How to check the uniqueness of an ODE ,when the function in the R.H.S of the ODE is not Locally lipschitz
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Let $$dot X=f(X)$$ $$X(0)=X_0$$ be an ODE. If $f$ is locally Lipschitz, then the uniqueness of the solution of the ODE is guaranteed. How to check the uniqueness if $f$ is not Locally Lipschitz? For example if $Xin mathbb R$ and $f(X)=3X^{(frac{2}{3})}$ and $X_0=0$ then the two solutions of the ODE is $$X(t)=t^3$$ and$$X(0)=0$$ but if we replace the initial condition by $X_0=1$then it seems to me the only solution is $$X(t)=(t+1)^3$$ Now how to check wheather there are others solution or not?
ordinary-differential-equations
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add a comment |
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Let $$dot X=f(X)$$ $$X(0)=X_0$$ be an ODE. If $f$ is locally Lipschitz, then the uniqueness of the solution of the ODE is guaranteed. How to check the uniqueness if $f$ is not Locally Lipschitz? For example if $Xin mathbb R$ and $f(X)=3X^{(frac{2}{3})}$ and $X_0=0$ then the two solutions of the ODE is $$X(t)=t^3$$ and$$X(0)=0$$ but if we replace the initial condition by $X_0=1$then it seems to me the only solution is $$X(t)=(t+1)^3$$ Now how to check wheather there are others solution or not?
ordinary-differential-equations
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1
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When you consider $(-varepsilon, varepsilon)times (1-delta, 1+delta)$ then $f(X) = 3X^{2/3}$ is Lipschitz.
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– Jacky Chong
Dec 19 '18 at 8:49
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There are more solutions, in fact uncountably many. Indeed, for $a>1$ the function $X(t)=(t+a)^3$ for $tin(-infty,-a)$, $X(t)=0$ for $tin[-a,-1]$ and $X(t)=(t+1)^3$ for $tin(-1,infty)$ satisfies the IVP. Further, $X(t)=0$ for $tin(-infty,-1]$, $X(t)=(t+1)^3$ satisfies the IVP, too.
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– user539887
Dec 19 '18 at 8:54
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Now is there any guarantee that if f is not locally Lipschitz, then there exists more than one solution of the IVP?
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– Samiron Parui
Dec 22 '18 at 6:46
add a comment |
$begingroup$
Let $$dot X=f(X)$$ $$X(0)=X_0$$ be an ODE. If $f$ is locally Lipschitz, then the uniqueness of the solution of the ODE is guaranteed. How to check the uniqueness if $f$ is not Locally Lipschitz? For example if $Xin mathbb R$ and $f(X)=3X^{(frac{2}{3})}$ and $X_0=0$ then the two solutions of the ODE is $$X(t)=t^3$$ and$$X(0)=0$$ but if we replace the initial condition by $X_0=1$then it seems to me the only solution is $$X(t)=(t+1)^3$$ Now how to check wheather there are others solution or not?
ordinary-differential-equations
$endgroup$
Let $$dot X=f(X)$$ $$X(0)=X_0$$ be an ODE. If $f$ is locally Lipschitz, then the uniqueness of the solution of the ODE is guaranteed. How to check the uniqueness if $f$ is not Locally Lipschitz? For example if $Xin mathbb R$ and $f(X)=3X^{(frac{2}{3})}$ and $X_0=0$ then the two solutions of the ODE is $$X(t)=t^3$$ and$$X(0)=0$$ but if we replace the initial condition by $X_0=1$then it seems to me the only solution is $$X(t)=(t+1)^3$$ Now how to check wheather there are others solution or not?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 19 '18 at 8:40
Samiron ParuiSamiron Parui
1808
1808
1
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When you consider $(-varepsilon, varepsilon)times (1-delta, 1+delta)$ then $f(X) = 3X^{2/3}$ is Lipschitz.
$endgroup$
– Jacky Chong
Dec 19 '18 at 8:49
$begingroup$
There are more solutions, in fact uncountably many. Indeed, for $a>1$ the function $X(t)=(t+a)^3$ for $tin(-infty,-a)$, $X(t)=0$ for $tin[-a,-1]$ and $X(t)=(t+1)^3$ for $tin(-1,infty)$ satisfies the IVP. Further, $X(t)=0$ for $tin(-infty,-1]$, $X(t)=(t+1)^3$ satisfies the IVP, too.
