Picard iteration for second order ODE












2












$begingroup$


Any help for the following two questions will be much appreciated:



Question 1:



Consder the ODE with IVP



$y''(t)+g(t,y)=0, y(0)=y_0, y'(0)=z_0$



where g is continues on some region $D$, $(0,y_0)in D$. I want to show that this is equivalent to the integral equation



$ begin{align}
y(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
end{align}
\
$



This is what I have done so far to solve this: I integra either side and get



$ int_0^t y''(s) ds =-int_0^t g(s,f(s)) ds$ I calculate this, integrate again and get



$y(s)=y_0+z_0t-int_0^t int_0^t g(s,f(s))ds$. My problem is that I don't know what to do with this last double integral.



Question 2:



Consider the same integral equation where $g(y,t)$, $frac{partial g}{ partial y} (t,y)$ are continuous on the rectangle $D={ (t,y)in mathbb{R}^2 : mid t mid leq a, mid y-y_0 mid leq b}$. Let $mid g(t,y) mid leq M$, $mid frac{partial g}{ partial y} (t,y) mid leq K$ for all $(t,y) in D$. We define $phi_0(t)=y_0$ and



$
phi_n(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
$



I want to show that $phi_n(t)$ is well defiend for $mid tmid leq min {a,frac{2b}{2mid z_0mid+2M}}$. Here I am quite lost. To show it is well-defined I need to show that the integral is never infinity. But I don't know how to prove this.






real-analysis ordinary-differential-equations







share|cite|improve this question











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  • $begingroup$
    How come you have not got an answer yet!
    $endgroup$
    – Mhenni Benghorbal
    Feb 13 '16 at 23:32












  • $begingroup$
    I don't know, hopefully some kind soul will come by soon.
    $endgroup$
    – Soren123
    Feb 14 '16 at 8:00
















2












$begingroup$


Any help for the following two questions will be much appreciated:



Question 1:



Consder the ODE with IVP



$y''(t)+g(t,y)=0, y(0)=y_0, y'(0)=z_0$



where g is continues on some region $D$, $(0,y_0)in D$. I want to show that this is equivalent to the integral equation



$ begin{align}
y(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
end{align}
\
$



This is what I have done so far to solve this: I integra either side and get



$ int_0^t y''(s) ds =-int_0^t g(s,f(s)) ds$ I calculate this, integrate again and get



$y(s)=y_0+z_0t-int_0^t int_0^t g(s,f(s))ds$. My problem is that I don't know what to do with this last double integral.



Question 2:



Consider the same integral equation where $g(y,t)$, $frac{partial g}{ partial y} (t,y)$ are continuous on the rectangle $D={ (t,y)in mathbb{R}^2 : mid t mid leq a, mid y-y_0 mid leq b}$. Let $mid g(t,y) mid leq M$, $mid frac{partial g}{ partial y} (t,y) mid leq K$ for all $(t,y) in D$. We define $phi_0(t)=y_0$ and



$
phi_n(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
$



I want to show that $phi_n(t)$ is well defiend for $mid tmid leq min {a,frac{2b}{2mid z_0mid+2M}}$. Here I am quite lost. To show it is well-defined I need to show that the integral is never infinity. But I don't know how to prove this.






real-analysis ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    How come you have not got an answer yet!
    $endgroup$
    – Mhenni Benghorbal
    Feb 13 '16 at 23:32












  • $begingroup$
    I don't know, hopefully some kind soul will come by soon.
    $endgroup$
    – Soren123
    Feb 14 '16 at 8:00














2












2








2


1



$begingroup$


Any help for the following two questions will be much appreciated:



Question 1:



Consder the ODE with IVP



$y''(t)+g(t,y)=0, y(0)=y_0, y'(0)=z_0$



where g is continues on some region $D$, $(0,y_0)in D$. I want to show that this is equivalent to the integral equation



$ begin{align}
y(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
end{align}
\
$



This is what I have done so far to solve this: I integra either side and get



$ int_0^t y''(s) ds =-int_0^t g(s,f(s)) ds$ I calculate this, integrate again and get



$y(s)=y_0+z_0t-int_0^t int_0^t g(s,f(s))ds$. My problem is that I don't know what to do with this last double integral.



Question 2:



Consider the same integral equation where $g(y,t)$, $frac{partial g}{ partial y} (t,y)$ are continuous on the rectangle $D={ (t,y)in mathbb{R}^2 : mid t mid leq a, mid y-y_0 mid leq b}$. Let $mid g(t,y) mid leq M$, $mid frac{partial g}{ partial y} (t,y) mid leq K$ for all $(t,y) in D$. We define $phi_0(t)=y_0$ and



$
phi_n(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
$



I want to show that $phi_n(t)$ is well defiend for $mid tmid leq min {a,frac{2b}{2mid z_0mid+2M}}$. Here I am quite lost. To show it is well-defined I need to show that the integral is never infinity. But I don't know how to prove this.






real-analysis ordinary-differential-equations







share|cite|improve this question











$endgroup$




Any help for the following two questions will be much appreciated:



Question 1:



Consder the ODE with IVP



$y''(t)+g(t,y)=0, y(0)=y_0, y'(0)=z_0$



where g is continues on some region $D$, $(0,y_0)in D$. I want to show that this is equivalent to the integral equation



$ begin{align}
y(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
end{align}
\
$



This is what I have done so far to solve this: I integra either side and get



$ int_0^t y''(s) ds =-int_0^t g(s,f(s)) ds$ I calculate this, integrate again and get



$y(s)=y_0+z_0t-int_0^t int_0^t g(s,f(s))ds$. My problem is that I don't know what to do with this last double integral.



