Justify that $ V(s)V'(s)=f(s,V(s))$ has locally exactly one solution












1












$begingroup$


Let $f in C^1(mathbb{R}^2)$



Let $V:(a,b)rightarrow mathbb{R}$ a solution of $ V(s)V'(s)=f(s,V(s))$ $(1)$



1.Justify that $(1)$ has locally exactly one solution with $V(s_{0})=V_{0}>0$.



2.If $(a,b)$ with $-infty leq a <s_{0} <b leq infty$ is the maximal existence interval, which possibilities exist for $V(s)$ if $s downarrow$ and if $s uparrow b$



I tried to integrate but i can´t solve the Integral $int_{0}^{s}f(t,V(t))dt $.



Another idea was that we divide by $V(s)$ in a neightbourhood $U$of $s_{0}$ and conclude with $1/V(s)=F(s,V(s)) in C^1(U)$ and use Picard-Lindelöf on $f*F in C^1(u)$.



I´m not sure how to use this for task $2$ if it is right. Can someone give me some hints or something.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your really wanted to write $V(s)V'(s)$ like in $frac12frac{d}{ds}(V(s)^2)$? Then set $y=V^2$ so that $y'=2f(t,sqrt{y})$ and proceed from there.
    $endgroup$
    – LutzL
    Dec 18 '18 at 1:38










  • $begingroup$
    What do u mean with procced from there? Integrating doesn't work.
    $endgroup$
    – mathbob
    Dec 18 '18 at 8:57










  • $begingroup$
    You are not expected to solve this problem, you only need to apply the existence and uniqueness theorems. Locally you can approximate $f$ by a linear function because of its differentiability. The solution with this linear right side gives qualitative insight into the shape of solutions in this neighborhood. This might be helpful around $V=0$.
    $endgroup$
    – LutzL
    Dec 18 '18 at 9:04












  • $begingroup$
    Remember that this is a dynamical system, its evolution can potentially lead to anywhere in the state space.
    $endgroup$
    – LutzL
    Dec 18 '18 at 11:22










  • $begingroup$
    I don't get it, what you try to explain. Maybe my english is too bad by writing task 1. We have $(1)$ and $V(s_{0})=V_{0}>0$ so we shoud justify with that two things the locally exactly one solution. In my opinioln this follows direct from Picard-L.( existence and uniqueness) . So why is V=0 important? In an neighbourhood of $s_{0}$ is $V neq 0$ then we get $V in C^1$
    $endgroup$
    – mathbob
    Dec 18 '18 at 13:26


















1












$begingroup$


Let $f in C^1(mathbb{R}^2)$



Let $V:(a,b)rightarrow mathbb{R}$ a solution of $ V(s)V'(s)=f(s,V(s))$ $(1)$



1.Justify that $(1)$ has locally exactly one solution with $V(s_{0})=V_{0}>0$.



2.If $(a,b)$ with $-infty leq a <s_{0} <b leq infty$ is the maximal existence interval, which possibilities exist for $V(s)$ if $s downarrow$ and if $s uparrow b$



I tried to integrate but i can´t solve the Integral $int_{0}^{s}f(t,V(t))dt $.



Another idea was that we divide by $V(s)$ in a neightbourhood $U$of $s_{0}$ and conclude with $1/V(s)=F(s,V(s)) in C^1(U)$ and use Picard-Lindelöf on $f*F in C^1(u)$.



I´m not sure how to use this for task $2$ if it is right. Can someone give me some hints or something.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your really wanted to write $V(s)V'(s)$ like in $frac12frac{d}{ds}(V(s)^2)$? Then set $y=V^2$ so that $y'=2f(t,sqrt{y})$ and proceed from there.
    $endgroup$
    – LutzL
    Dec 18 '18 at 1:38










  • $begingroup$
    What do u mean with procced from there? Integrating doesn't work.
    $endgroup$
    – mathbob
    Dec 18 '18 at 8:57










  • $begingroup$
    You are not expected to solve this problem, you only need to apply the existence and uniqueness theorems. Locally you can approximate $f$ by a linear function because of its differentiability. The solution with this linear right side gives qualitative insight into the shape of solutions in this neighborhood. This might be helpful around $V=0$.
    $endgroup$
    – LutzL
    Dec 18 '18 at 9:04












