Uniqueness of $y'=cos(y)$ solutions












1












$begingroup$


Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32
















1












$begingroup$


Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32














1












1








1





$begingroup$


Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations







share|cite|improve this question









$endgroup$




Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 18:51









AtmosAtmos

4,832420




4,832420








  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32














  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32








2




2




$begingroup$
If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
$endgroup$
– Giuseppe Negro
Jan 24 at 19:38




$begingroup$
If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
$endgroup$
– Giuseppe Negro
Jan 24 at 19:38












$begingroup$
@GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
$endgroup$
– Atmos
Jan 25 at 17:32




$begingroup$
@GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
$endgroup$
– Atmos
Jan 25 at 17:32










1 Answer
1






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1












$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51











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1 Answer
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1 Answer
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active

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1












$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51
















1












$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51














1












1








1





$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$



In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 15:07

























answered Jan 24 at 19:03









Christian BlatterChristian Blatter

175k8115327




175k8115327












  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51


















  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51
















$begingroup$
I found that $u'(x)=-sin(y(x))$
$endgroup$
– Atmos
Jan 24 at 20:51




$begingroup$
I found that $u'(x)=-sin(y(x))$
$endgroup$
– Atmos
Jan 24 at 20:51


















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