Question 2:



Consider the same integral equation where $g(y,t)$, $frac{partial g}{ partial y} (t,y)$ are continuous on the rectangle $D={ (t,y)in mathbb{R}^2 : mid t mid leq a, mid y-y_0 mid leq b}$. Let $mid g(t,y) mid leq M$, $mid frac{partial g}{ partial y} (t,y) mid leq K$ for all $(t,y) in D$. We define $phi_0(t)=y_0$ and



$
phi_n(t)=y_0+z_0t-int_0^t (t-s)g(s,y(s))ds
$



I want to show that $phi_n(t)$ is well defiend for $mid tmid leq min {a,frac{2b}{2mid z_0mid+2M}}$. Here I am quite lost. To show it is well-defined I need to show that the integral is never infinity. But I don't know how to prove this.






real-analysis ordinary-differential-equations




real-analysis ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 14 '16 at 7:59







Soren123

















asked Feb 13 '16 at 20:27









Soren123Soren123

9311




9311












  • $begingroup$
    How come you have not got an answer yet!
    $endgroup$
    – Mhenni Benghorbal
    Feb 13 '16 at 23:32












  • $begingroup$
    I don't know, hopefully some kind soul will come by soon.
    $endgroup$
    – Soren123
    Feb 14 '16 at 8:00


















  • $begingroup$
    How come you have not got an answer yet!
    $endgroup$
    – Mhenni Benghorbal
    Feb 13 '16 at 23:32












  • $begingroup$
    I don't know, hopefully some kind soul will come by soon.
    $endgroup$
    – Soren123
    Feb 14 '16 at 8:00
















$begingroup$
How come you have not got an answer yet!
$endgroup$
– Mhenni Benghorbal
Feb 13 '16 at 23:32






$begingroup$
How come you have not got an answer yet!
$endgroup$
– Mhenni Benghorbal
Feb 13 '16 at 23:32














$begingroup$
I don't know, hopefully some kind soul will come by soon.
$endgroup$
– Soren123
Feb 14 '16 at 8:00




$begingroup$
I don't know, hopefully some kind soul will come by soon.
$endgroup$
– Soren123
Feb 14 '16 at 8:00










1 Answer
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Try showing $r_j(t)=|phi_{j+1}(t)-phi_j(t)| leq frac{M(|z_0|+frac{Ma}{2})K^{j}t^{2j+1}}{(2j+1)!}$.



If K<1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})t^{2j+1}}{(2j+1)!}$
If K>1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})K^{2j+1}t^{2j+1}}{(2j+1)!}$ which are the terms of hyperbolic sin function, which converges.






share|cite|improve this answer









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    $begingroup$

    Try showing $r_j(t)=|phi_{j+1}(t)-phi_j(t)| leq frac{M(|z_0|+frac{Ma}{2})K^{j}t^{2j+1}}{(2j+1)!}$.



    If K<1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})t^{2j+1}}{(2j+1)!}$
    If K>1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})K^{2j+1}t^{2j+1}}{(2j+1)!}$ which are the terms of hyperbolic sin function, which converges.






    share|cite|improve this answer









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      0












      $begingroup$

      Try showing $r_j(t)=|phi_{j+1}(t)-phi_j(t)| leq frac{M(|z_0|+frac{Ma}{2})K^{j}t^{2j+1}}{(2j+1)!}$.



      If K<1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})t^{2j+1}}{(2j+1)!}$
      If K>1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})K^{2j+1}t^{2j+1}}{(2j+1)!}$ which are the terms of hyperbolic sin function, which converges.






      share|cite|improve this answer









      $endgroup$
















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        0





        $begingroup$

        Try showing $r_j(t)=|phi_{j+1}(t)-phi_j(t)| leq frac{M(|z_0|+frac{Ma}{2})K^{j}t^{2j+1}}{(2j+1)!}$.



        If K<1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})t^{2j+1}}{(2j+1)!}$
        If K>1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})K^{2j+1}t^{2j+1}}{(2j+1)!}$ which are the terms of hyperbolic sin function, which converges.






        share|cite|improve this answer









        $endgroup$



        Try showing $r_j(t)=|phi_{j+1}(t)-phi_j(t)| leq frac{M(|z_0|+frac{Ma}{2})K^{j}t^{2j+1}}{(2j+1)!}$.



        If K<1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})t^{2j+1}}{(2j+1)!}$
        If K>1, then $r_j(t) leq frac{M(|z_0|+frac{Ma}{2})K^{2j+1}t^{2j+1}}{(2j+1)!}$ which are the terms of hyperbolic sin function, which converges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 0:42









        JungleshrimpJungleshrimp

        314111




        314111






























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