if f is linearly bounded in its second argument then each solution to u'(t)=f(t,u(t)) is bounded.
Let $(X,Vert Vert)$ be a Banach space and $f:mathbb{R} times X rightarrow X$ continuous.
If f is linearly bounded in its second argument, i.e. there exist $alpha, beta in C(mathbb{R} ; mathbb{R}_0^+) cap L^1(mathbb{R})$ such that for arbitrary $(t,v) in mathbb{R} times X$ we have:
$Vert f(t,v) Vert leq alpha (t) + beta (t) Vert v Vert$
then every solution of $u'(t) = f(t,u(t)), t in mathbb{R}$ with $u(t_0) = u_0 in X, t_0 in mathbb{R}$ is bounded.
My attempt:
Given an arbitrary bounded interval, say wlog $V := (t_1, t_2) subset mathbb{R}$ we try to show that $u(v) subset overline{B(u_0, r)}$.
Let $t in V$ be arbitrary. The solution u has the form $u(t) = u_0 + int_{t_0}^t f(x, u(x)) dx$, therefore:
$Vert u(t) - u_0 Vert = Vert u_0 + int_{t_0}^t f(x, u(x)) dx - u_0 Vert leq int_{t_0}^t f(x,u(x)) dx Vert leq int_{t_0}^t alpha (x) dx + Vert u Vert int_{t_0}^t beta (x) dx $
I wanted to use that $alpha, beta in L^1(mathbb{R})$ and then choose the radius the be the supremum over all radii for all t in $(t_1,t_2)$, but the problem is that I get $Vert u Vert $ on the right hand side, which is the term I want to show to be bounded in the first place. I'm also not sure whether I can assume the interval $(t_1, t_2)$ to be of this particular form. Can anybody help out here please?
real-analysis differential-equations banach-spaces
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
add a comment |
Let $(X,Vert Vert)$ be a Banach space and $f:mathbb{R} times X rightarrow X$ continuous.
If f is linearly bounded in its second argument, i.e. there exist $alpha, beta in C(mathbb{R} ; mathbb{R}_0^+) cap L^1(mathbb{R})$ such that for arbitrary $(t,v) in mathbb{R} times X$ we have:
$Vert f(t,v) Vert leq alpha (t) + beta (t) Vert v Vert$
then every solution of $u'(t) = f(t,u(t)), t in mathbb{R}$ with $u(t_0) = u_0 in X, t_0 in mathbb{R}$ is bounded.
My attempt:
Given an arbitrary bounded interval, say wlog $V := (t_1, t_2) subset mathbb{R}$ we try to show that $u(v) subset overline{B(u_0, r)}$.
Let $t in V$ be arbitrary. The solution u has the form $u(t) = u_0 + int_{t_0}^t f(x, u(x)) dx$, therefore:
$Vert u(t) - u_0 Vert = Vert u_0 + int_{t_0}^t f(x, u(x)) dx - u_0 Vert leq int_{t_0}^t f(x,u(x)) dx Vert leq int_{t_0}^t alpha (x) dx + Vert u Vert int_{t_0}^t beta (x) dx $
I wanted to use that $alpha, beta in L^1(mathbb{R})$ and then choose the radius the be the supremum over all radii for all t in $(t_1,t_2)$, but the problem is that I get $Vert u Vert $ on the right hand side, which is the term I want to show to be bounded in the first place. I'm also not sure whether I can assume the interval $(t_1, t_2)$ to be of this particular form. Can anybody help out here please?
real-analysis differential-equations banach-spaces
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
1
Grönwall's inequality
– user539887
Nov 27 '18 at 21:09
add a comment |
Let $(X,Vert Vert)$ be a Banach space and $f:mathbb{R} times X rightarrow X$ continuous.
If f is linearly bounded in its second argument, i.e. there exist $alpha, beta in C(mathbb{R} ; mathbb{R}_0^+) cap L^1(mathbb{R})$ such that for arbitrary $(t,v) in mathbb{R} times X$ we have:
$Vert f(t,v) Vert leq alpha (t) + beta (t) Vert v Vert$
then every solution of $u'(t) = f(t,u(t)), t in mathbb{R}$ with $u(t_0) = u_0 in X, t_0 in mathbb{R}$ is bounded.
My attempt:
Given an arbitrary bounded interval, say wlog $V := (t_1, t_2) subset mathbb{R}$ we try to show that $u(v) subset overline{B(u_0, r)}$.
Let $t in V$ be arbitrary. The solution u has the form $u(t) = u_0 + int_{t_0}^t f(x, u(x)) dx$, therefore:
$Vert u(t) - u_0 Vert = Vert u_0 + int_{t_0}^t f(x, u(x)) dx - u_0 Vert leq int_{t_0}^t f(x,u(x)) dx Vert leq int_{t_0}^t alpha (x) dx + Vert u Vert int_{t_0}^t beta (x) dx $
I wanted to use that $alpha, beta in L^1(mathbb{R})$ and then choose the radius the be the supremum over all radii for all t in $(t_1,t_2)$, but the problem is that I get $Vert u Vert $ on the right hand side, which is the term I want to show to be bounded in the first place. I'm also not sure whether I can assume the interval $(t_1, t_2)$ to be of this particular form. Can anybody help out here please?
real-analysis differential-equations banach-spaces
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
Let $(X,Vert Vert)$ be a Banach space and $f:mathbb{R} times X rightarrow X$ continuous.
If f is linearly bounded in its second argument, i.e. there exist $alpha, beta in C(mathbb{R} ; mathbb{R}_0^+) cap L^1(mathbb{R})$ such that for arbitrary $(t,v) in mathbb{R} times X$ we have:
$Vert f(t,v) Vert leq alpha (t) + beta (t) Vert v Vert$
then every solution of $u'(t) = f(t,u(t)), t in mathbb{R}$ with $u(t_0) = u_0 in X, t_0 in mathbb{R}$ is bounded.
My attempt:
Given an arbitrary bounded interval, say wlog $V := (t_1, t_2) subset mathbb{R}$ we try to show that $u(v) subset overline{B(u_0, r)}$.
Let $t in V$ be arbitrary. The solution u has the form $u(t) = u_0 + int_{t_0}^t f(x, u(x)) dx$, therefore:
$Vert u(t) - u_0 Vert = Vert u_0 + int_{t_0}^t f(x, u(x)) dx - u_0 Vert leq int_{t_0}^t f(x,u(x)) dx Vert leq int_{t_0}^t alpha (x) dx + Vert u Vert int_{t_0}^t beta (x) dx $
I wanted to use that $alpha, beta in L^1(mathbb{R})$ and then choose the radius the be the supremum over all radii for all t in $(t_1,t_2)$, but the problem is that I get $Vert u Vert $ on the right hand side, which is the term I want to show to be bounded in the first place. I'm also not sure whether I can assume the interval $(t_1, t_2)$ to be of this particular form. Can anybody help out here please?
real-analysis differential-equations banach-spaces
real-analysis differential-equations banach-spaces
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
asked Nov 27 '18 at 20:15
eager2learneager2learn
1,23711430
1,23711430
1
Grönwall's inequality
– user539887
Nov 27 '18 at 21:09
add a comment |
1
Grönwall's inequality
– user539887
Nov 27 '18 at 21:09
1
1
Grönwall's inequality
– user539887
Nov 27 '18 at 21:09
Grönwall's inequality
– user539887
Nov 27 '18 at 21:09
add a comment |
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Grönwall's inequality
– user539887
Nov 27 '18 at 21:09