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Does a strong solution to a SDE imply lipschitz condition?
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$begingroup$
Consider $dX_t=b(X_t,t)dt+sigma(X_t,t)dB_t$. I know that,
$|b(x,t)-b(y,t)|+|sigma(x,t)-sigma(y,t)|leq D|x-y|$ for some constant D implies the existence and uniqueness of a strong solution. However in Oksendal, to show that $dX_t=sign(X_t)dB_t$ doesn't have a strong solution another proof is presented, even though $sign(x)$ fails the lipschitz condition. I have seen it at a couple of more places for this equation.
My question: Is Lipschitz condition an iff condition for a strong solution?
stochastic-calculus stochastic-integrals stochastic-analysis
asked 4 hours ago
Who caresWho cares
16713
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add a comment |
$begingroup$
Consider $dX_t=b(X_t,t)dt+sigma(X_t,t)dB_t$. I know that,
$|b(x,t)-b(y,t)|+|sigma(x,t)-sigma(y,t)|leq D|x-y|$ for some constant D implies the existence and uniqueness of a strong solution. However in Oksendal, to show that $dX_t=sign(X_t)dB_t$ doesn't have a strong solution another proof is presented, even though $sign(x)$ fails the lipschitz condition. I have seen it at a couple of more places for this equation.
My question: Is Lipschitz condition an iff condition for a strong solution?
stochastic-calculus stochastic-integrals stochastic-analysis
asked 4 hours ago
Who caresWho cares
16713
$endgroup$
add a comment |
$begingroup$
Consider $dX_t=b(X_t,t)dt+sigma(X_t,t)dB_t$. I know that,
$|b(x,t)-b(y,t)|+|sigma(x,t)-sigma(y,t)|leq D|x-y|$ for some constant D implies the existence and uniqueness of a strong solution. However in Oksendal, to show that $dX_t=sign(X_t)dB_t$ doesn't have a strong solution another proof is presented, even though $sign(x)$ fails the lipschitz condition. I have seen it at a couple of more places for this equation.
My question: Is Lipschitz condition an iff condition for a strong solution?
stochastic-calculus stochastic-integrals stochastic-analysis
asked 4 hours ago
Who caresWho cares
16713
$endgroup$
Consider $dX_t=b(X_t,t)dt+sigma(X_t,t)dB_t$. I know that,
$|b(x,t)-b(y,t)|+|sigma(x,t)-sigma(y,t)|leq D|x-y|$ for some constant D implies the existence and uniqueness of a strong solution. However in Oksendal, to show that $dX_t=sign(X_t)dB_t$ doesn't have a strong solution another proof is presented, even though $sign(x)$ fails the lipschitz condition. I have seen it at a couple of more places for this equation.
My question: Is Lipschitz condition an iff condition for a strong solution?
stochastic-calculus stochastic-integrals stochastic-analysis
stochastic-calculus stochastic-integrals stochastic-analysis
asked 4 hours ago
Who caresWho cares
16713
asked 4 hours ago
Who caresWho cares
16713
edited 4 hours ago
Who cares
asked 4 hours ago
Who caresWho cares
16713
asked 4 hours ago
Who caresWho cares
16713
asked 4 hours ago
Who caresWho cares
16713
16713
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1 Answer
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No, Lipschitz continuity is not a necessary condition for the existence of a strong solution. There is, for instance, the following general result which goes back to Zvonkin:
Let $(B_t)_{t geq 0}$ be a one-dimensional Bownian motion. If the coefficients of the SDE $$dX_t = b(t,X_t) , dt + sigma(t,X_t) , dB_t, qquad X_0 = x_0, tag{1}$$ are bounded, $b$ is measurable, $sigma$ is continuous and there exist constants $C>0$ and $epsilon>0$ such that $$|sigma(t,x)-sigma(t,y)| leq C sqrt{|x-y|} quad text{and} quad |sigma(t,x)| geq epsilon$$ for all $t geq 0$, $x,y in mathbb{R}$, then the SDE $(1)$ has a (unique) strong solution.
Note that the result does not require any regularity assumptions on the drift $b$ (except from measurability) and only Hölder continuity of order $1/2$ for the diffusion coefficient $sigma$.
