Global existence of IVP on $mathbb R$
$begingroup$
Suppose $f(t, x)in C^1(mathbb R^2)$ and satisfies $|f(t, x) |leq 1+|x|$ for any $(t, x) inmathbb R^2$. Then prove that the IVP $x'(t) =f(t, x)$ and $x(0)=0$ has a solution defined on whole $mathbb R$.
I tried to find local solutions by restricting $f$ to rectangles. But it seems like it couldn't be guaranteed that the solution even exists on the whole interval. Another idea is to try to prove it by approximating $f$ uniformly by polynomials. I'm sure how to make it work. I'd like to get hints on how to solve this. Thanks in advance.
ordinary-differential-equations metric-spaces
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
$endgroup$
add a comment |
$begingroup$
Suppose $f(t, x)in C^1(mathbb R^2)$ and satisfies $|f(t, x) |leq 1+|x|$ for any $(t, x) inmathbb R^2$. Then prove that the IVP $x'(t) =f(t, x)$ and $x(0)=0$ has a solution defined on whole $mathbb R$.
I tried to find local solutions by restricting $f$ to rectangles. But it seems like it couldn't be guaranteed that the solution even exists on the whole interval. Another idea is to try to prove it by approximating $f$ uniformly by polynomials. I'm sure how to make it work. I'd like to get hints on how to solve this. Thanks in advance.
ordinary-differential-equations metric-spaces
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
$endgroup$
add a comment |
$begingroup$
Suppose $f(t, x)in C^1(mathbb R^2)$ and satisfies $|f(t, x) |leq 1+|x|$ for any $(t, x) inmathbb R^2$. Then prove that the IVP $x'(t) =f(t, x)$ and $x(0)=0$ has a solution defined on whole $mathbb R$.
I tried to find local solutions by restricting $f$ to rectangles. But it seems like it couldn't be guaranteed that the solution even exists on the whole interval. Another idea is to try to prove it by approximating $f$ uniformly by polynomials. I'm sure how to make it work. I'd like to get hints on how to solve this. Thanks in advance.
ordinary-differential-equations metric-spaces
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
$endgroup$
Suppose $f(t, x)in C^1(mathbb R^2)$ and satisfies $|f(t, x) |leq 1+|x|$ for any $(t, x) inmathbb R^2$. Then prove that the IVP $x'(t) =f(t, x)$ and $x(0)=0$ has a solution defined on whole $mathbb R$.
I tried to find local solutions by restricting $f$ to rectangles. But it seems like it couldn't be guaranteed that the solution even exists on the whole interval. Another idea is to try to prove it by approximating $f$ uniformly by polynomials. I'm sure how to make it work. I'd like to get hints on how to solve this. Thanks in advance.
ordinary-differential-equations metric-spaces
ordinary-differential-equations metric-spaces
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
asked Dec 5 '18 at 13:59
lEmlEm
3,2621719
3,2621719
add a comment |
add a comment |
1 Answer
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You can use the extensibility theorem - see document, Theorem 1.20 and its corollary about maximal solutions to prove what you want. Or mimic the proof to your particular example.
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
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1 Answer
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1 Answer
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$begingroup$
You can use the extensibility theorem - see document, Theorem 1.20 and its corollary about maximal solutions to prove what you want. Or mimic the proof to your particular example.
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
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add a comment |
$begingroup$
You can use the extensibility theorem - see document, Theorem 1.20 and its corollary about maximal solutions to prove what you want. Or mimic the proof to your particular example.
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
add a comment |
$begingroup$
You can use the extensibility theorem - see document, Theorem 1.20 and its corollary about maximal solutions to prove what you want. Or mimic the proof to your particular example.
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
$endgroup$
You can use the extensibility theorem - see document, Theorem 1.20 and its corollary about maximal solutions to prove what you want. Or mimic the proof to your particular example.
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
answered Dec 5 '18 at 14:48
mathcounterexamples.netmathcounterexamples.net
26k21955
26k21955
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