  • $begingroup$
    Remember that this is a dynamical system, its evolution can potentially lead to anywhere in the state space.
    $endgroup$
    – LutzL
    Dec 18 '18 at 11:22










  • $begingroup$
    I don't get it, what you try to explain. Maybe my english is too bad by writing task 1. We have $(1)$ and $V(s_{0})=V_{0}>0$ so we shoud justify with that two things the locally exactly one solution. In my opinioln this follows direct from Picard-L.( existence and uniqueness) . So why is V=0 important? In an neighbourhood of $s_{0}$ is $V neq 0$ then we get $V in C^1$
    $endgroup$
    – mathbob
    Dec 18 '18 at 13:26
















1












1








1


1



$begingroup$


Let $f in C^1(mathbb{R}^2)$



Let $V:(a,b)rightarrow mathbb{R}$ a solution of $ V(s)V'(s)=f(s,V(s))$ $(1)$



1.Justify that $(1)$ has locally exactly one solution with $V(s_{0})=V_{0}>0$.



2.If $(a,b)$ with $-infty leq a <s_{0} <b leq infty$ is the maximal existence interval, which possibilities exist for $V(s)$ if $s downarrow$ and if $s uparrow b$



I tried to integrate but i can´t solve the Integral $int_{0}^{s}f(t,V(t))dt $.



Another idea was that we divide by $V(s)$ in a neightbourhood $U$of $s_{0}$ and conclude with $1/V(s)=F(s,V(s)) in C^1(U)$ and use Picard-Lindelöf on $f*F in C^1(u)$.



I´m not sure how to use this for task $2$ if it is right. Can someone give me some hints or something.










share|cite|improve this question









$endgroup$




Let $f in C^1(mathbb{R}^2)$



Let $V:(a,b)rightarrow mathbb{R}$ a solution of $ V(s)V'(s)=f(s,V(s))$ $(1)$



1.Justify that $(1)$ has locally exactly one solution with $V(s_{0})=V_{0}>0$.



2.If $(a,b)$ with $-infty leq a <s_{0} <b leq infty$ is the maximal existence interval, which possibilities exist for $V(s)$ if $s downarrow$ and if $s uparrow b$



I tried to integrate but i can´t solve the Integral $int_{0}^{s}f(t,V(t))dt $.



Another idea was that we divide by $V(s)$ in a neightbourhood $U$of $s_{0}$ and conclude with $1/V(s)=F(s,V(s)) in C^1(U)$ and use Picard-Lindelöf on $f*F in C^1(u)$.



I´m not sure how to use this for task $2$ if it is right. Can someone give me some hints or something.







ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 0:23









mathbobmathbob

649




649








  • 1




    $begingroup$
    Your really wanted to write $V(s)V'(s)$ like in $frac12frac{d}{ds}(V(s)^2)$? Then set $y=V^2$ so that $y'=2f(t,sqrt{y})$ and proceed from there.
    $endgroup$
    – LutzL
    Dec 18 '18 at 1:38










  • $begingroup$
    What do u mean with procced from there? Integrating doesn't work.
    $endgroup$
    – mathbob
    Dec 18 '18 at 8:57










  • $begingroup$
    You are not expected to solve this problem, you only need to apply the existence and uniqueness theorems. Locally you can approximate $f$ by a linear function because of its differentiability. The solution with this linear right side gives qualitative insight into the shape of solutions in this neighborhood. This might be helpful around $V=0$.
    $endgroup$
    – LutzL
    Dec 18 '18 at 9:04












  • $begingroup$
    Remember that this is a dynamical system, its evolution can potentially lead to anywhere in the state space.
    $endgroup$
    – LutzL
    Dec 18 '18 at 11:22










  • $begingroup$
    I don't get it, what you try to explain. Maybe my english is too bad by writing task 1. We have $(1)$ and $V(s_{0})=V_{0}>0$ so we shoud justify with that two things the locally exactly one solution. In my opinioln this follows direct from Picard-L.( existence and uniqueness) . So why is V=0 important? In an neighbourhood of $s_{0}$ is $V neq 0$ then we get $V in C^1$
    $endgroup$
    – mathbob
    Dec 18 '18 at 13:26
















