Find a set of solutions for $y'=sqrt{y^2-1}, y(0)=1$.












0












$begingroup$


For the IVP $y'=sqrt{y^2-1}, y(0)=1$ I am supposed to find a set of solutions depending on $2$ parameters.



While I can easily find 2 different solutions $y_1(x)=1$ and $y_2(x)=frac{1}{2}(e^{-x}+e^x)$ that solve the IVP I don't know how to bring parameters into this problem. How can any solution depend on a parameter when the starting value is given? And if no solution depends on any parameter, what's the trick to create a set of solutions depending on 2 parameters?






ordinary-differential-equations initial-value-problems







share|cite|improve this question









$endgroup$












  • $begingroup$
    The RHS of the equation is not Lipschitz continuous at $y=0$, therefore, the solution is not assured to be unique. Maybe you can find a family of solutions satisfying the IVP based on this.
    $endgroup$
    – rafa11111
    Dec 10 '18 at 15:20
















0












$begingroup$


For the IVP $y'=sqrt{y^2-1}, y(0)=1$ I am supposed to find a set of solutions depending on $2$ parameters.



While I can easily find 2 different solutions $y_1(x)=1$ and $y_2(x)=frac{1}{2}(e^{-x}+e^x)$ that solve the IVP I don't know how to bring parameters into this problem. How can any solution depend on a parameter when the starting value is given? And if no solution depends on any parameter, what's the trick to create a set of solutions depending on 2 parameters?






ordinary-differential-equations initial-value-problems







share|cite|improve this question









$endgroup$












  • $begingroup$
    The RHS of the equation is not Lipschitz continuous at $y=0$, therefore, the solution is not assured to be unique. Maybe you can find a family of solutions satisfying the IVP based on this.
    $endgroup$
    – rafa11111
    Dec 10 '18 at 15:20














0












0








0





$begingroup$


For the IVP $y'=sqrt{y^2-1}, y(0)=1$ I am supposed to find a set of solutions depending on $2$ parameters.



While I can easily find 2 different solutions $y_1(x)=1$ and $y_2(x)=frac{1}{2}(e^{-x}+e^x)$ that solve the IVP I don't know how to bring parameters into this problem. How can any solution depend on a parameter when the starting value is given? And if no solution depends on any parameter, what's the trick to create a set of solutions depending on 2 parameters?






ordinary-differential-equations initial-value-problems







share|cite|improve this question









$endgroup$




For the IVP $y'=sqrt{y^2-1}, y(0)=1$ I am supposed to find a set of solutions depending on $2$ parameters.



While I can easily find 2 different solutions $y_1(x)=1$ and $y_2(x)=frac{1}{2}(e^{-x}+e^x)$ that solve the IVP I don't know how to bring parameters into this problem. How can any solution depend on a parameter when the starting value is given? And if no solution depends on any parameter, what's the trick to create a set of solutions depending on 2 parameters?






ordinary-differential-equations initial-value-problems




ordinary-differential-equations initial-value-problems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 15:12









RedLanternRedLantern

415




415












  • $begingroup$
    The RHS of the equation is not Lipschitz continuous at $y=0$, therefore, the solution is not assured to be unique. Maybe you can find a family of solutions satisfying the IVP based on this.
    $endgroup$
    – rafa11111
    Dec 10 '18 at 15:20


















  • $begingroup$
    The RHS of the equation is not Lipschitz continuous at $y=0$, therefore, the solution is not assured to be unique. Maybe you can find a family of solutions satisfying the IVP based on this.
    $endgroup$
    – rafa11111
    Dec 10 '18 at 15:20
















$begingroup$
The RHS of the equation is not Lipschitz continuous at $y=0$, therefore, the solution is not assured to be unique. Maybe you can find a family of solutions satisfying the IVP based on this.
$endgroup$
– rafa11111
Dec 10 '18 at 15:20




$begingroup$
The RHS of the equation is not Lipschitz continuous at $y=0$, therefore, the solution is not assured to be unique. Maybe you can find a family of solutions satisfying the IVP based on this.
$endgroup$
– rafa11111
Dec 10 '18 at 15:20










2 Answers
2






active

oldest

votes


















0












$begingroup$

If $y=1$, $y'=0$ yields $y=1$, is consistent with the initial condition.



Then if $yne1$,$$frac{y'}{sqrt{y^2-1}}=1$$ and



$$text{arcosh }y=x+c,$$ $$y=cosh(x+c).$$



As said by Robert, we can switch (twice) between the two solutions:



$$y=begin{cases}x<c_-tocosh(x-c_-),\c_-<x<c_+to1,\x>c_+tocosh(x-c_+)end{cases}.$$



where $c_-le 0le c_+$. These are your two constants.



Solutions with a single or with no switch are also possible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:33










  • $begingroup$
    @RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:25





















2












$begingroup$

Hint: your solution can be $1$ for a while and then switch to a translate of $y_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:26












  • $begingroup$
    Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 15:27












  • $begingroup$
    But then they can only switch at the point $x=0$. How do I bring parameters in?
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:28











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

If $y=1$, $y'=0$ yields $y=1$, is consistent with the initial condition.



Then if $yne1$,$$frac{y'}{sqrt{y^2-1}}=1$$ and



$$text{arcosh }y=x+c,$$ $$y=cosh(x+c).$$



As said by Robert, we can switch (twice) between the two solutions:



$$y=begin{cases}x<c_-tocosh(x-c_-),\c_-<x<c_+to1,\x>c_+tocosh(x-c_+)end{cases}.$$



where $c_-le 0le c_+$. These are your two constants.