$endgroup$
– user539887
Dec 19 '18 at 8:54
$begingroup$
Now is there any guarantee that if f is not locally Lipschitz, then there exists more than one solution of the IVP?
$endgroup$
– Samiron Parui
Dec 22 '18 at 6:46
add a comment |
1
$begingroup$
When you consider $(-varepsilon, varepsilon)times (1-delta, 1+delta)$ then $f(X) = 3X^{2/3}$ is Lipschitz.
$endgroup$
– Jacky Chong
Dec 19 '18 at 8:49
$begingroup$
There are more solutions, in fact uncountably many. Indeed, for $a>1$ the function $X(t)=(t+a)^3$ for $tin(-infty,-a)$, $X(t)=0$ for $tin[-a,-1]$ and $X(t)=(t+1)^3$ for $tin(-1,infty)$ satisfies the IVP. Further, $X(t)=0$ for $tin(-infty,-1]$, $X(t)=(t+1)^3$ satisfies the IVP, too.
$endgroup$
– user539887
Dec 19 '18 at 8:54
$begingroup$
Now is there any guarantee that if f is not locally Lipschitz, then there exists more than one solution of the IVP?
$endgroup$
– Samiron Parui
Dec 22 '18 at 6:46
1
1
$begingroup$
When you consider $(-varepsilon, varepsilon)times (1-delta, 1+delta)$ then $f(X) = 3X^{2/3}$ is Lipschitz.
$endgroup$
– Jacky Chong
Dec 19 '18 at 8:49
$begingroup$
When you consider $(-varepsilon, varepsilon)times (1-delta, 1+delta)$ then $f(X) = 3X^{2/3}$ is Lipschitz.
$endgroup$
– Jacky Chong
Dec 19 '18 at 8:49
$begingroup$
There are more solutions, in fact uncountably many. Indeed, for $a>1$ the function $X(t)=(t+a)^3$ for $tin(-infty,-a)$, $X(t)=0$ for $tin[-a,-1]$ and $X(t)=(t+1)^3$ for $tin(-1,infty)$ satisfies the IVP. Further, $X(t)=0$ for $tin(-infty,-1]$, $X(t)=(t+1)^3$ satisfies the IVP, too.
$endgroup$
– user539887
Dec 19 '18 at 8:54
$begingroup$
There are more solutions, in fact uncountably many. Indeed, for $a>1$ the function $X(t)=(t+a)^3$ for $tin(-infty,-a)$, $X(t)=0$ for $tin[-a,-1]$ and $X(t)=(t+1)^3$ for $tin(-1,infty)$ satisfies the IVP. Further, $X(t)=0$ for $tin(-infty,-1]$, $X(t)=(t+1)^3$ satisfies the IVP, too.
$endgroup$
– user539887
Dec 19 '18 at 8:54
$begingroup$
Now is there any guarantee that if f is not locally Lipschitz, then there exists more than one solution of the IVP?
$endgroup$
– Samiron Parui
Dec 22 '18 at 6:46
$begingroup$
Now is there any guarantee that if f is not locally Lipschitz, then there exists more than one solution of the IVP?
$endgroup$
– Samiron Parui
Dec 22 '18 at 6:46
add a comment |
0
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$begingroup$
When you consider $(-varepsilon, varepsilon)times (1-delta, 1+delta)$ then $f(X) = 3X^{2/3}$ is Lipschitz.
$endgroup$
– Jacky Chong
Dec 19 '18 at 8:49
$begingroup$
There are more solutions, in fact uncountably many. Indeed, for $a>1$ the function $X(t)=(t+a)^3$ for $tin(-infty,-a)$, $X(t)=0$ for $tin[-a,-1]$ and $X(t)=(t+1)^3$ for $tin(-1,infty)$ satisfies the IVP. Further, $X(t)=0$ for $tin(-infty,-1]$, $X(t)=(t+1)^3$ satisfies the IVP, too.
$endgroup$
– user539887
Dec 19 '18 at 8:54
$begingroup$
Now is there any guarantee that if f is not locally Lipschitz, then there exists more than one solution of the IVP?
$endgroup$
– Samiron Parui
Dec 22 '18 at 6:46