Existence and Uniqueness Theorems for First-Order ODE's The 2019 Stack Overflow Developer Survey Results Are InA version of existence and uniqueness of solution for an initival value problem of ODEElementary Proof Involving Wronskian and Existence and Uniquenessguarantee existence but not uniquenessExistence And Uniqueness Theorem QuestionODE) Uniqueness-Existence ProblemWhy do the Existence and Uniqueness Theorem and The Principle of Superposition not contradict each other?Using the uniqueness theorem for differential equations without initial value?“Existence and Uniqueness” Theorems for first order IVP: two or just one?Existence and uniqueness of solutionQuestion regarding Existence and Uniqueness of a Differential Equation
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Existence and Uniqueness Theorems for First-Order ODE's
The 2019 Stack Overflow Developer Survey Results Are InA version of existence and uniqueness of solution for an initival value problem of ODEElementary Proof Involving Wronskian and Existence and Uniquenessguarantee existence but not uniquenessExistence And Uniqueness Theorem QuestionODE) Uniqueness-Existence ProblemWhy do the Existence and Uniqueness Theorem and The Principle of Superposition not contradict each other?Using the uniqueness theorem for differential equations without initial value?“Existence and Uniqueness” Theorems for first order IVP: two or just one?Existence and uniqueness of solutionQuestion regarding Existence and Uniqueness of a Differential Equation
$begingroup$
There is a theorem regarding the existence and uniqueness of solutions to first-order ODE's:
Thm.: Consider the initial value problem $y'=f(t,y)$, $y(t_0)=y_0$. Suppose $f$ and $frac∂f∂y$ are continuous on some open rectangle $(t, y)∈(a, b)×(c, d)$ containing the point $(t_0, y_0)$. Then in some interval $t_0-h, t_0+h)⊆(a, b)$, there exists a unique solution $y = g(t)$ that satisfies the initial value problem.
Now if we consider this example: $t^2 y'+2ty-y^3=0$, where $t>0$, the general solution is $y = ±sqrtfrac5t2+Ct^5$. But $y=0$ is also a solution and I am wondering whether the above theorem can help me find such solutions which are not contained in the family of general solutions.
So, dividing the whole equation by $t^2$ gives $y'=fracy^3t^2-frac2yt$. Let $f(t,y)=fracy^3t^2-frac2yt$, then $frac∂f∂y = frac3y^2t^2-frac2t$. As long as $t≠0$, then $f$ and $frac∂f∂y$ are continuous, so I don't understand why we cannot identify anything wrong with $y=0$ using this theorem. To my understanding, since the solution $y=0$ is not contained in the general solutions, it should violate the conditions stated in the theorem, so I am quite confused. Thank you very much for answering.
ordinary-differential-equations
edited Mar 30 at 15:23
LutzL
60.4k42057
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
$endgroup$
add a comment |
$begingroup$
There is a theorem regarding the existence and uniqueness of solutions to first-order ODE's:
Thm.: Consider the initial value problem $y'=f(t,y)$, $y(t_0)=y_0$. Suppose $f$ and $frac∂f∂y$ are continuous on some open rectangle $(t, y)∈(a, b)×(c, d)$ containing the point $(t_0, y_0)$. Then in some interval $t_0-h, t_0+h)⊆(a, b)$, there exists a unique solution $y = g(t)$ that satisfies the initial value problem.
Now if we consider this example: $t^2 y'+2ty-y^3=0$, where $t>0$, the general solution is $y = ±sqrtfrac5t2+Ct^5$. But $y=0$ is also a solution and I am wondering whether the above theorem can help me find such solutions which are not contained in the family of general solutions.
So, dividing the whole equation by $t^2$ gives $y'=fracy^3t^2-frac2yt$. Let $f(t,y)=fracy^3t^2-frac2yt$, then $frac∂f∂y = frac3y^2t^2-frac2t$. As long as $t≠0$, then $f$ and $frac∂f∂y$ are continuous, so I don't understand why we cannot identify anything wrong with $y=0$ using this theorem. To my understanding, since the solution $y=0$ is not contained in the general solutions, it should violate the conditions stated in the theorem, so I am quite confused. Thank you very much for answering.
ordinary-differential-equations
edited Mar 30 at 15:23
LutzL
60.4k42057
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
$endgroup$
$begingroup$
@Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity?
