Prove that maximal solution of a Cauchy problem has domain $(-infty,b)$
$begingroup$
Given this Cauchy problem
$$
(P):
begin{cases}
y'=f(y,t)=y|y|-t^2 \
y(0)=0
end{cases}
$$
I have to prove that , called $overline y$ the maximal solution of $(P)$, $dom(overline y)=(-infty,b)$ for some $b>0$. Moreover i need to demonstrate that $$lim_{tto-infty}=+infty text{ and }lim_{tto b^{-}}=-infty$$ I was able to demonstrate that the solution exist and is unique in a neighborhood of $t=0$ because $displaystyle frac{partial f}{partial y}=2|y|$ is continuous. I have found that $overline y$ must be decreasing in all its domain (because $y'geq0 iff ygeq|t|$) but still I can't proceed to demonstrate none of the requests. I believe that I should use some sort of comparison but the only thing I came up with was this disequation: $$y'=y|y|-t^2leq-y^2 quad text{for } t>0$$
and I know that a Cauchy problem with $x'=-x^2$ blow up in finite time but still to apply the comparison theorem I know I need to have $x(0)geq y(0)=0$. The problem is that if $x(0) geq 0$ the problem blow-up to $+infty$ and this give me nothing about $overline y$. I hope I have explained my problem well, thanks in advance for the help.
ordinary-differential-equations
asked Jan 20 at 19:33
edo1998edo1998
419113
$endgroup$
add a comment |
$begingroup$
Given this Cauchy problem
$$
(P):
begin{cases}
y'=f(y,t)=y|y|-t^2 \
y(0)=0
end{cases}
$$
I have to prove that , called $overline y$ the maximal solution of $(P)$, $dom(overline y)=(-infty,b)$ for some $b>0$. Moreover i need to demonstrate that $$lim_{tto-infty}=+infty text{ and }lim_{tto b^{-}}=-infty$$ I was able to demonstrate that the solution exist and is unique in a neighborhood of $t=0$ because $displaystyle frac{partial f}{partial y}=2|y|$ is continuous. I have found that $overline y$ must be decreasing in all its domain (because $y'geq0 iff ygeq|t|$) but still I can't proceed to demonstrate none of the requests. I believe that I should use some sort of comparison but the only thing I came up with was this disequation: $$y'=y|y|-t^2leq-y^2 quad text{for } t>0$$
and I know that a Cauchy problem with $x'=-x^2$ blow up in finite time but still to apply the comparison theorem I know I need to have $x(0)geq y(0)=0$. The problem is that if $x(0) geq 0$ the problem blow-up to $+infty$ and this give me nothing about $overline y$. I hope I have explained my problem well, thanks in advance for the help.
ordinary-differential-equations
asked Jan 20 at 19:33
edo1998edo1998
419113
$endgroup$
add a comment |
$begingroup$
Given this Cauchy problem
$$
(P):
begin{cases}
y'=f(y,t)=y|y|-t^2 \
y(0)=0
end{cases}
$$
I have to prove that , called $overline y$ the maximal solution of $(P)$, $dom(overline y)=(-infty,b)$ for some $b>0$. Moreover i need to demonstrate that $$lim_{tto-infty}=+infty text{ and }lim_{tto b^{-}}=-infty$$ I was able to demonstrate that the solution exist and is unique in a neighborhood of $t=0$ because $displaystyle frac{partial f}{partial y}=2|y|$ is continuous. I have found that $overline y$ must be decreasing in all its domain (because $y'geq0 iff ygeq|t|$) but still I can't proceed to demonstrate none of the requests. I believe that I should use some sort of comparison but the only thing I came up with was this disequation: $$y'=y|y|-t^2leq-y^2 quad text{for } t>0$$
and I know that a Cauchy problem with $x'=-x^2$ blow up in finite time but still to apply the comparison theorem I know I need to have $x(0)geq y(0)=0$. The problem is that if $x(0) geq 0$ the problem blow-up to $+infty$ and this give me nothing about $overline y$. I hope I have explained my problem well, thanks in advance for the help.