A nice overview on known existence and uniqueness results for SDEs can be found in the book Singular Stochastic Differential Equations by Cherny & Engelbert.
answered 2 hours ago
sazsaz
83.1k863132
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$begingroup$
No, Lipschitz continuity is not a necessary condition for the existence of a strong solution. There is, for instance, the following general result which goes back to Zvonkin:
Let $(B_t)_{t geq 0}$ be a one-dimensional Bownian motion. If the coefficients of the SDE $$dX_t = b(t,X_t) , dt + sigma(t,X_t) , dB_t, qquad X_0 = x_0, tag{1}$$ are bounded, $b$ is measurable, $sigma$ is continuous and there exist constants $C>0$ and $epsilon>0$ such that $$|sigma(t,x)-sigma(t,y)| leq C sqrt{|x-y|} quad text{and} quad |sigma(t,x)| geq epsilon$$ for all $t geq 0$, $x,y in mathbb{R}$, then the SDE $(1)$ has a (unique) strong solution.
Note that the result does not require any regularity assumptions on the drift $b$ (except from measurability) and only Hölder continuity of order $1/2$ for the diffusion coefficient $sigma$.
A nice overview on known existence and uniqueness results for SDEs can be found in the book Singular Stochastic Differential Equations by Cherny & Engelbert.
answered 2 hours ago
sazsaz
83.1k863132
$endgroup$
add a comment |
$begingroup$
No, Lipschitz continuity is not a necessary condition for the existence of a strong solution. There is, for instance, the following general result which goes back to Zvonkin:
Let $(B_t)_{t geq 0}$ be a one-dimensional Bownian motion. If the coefficients of the SDE $$dX_t = b(t,X_t) , dt + sigma(t,X_t) , dB_t, qquad X_0 = x_0, tag{1}$$ are bounded, $b$ is measurable, $sigma$ is continuous and there exist constants $C>0$ and $epsilon>0$ such that $$|sigma(t,x)-sigma(t,y)| leq C sqrt{|x-y|} quad text{and} quad |sigma(t,x)| geq epsilon$$ for all $t geq 0$, $x,y in mathbb{R}$, then the SDE $(1)$ has a (unique) strong solution.
Note that the result does not require any regularity assumptions on the drift $b$ (except from measurability) and only Hölder continuity of order $1/2$ for the diffusion coefficient $sigma$.
A nice overview on known existence and uniqueness results for SDEs can be found in the book Singular Stochastic Differential Equations by Cherny & Engelbert.
answered 2 hours ago
sazsaz
83.1k863132
$endgroup$
add a comment |
$begingroup$
No, Lipschitz continuity is not a necessary condition for the existence of a strong solution. There is, for instance, the following general result which goes back to Zvonkin:
Let $(B_t)_{t geq 0}$ be a one-dimensional Bownian motion. If the coefficients of the SDE $$dX_t = b(t,X_t) , dt + sigma(t,X_t) , dB_t, qquad X_0 = x_0, tag{1}$$ are bounded, $b$ is measurable, $sigma$ is continuous and there exist constants $C>0$ and $epsilon>0$ such that $$|sigma(t,x)-sigma(t,y)| leq C sqrt{|x-y|} quad text{and} quad |sigma(t,x)| geq epsilon$$ for all $t geq 0$, $x,y in mathbb{R}$, then the SDE $(1)$ has a (unique) strong solution.
Note that the result does not require any regularity assumptions on the drift $b$ (except from measurability) and only Hölder continuity of order $1/2$ for the diffusion coefficient $sigma$.
A nice overview on known existence and uniqueness results for SDEs can be found in the book Singular Stochastic Differential Equations by Cherny & Engelbert.
answered 2 hours ago
sazsaz
83.1k863132
$endgroup$
No, Lipschitz continuity is not a necessary condition for the existence of a strong solution. There is, for instance, the following general result which goes back to Zvonkin:
Let $(B_t)_{t geq 0}$ be a one-dimensional Bownian motion. If the coefficients of the SDE $$dX_t = b(t,X_t) , dt + sigma(t,X_t) , dB_t, qquad X_0 = x_0, tag{1}$$ are bounded, $b$ is measurable, $sigma$ is continuous and there exist constants $C>0$ and $epsilon>0$ such that $$|sigma(t,x)-sigma(t,y)| leq C sqrt{|x-y|} quad text{and} quad |sigma(t,x)| geq epsilon$$ for all $t geq 0$, $x,y in mathbb{R}$, then the SDE $(1)$ has a (unique) strong solution.
Note that the result does not require any regularity assumptions on the drift $b$ (except from measurability) and only Hölder continuity of order $1/2$ for the diffusion coefficient $sigma$.
A nice overview on known existence and uniqueness results for SDEs can be found in the book Singular Stochastic Differential Equations by Cherny & Engelbert.
answered 2 hours ago
sazsaz
83.1k863132
answered 2 hours ago
sazsaz
83.1k863132
answered 2 hours ago
sazsaz
83.1k863132
answered 2 hours ago
sazsaz
83.1k863132
83.1k863132
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