  • 1




    $begingroup$
    Your really wanted to write $V(s)V'(s)$ like in $frac12frac{d}{ds}(V(s)^2)$? Then set $y=V^2$ so that $y'=2f(t,sqrt{y})$ and proceed from there.
    $endgroup$
    – LutzL
    Dec 18 '18 at 1:38










  • $begingroup$
    What do u mean with procced from there? Integrating doesn't work.
    $endgroup$
    – mathbob
    Dec 18 '18 at 8:57










  • $begingroup$
    You are not expected to solve this problem, you only need to apply the existence and uniqueness theorems. Locally you can approximate $f$ by a linear function because of its differentiability. The solution with this linear right side gives qualitative insight into the shape of solutions in this neighborhood. This might be helpful around $V=0$.
    $endgroup$
    – LutzL
    Dec 18 '18 at 9:04












  • $begingroup$
    Remember that this is a dynamical system, its evolution can potentially lead to anywhere in the state space.
    $endgroup$
    – LutzL
    Dec 18 '18 at 11:22










  • $begingroup$
    I don't get it, what you try to explain. Maybe my english is too bad by writing task 1. We have $(1)$ and $V(s_{0})=V_{0}>0$ so we shoud justify with that two things the locally exactly one solution. In my opinioln this follows direct from Picard-L.( existence and uniqueness) . So why is V=0 important? In an neighbourhood of $s_{0}$ is $V neq 0$ then we get $V in C^1$
    $endgroup$
    – mathbob
    Dec 18 '18 at 13:26










1




1




$begingroup$
Your really wanted to write $V(s)V'(s)$ like in $frac12frac{d}{ds}(V(s)^2)$? Then set $y=V^2$ so that $y'=2f(t,sqrt{y})$ and proceed from there.
$endgroup$
– LutzL
Dec 18 '18 at 1:38




$begingroup$
Your really wanted to write $V(s)V'(s)$ like in $frac12frac{d}{ds}(V(s)^2)$? Then set $y=V^2$ so that $y'=2f(t,sqrt{y})$ and proceed from there.
$endgroup$
– LutzL
Dec 18 '18 at 1:38












$begingroup$
What do u mean with procced from there? Integrating doesn't work.
$endgroup$
– mathbob
Dec 18 '18 at 8:57




$begingroup$
What do u mean with procced from there? Integrating doesn't work.
$endgroup$
– mathbob
Dec 18 '18 at 8:57












$begingroup$
You are not expected to solve this problem, you only need to apply the existence and uniqueness theorems. Locally you can approximate $f$ by a linear function because of its differentiability. The solution with this linear right side gives qualitative insight into the shape of solutions in this neighborhood. This might be helpful around $V=0$.
$endgroup$
– LutzL
Dec 18 '18 at 9:04






$begingroup$
You are not expected to solve this problem, you only need to apply the existence and uniqueness theorems. Locally you can approximate $f$ by a linear function because of its differentiability. The solution with this linear right side gives qualitative insight into the shape of solutions in this neighborhood. This might be helpful around $V=0$.
$endgroup$
– LutzL
Dec 18 '18 at 9:04














$begingroup$
Remember that this is a dynamical system, its evolution can potentially lead to anywhere in the state space.
$endgroup$
– LutzL
Dec 18 '18 at 11:22




$begingroup$
Remember that this is a dynamical system, its evolution can potentially lead to anywhere in the state space.
$endgroup$
– LutzL
Dec 18 '18 at 11:22












$begingroup$
I don't get it, what you try to explain. Maybe my english is too bad by writing task 1. We have $(1)$ and $V(s_{0})=V_{0}>0$ so we shoud justify with that two things the locally exactly one solution. In my opinioln this follows direct from Picard-L.( existence and uniqueness) . So why is V=0 important? In an neighbourhood of $s_{0}$ is $V neq 0$ then we get $V in C^1$
$endgroup$
– mathbob
Dec 18 '18 at 13:26






$begingroup$
I don't get it, what you try to explain. Maybe my english is too bad by writing task 1. We have $(1)$ and $V(s_{0})=V_{0}>0$ so we shoud justify with that two things the locally exactly one solution. In my opinioln this follows direct from Picard-L.( existence and uniqueness) . So why is V=0 important? In an neighbourhood of $s_{0}$ is $V neq 0$ then we get $V in C^1$
$endgroup$
– mathbob
Dec 18 '18 at 13:26












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