Unique solution of an ODE with a boundedly positive right-hand-side

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I want to solve an initial value problem $dot x(t) = F(x)$ with given $x(0)$, or more precisely the integral equation $$x(t)=x(0)+int_0^t F(x(s))ds.$$ $F$ may be discontinuous and $m < F(x) < M$ for all $x$.



Note that any solution to this integral equation must have $x(s)$ Lipschitz-continuous, so the integral is well-defined.



The common counter-examples to uniqueness (or existence) of ODEs (or their associated integral equations) seem to rely on $F$ switching sign, or being close to $0$, and my intuition is that the lower bound $m<F(x)$ should imply existence of a unique solution, but standard textbooks seem to always assume that $F$ is at least continuous.




differential-equations




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  • There are many issues here, what does a solution mean in the above context?
    – copper.hat
    Aug 3 at 2:29














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I want to solve an initial value problem $dot x(t) = F(x)$ with given $x(0)$, or more precisely the integral equation $$x(t)=x(0)+int_0^t F(x(s))ds.$$ $F$ may be discontinuous and $m < F(x) < M$ for all $x$.



Note that any solution to this integral equation must have $x(s)$ Lipschitz-continuous, so the integral is well-defined.



The common counter-examples to uniqueness (or existence) of ODEs (or their associated integral equations) seem to rely on $F$ switching sign, or being close to $0$, and my intuition is that the lower bound $m<F(x)$ should imply existence of a unique solution, but standard textbooks seem to always assume that $F$ is at least continuous.




differential-equations




share|cite|improve this question





















  • There are many issues here, what does a solution mean in the above context?
    – copper.hat
    Aug 3 at 2:29












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
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1





I want to solve an initial value problem $dot x(t) = F(x)$ with given $x(0)$, or more precisely the integral equation $$x(t)=x(0)+int_0^t F(x(s))ds.$$ $F$ may be discontinuous and $m < F(x) < M$ for all $x$.



Note that any solution to this integral equation must have $x(s)$ Lipschitz-continuous, so the integral is well-defined.



The common counter-examples to uniqueness (or existence) of ODEs (or their associated integral equations) seem to rely on $F$ switching sign, or being close to $0$, and my intuition is that the lower bound $m<F(x)$ should imply existence of a unique solution, but standard textbooks seem to always assume that $F$ is at least continuous.




differential-equations




share|cite|improve this question













I want to solve an initial value problem $dot x(t) = F(x)$ with given $x(0)$, or more precisely the integral equation $$x(t)=x(0)+int_0^t F(x(s))ds.$$ $F$ may be discontinuous and $m < F(x) < M$ for all $x$.



Note that any solution to this integral equation must have $x(s)$ Lipschitz-continuous, so the integral is well-defined.



The common counter-examples to uniqueness (or existence) of ODEs (or their associated integral equations) seem to rely on $F$ switching sign, or being close to $0$, and my intuition is that the lower bound $m<F(x)$ should imply existence of a unique solution, but standard textbooks seem to always assume that $F$ is at least continuous.





differential-equations





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









Armando j18eos

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asked Aug 2 at 23:01









Moritz Meyer-ter-Vehn

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  • There are many issues here, what does a solution mean in the above context?
    – copper.hat
    Aug 3 at 2:29
















  • There are many issues here, what does a solution mean in the above context?
    – copper.hat
    Aug 3 at 2:29















There are many issues here, what does a solution mean in the above context?
– copper.hat
Aug 3 at 2:29




There are many issues here, what does a solution mean in the above context?
– copper.hat
Aug 3 at 2:29















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