ordinary-differential-equations
asked Jan 20 at 19:33
edo1998edo1998
419113
$endgroup$
Given this Cauchy problem
$$
(P):
begin{cases}
y'=f(y,t)=y|y|-t^2 \
y(0)=0
end{cases}
$$
I have to prove that , called $overline y$ the maximal solution of $(P)$, $dom(overline y)=(-infty,b)$ for some $b>0$. Moreover i need to demonstrate that $$lim_{tto-infty}=+infty text{ and }lim_{tto b^{-}}=-infty$$ I was able to demonstrate that the solution exist and is unique in a neighborhood of $t=0$ because $displaystyle frac{partial f}{partial y}=2|y|$ is continuous. I have found that $overline y$ must be decreasing in all its domain (because $y'geq0 iff ygeq|t|$) but still I can't proceed to demonstrate none of the requests. I believe that I should use some sort of comparison but the only thing I came up with was this disequation: $$y'=y|y|-t^2leq-y^2 quad text{for } t>0$$
and I know that a Cauchy problem with $x'=-x^2$ blow up in finite time but still to apply the comparison theorem I know I need to have $x(0)geq y(0)=0$. The problem is that if $x(0) geq 0$ the problem blow-up to $+infty$ and this give me nothing about $overline y$. I hope I have explained my problem well, thanks in advance for the help.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 20 at 19:33
edo1998edo1998
419113
asked Jan 20 at 19:33
edo1998edo1998
419113
asked Jan 20 at 19:33
edo1998edo1998
419113
asked Jan 20 at 19:33
edo1998edo1998
419113
asked Jan 20 at 19:33
edo1998edo1998
419113
419113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You know that $y(t)<0$ for $t>0$. Then you can use for some small time
$$
y'le-t^2implies y(t)le-frac{t^3}3
$$
so that for instance $y(1)le-frac13$ which you can then use with your original inequality or strengthen it to
$$
y'le -y^2-1 ~~text{ for }~~ tge 1
$$
giving a shifted tangent function as upper bound.
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
$endgroup$
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
1
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
add a comment |
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1 Answer
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$begingroup$
You know that $y(t)<0$ for $t>0$. Then you can use for some small time
$$
y'le-t^2implies y(t)le-frac{t^3}3
$$
so that for instance $y(1)le-frac13$ which you can then use with your original inequality or strengthen it to
$$
y'le -y^2-1 ~~text{ for }~~ tge 1
$$
giving a shifted tangent function as upper bound.
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
$endgroup$
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
1
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
add a comment |
$begingroup$
You know that $y(t)<0$ for $t>0$. Then you can use for some small time
$$
y'le-t^2implies y(t)le-frac{t^3}3
$$
so that for instance $y(1)le-frac13$ which you can then use with your original inequality or strengthen it to
$$
y'le -y^2-1 ~~text{ for }~~ tge 1
$$
giving a shifted tangent function as upper bound.
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
$endgroup$
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
1
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
add a comment |
$begingroup$
You know that $y(t)<0$ for $t>0$. Then you can use for some small time
$$
y'le-t^2implies y(t)le-frac{t^3}3
$$
so that for instance $y(1)le-frac13$ which you can then use with your original inequality or strengthen it to
$$
y'le -y^2-1 ~~text{ for }~~ tge 1
$$
giving a shifted tangent function as upper bound.
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
$endgroup$
You know that $y(t)<0$ for $t>0$. Then you can use for some small time
$$
y'le-t^2implies y(t)le-frac{t^3}3
$$
so that for instance $y(1)le-frac13$ which you can then use with your original inequality or strengthen it to
$$
y'le -y^2-1 ~~text{ for }~~ tge 1
$$
giving a shifted tangent function as upper bound.
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
answered Jan 20 at 19:51
LutzLLutzL
58.8k42055
58.8k42055
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
1
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
add a comment |
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
1
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
$begingroup$
Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help
$endgroup$
– edo1998
Jan 20 at 20:01
1
1
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=pmfrac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question.
$endgroup$
– LutzL
Jan 20 at 20:05
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
$begingroup$
Yes, I understand. I will check the Riccati equations because I don't know them
$endgroup$
– edo1998
Jan 20 at 20:07
add a comment |
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