Solutions with a single or with no switch are also possible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:33










  • $begingroup$
    @RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:25


















0












$begingroup$

If $y=1$, $y'=0$ yields $y=1$, is consistent with the initial condition.



Then if $yne1$,$$frac{y'}{sqrt{y^2-1}}=1$$ and



$$text{arcosh }y=x+c,$$ $$y=cosh(x+c).$$



As said by Robert, we can switch (twice) between the two solutions:



$$y=begin{cases}x<c_-tocosh(x-c_-),\c_-<x<c_+to1,\x>c_+tocosh(x-c_+)end{cases}.$$



where $c_-le 0le c_+$. These are your two constants.



Solutions with a single or with no switch are also possible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:33










  • $begingroup$
    @RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:25
















0












0








0





$begingroup$

If $y=1$, $y'=0$ yields $y=1$, is consistent with the initial condition.



Then if $yne1$,$$frac{y'}{sqrt{y^2-1}}=1$$ and



$$text{arcosh }y=x+c,$$ $$y=cosh(x+c).$$



As said by Robert, we can switch (twice) between the two solutions:



$$y=begin{cases}x<c_-tocosh(x-c_-),\c_-<x<c_+to1,\x>c_+tocosh(x-c_+)end{cases}.$$



where $c_-le 0le c_+$. These are your two constants.



Solutions with a single or with no switch are also possible.






share|cite|improve this answer









$endgroup$



If $y=1$, $y'=0$ yields $y=1$, is consistent with the initial condition.



Then if $yne1$,$$frac{y'}{sqrt{y^2-1}}=1$$ and



$$text{arcosh }y=x+c,$$ $$y=cosh(x+c).$$



As said by Robert, we can switch (twice) between the two solutions:



$$y=begin{cases}x<c_-tocosh(x-c_-),\c_-<x<c_+to1,\x>c_+tocosh(x-c_+)end{cases}.$$



where $c_-le 0le c_+$. These are your two constants.



Solutions with a single or with no switch are also possible.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 15:30









Yves DaoustYves Daoust

128k674226




128k674226












  • $begingroup$
    Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:33










  • $begingroup$
    @RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:25




















  • $begingroup$
    Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:33










  • $begingroup$
    @RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
    $endgroup$
    – Yves Daoust
    Dec 10 '18 at 17:25


















$begingroup$
Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
$endgroup$
– RedLantern
Dec 10 '18 at 15:33




$begingroup$
Oh, I think I understand what is going on here! So you just ignore the IVP for the solution generated by separation of variables which is okay because we glue it together with the constant solution in the necessary area.
$endgroup$
– RedLantern
Dec 10 '18 at 15:33












$begingroup$
@RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
$endgroup$
– Yves Daoust
Dec 10 '18 at 17:25






$begingroup$
@RedLantern: as you can check, all these solutions verity the ODE and the initial condition.
$endgroup$
– Yves Daoust
Dec 10 '18 at 17:25













2












$begingroup$

Hint: your solution can be $1$ for a while and then switch to a translate of $y_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:26












  • $begingroup$
    Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 15:27












  • $begingroup$
    But then they can only switch at the point $x=0$. How do I bring parameters in?
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:28
















2












$begingroup$

Hint: your solution can be $1$ for a while and then switch to a translate of $y_2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:26












  • $begingroup$
    Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 15:27












  • $begingroup$
    But then they can only switch at the point $x=0$. How do I bring parameters in?
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:28














2












2








2





$begingroup$

Hint: your solution can be $1$ for a while and then switch to a translate of $y_2$.






share|cite|improve this answer









$endgroup$



Hint: your solution can be $1$ for a while and then switch to a translate of $y_2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 15:19









Robert IsraelRobert Israel

323k23213467




323k23213467












  • $begingroup$
    I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:26












  • $begingroup$
    Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 15:27












  • $begingroup$
    But then they can only switch at the point $x=0$. How do I bring parameters in?
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:28


















  • $begingroup$
    I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:26












  • $begingroup$
    Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 15:27












  • $begingroup$
    But then they can only switch at the point $x=0$. How do I bring parameters in?
    $endgroup$
    – RedLantern
    Dec 10 '18 at 15:28
















$begingroup$
I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
$endgroup$
– RedLantern
Dec 10 '18 at 15:26






$begingroup$
I thought of this too. But doesn't have the solution to be continuous? Otherwise I could basically just write down $$y_{a,b}(x)=begin{cases}y_1(x), & xin[a,b]\y_2(x),& text{ else }end{cases}$$.
$endgroup$
– RedLantern
Dec 10 '18 at 15:26














$begingroup$
Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
$endgroup$
– Robert Israel
Dec 10 '18 at 15:27






$begingroup$
Yes, you should make sure the solution is continuous. That tells you which translate of $y_2$ to use.
$endgroup$
– Robert Israel
Dec 10 '18 at 15:27














$begingroup$
But then they can only switch at the point $x=0$. How do I bring parameters in?
$endgroup$
– RedLantern
Dec 10 '18 at 15:28




$begingroup$
But then they can only switch at the point $x=0$. How do I bring parameters in?
$endgroup$
– RedLantern
Dec 10 '18 at 15:28


















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