,2([0,T],V)$ can be indentified, then, every solution to (*) is of regularity $W^1,2(0,T;V,H)$. Also, a solution of $(*)$ is also a solution to $(**)$. The problem is and it's not clear to me, that if $uin W(0,T;V,H)$, then $u'in L^2(0,T;V^*)$. So, now $u'(t)in V^*$ while from $(*)$ $u'(t)in V$. I think, using the framework of evolution triples, we may understand it two ways - it depends on what we want ???
    What is the regularity of $tto int_0^tf(s,u(s)),ds$?



    Conversly, suppose that $u$ is a solution to $(**)$. Then, $u$ is absolutely continuous and hence differentiable almost everywhere. So every solution of $(**)$ solves also $(*)$. Once again, the regularity of the solution is $W^1,2(0,T;V,H)$.



    Having this, do you think I can repeat fixed point argumentation from Pickard's theorem and prove the existance of $(*)$ adding some hypothesis on $f$?







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have a few questions about a basic differetial equation. Below, there are some of my considerations. I would be very greateful for your help.



      Let $fcolon[0,T]timesmathbbRto mathbbR$ be continuous and Lipschitz with respect to the second argument. Consider the following Cauchy problem:
      $$u'(t)=f(t,x(t)),qquad u(0)=u_0.$$
      Then, Pickard's theorem gives unique solution with the following regularity $C([0,T])$. I don't want to look at this problem this way anymore. Hence, I introduce the framework of evolution triples $Vsubset Hsubset V^*$. In what follows, I assume that $fcolon [0,T]times Vto V$ and for every $vin V$ $f(cdot,,v)in L^2(0,T,V)$. Consider the following problem:



      Find $uin W^1,2(0,T;V,H)$ such that
      $$u'(t)=f(t,x(t))quadtexta.e.quad tin(0,T),qquad u(0)=u_0.qquad (*)$$



      Suppose that $u$ is a solution to $(*)$. Since $f$ is integrable, we have that
      $$int_0^tu'(s),ds=int_0^tf(s, u(s)),ds,$$
      hence
      $$u(t)=u(0)+int_0^tf(s, u(s)),ds.qquad (**)$$
      Now, I want to apply the following proposition:



      Proposition. If $f,gin L^1([0,T],X)$, then the following conditions are equivalent:



      a) $f(t)=v+int_0^tg(s),ds$, for almost all $tin [0,T]$,



      b) $int_0^tf(s)varphi'(s),ds=-int_0^tg(s)varphi(s),ds$,



      c) for every $v^*in V^*$,
      $$fracddtlangle v^*, f(cdot)rangle_V=langle v^*, g(cdot)rangle_V$$



      in the distributional sense on $(0,T)$.



      This gives that $u$ is absolutely continuous ($uin AC^1,2([0,T],V)$). Since the spaces $W^1,2(0,T;V,H)$ and $AC^1,2([0,T],V)$ can be indentified, then, every solution to (*) is of regularity $W^1,2(0,T;V,H)$. Also, a solution of $(*)$ is also a solution to $(**)$. The problem is and it's not clear to me, that if $uin W(0,T;V,H)$, then $u'in L^2(0,T;V^*)$. So, now $u'(t)in V^*$ while from $(*)$ $u'(t)in V$. I think, using the framework of evolution triples, we may understand it two ways - it depends on what we want ???
      What is the regularity of $tto int_0^tf(s,u(s)),ds$?



      Conversly, suppose that $u$ is a solution to $(**)$. Then, $u$ is absolutely continuous and hence differentiable almost everywhere. So every solution of $(**)$ solves also $(*)$. Once again, the regularity of the solution is $W^1,2(0,T;V,H)$.



      Having this, do you think I can repeat fixed point argumentation from Pickard's theorem and prove the existance of $(*)$ adding some hypothesis on $f$?







      share|cite|improve this question














      I have a few questions about a basic differetial equation. Below, there are some of my considerations. I would be very greateful for your help.



      Let $fcolon[0,T]timesmathbbRto mathbbR$ be continuous and Lipschitz with respect to the second argument. Consider the following Cauchy problem:
      $$u'(t)=f(t,x(t)),qquad u(0)=u_0.$$
      Then, Pickard's theorem gives unique solution with the following regularity $C([0,T])$. I don't want to look at this problem this way anymore. Hence, I introduce the framework of evolution triples $Vsubset Hsubset V^*$. In what follows, I assume that $fcolon [0,T]times Vto V$ and for every $vin V$ $f(cdot,,v)in L^2(0,T,V)$. Consider the following problem:



      Find $uin W^1,2(0,T;V,H)$ such that
      $$u'(t)=f(t,x(t))quadtexta.e.quad tin(0,T),qquad u(0)=u_0.qquad (*)$$



      Suppose that $u$ is a solution to $(*)$. Since $f$ is integrable, we have that
      $$int_0^tu'(s),ds=int_0^tf(s, u(s)),ds,$$
      hence
      $$u(t)=u(0)+int_0^tf(s, u(s)),ds.qquad (**)$$
      Now, I want to apply the following proposition:



      Proposition. If $f,gin L^1([0,T],X)$, then the following conditions are equivalent:



      a) $f(t)=v+int_0^tg(s),ds$, for almost all $tin [0,T]$,



      b) $int_0^tf(s)varphi'(s),ds=-int_0^tg(s)varphi(s),ds$,



      c) for every $v^*in V^*$,
      $$fracddtlangle v^*, f(cdot)rangle_V=langle v^*, g(cdot)rangle_V$$



      in the distributional sense on $(0,T)$.



      This gives that $u$ is absolutely continuous ($uin AC^1,2([0,T],V)$). Since the spaces $W^1,2(0,T;V,H)$ and $AC^1,2([0,T],V)$ can be indentified, then, every solution to (*) is of regularity $W^1,2(0,T;V,H)$. Also, a solution of $(*)$ is also a solution to $(**)$. The problem is and it's not clear to me, that if $uin W(0,T;V,H)$, then $u'in L^2(0,T;V^*)$. So, now $u'(t)in V^*$ while from $(*)$ $u'(t)in V$. I think, using the framework of evolution triples, we may understand it two ways - it depends on what we want ???
      What is the regularity of $tto int_0^tf(s,u(s)),ds$?



      Conversly, suppose that $u$ is a solution to $(**)$. Then, $u$ is absolutely continuous and hence differentiable almost everywhere. So every solution of $(**)$ solves also $(*)$. Once again, the regularity of the solution is $W^1,2(0,T;V,H)$.



      Having this, do you think I can repeat fixed point argumentation from Pickard's theorem and prove the existance of $(*)$ adding some hypothesis on $f$?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 9 at 20:48

























      asked Aug 9 at 20:40









      zorro47

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