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 $f$?
real-analysis functional-analysis differential-equations sobolev-spaces
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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 $f$?
real-analysis functional-analysis differential-equations sobolev-spaces
add a comment |Â
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 $f$?
real-analysis functional-analysis differential-equations sobolev-spaces
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 $f$?
real-analysis functional-analysis differential-equations sobolev-spaces
edited Aug 9 at 20:48
asked Aug 9 at 20:40
zorro47
497412
497412
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