$endgroup$
– John Lei
Sep 13 '17 at 3:14
add a comment |
$begingroup$
There is a theorem regarding the existence and uniqueness of solutions to first-order ODE's:
Thm.: Consider the initial value problem $y'=f(t,y)$, $y(t_0)=y_0$. Suppose $f$ and $frac∂f∂y$ are continuous on some open rectangle $(t, y)∈(a, b)×(c, d)$ containing the point $(t_0, y_0)$. Then in some interval $t_0-h, t_0+h)⊆(a, b)$, there exists a unique solution $y = g(t)$ that satisfies the initial value problem.
Now if we consider this example: $t^2 y'+2ty-y^3=0$, where $t>0$, the general solution is $y = ±sqrtfrac5t2+Ct^5$. But $y=0$ is also a solution and I am wondering whether the above theorem can help me find such solutions which are not contained in the family of general solutions.
So, dividing the whole equation by $t^2$ gives $y'=fracy^3t^2-frac2yt$. Let $f(t,y)=fracy^3t^2-frac2yt$, then $frac∂f∂y = frac3y^2t^2-frac2t$. As long as $t≠0$, then $f$ and $frac∂f∂y$ are continuous, so I don't understand why we cannot identify anything wrong with $y=0$ using this theorem. To my understanding, since the solution $y=0$ is not contained in the general solutions, it should violate the conditions stated in the theorem, so I am quite confused. Thank you very much for answering.
ordinary-differential-equations
edited Mar 30 at 15:23
LutzL
60.4k42057
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
$endgroup$
There is a theorem regarding the existence and uniqueness of solutions to first-order ODE's:
Thm.: Consider the initial value problem $y'=f(t,y)$, $y(t_0)=y_0$. Suppose $f$ and $frac∂f∂y$ are continuous on some open rectangle $(t, y)∈(a, b)×(c, d)$ containing the point $(t_0, y_0)$. Then in some interval $t_0-h, t_0+h)⊆(a, b)$, there exists a unique solution $y = g(t)$ that satisfies the initial value problem.
Now if we consider this example: $t^2 y'+2ty-y^3=0$, where $t>0$, the general solution is $y = ±sqrtfrac5t2+Ct^5$. But $y=0$ is also a solution and I am wondering whether the above theorem can help me find such solutions which are not contained in the family of general solutions.
So, dividing the whole equation by $t^2$ gives $y'=fracy^3t^2-frac2yt$. Let $f(t,y)=fracy^3t^2-frac2yt$, then $frac∂f∂y = frac3y^2t^2-frac2t$. As long as $t≠0$, then $f$ and $frac∂f∂y$ are continuous, so I don't understand why we cannot identify anything wrong with $y=0$ using this theorem. To my understanding, since the solution $y=0$ is not contained in the general solutions, it should violate the conditions stated in the theorem, so I am quite confused. Thank you very much for answering.
ordinary-differential-equations
ordinary-differential-equations
edited Mar 30 at 15:23
LutzL
60.4k42057
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
edited Mar 30 at 15:23
LutzL
60.4k42057
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
edited Mar 30 at 15:23
LutzL
60.4k42057
edited Mar 30 at 15:23
LutzL
60.4k42057
edited Mar 30 at 15:23
LutzL
60.4k42057
60.4k42057
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
asked Sep 12 '17 at 19:43
John LeiJohn Lei
445
445
$begingroup$
@Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity?
$endgroup$
– John Lei
Sep 13 '17 at 3:14
add a comment |
$begingroup$
@Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity?
$endgroup$
– John Lei
Sep 13 '17 at 3:14
$begingroup$
@Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity?
$endgroup$
– John Lei
Sep 13 '17 at 3:14
$begingroup$
@Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity?
$endgroup$
– John Lei
Sep 13 '17 at 3:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It often happens that there are particular solutions that are not covered by the general solution, but may be a limit of the general solution as some parameter goes to $+infty$ or $-infty$. In this case taking $C to infty$ gives you $y=0$.
There is no violation of the Existence and Uniqueness Theorem here. If $t_0 > 0$ and $y_0 > 0$, there is a unique $C$ such that $y_0 = sqrtfrac5 t_02+C t_0^5$, and the unique solution with this initial condition is $y = sqrtfrac5 t2+C t^5$ with that $C$. Similarly for $y_0 < 0$ with
$y_0 = -sqrtfrac5 t_02+C t_0^5$. And for $y_0 = 0$, the unique solution is $y = 0$.
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
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$begingroup$
It often happens that there are particular solutions that are not covered by the general solution, but may be a limit of the general solution as some parameter goes to $+infty$ or $-infty$. In this case taking $C to infty$ gives you $y=0$.
There is no violation of the Existence and Uniqueness Theorem here. If $t_0 > 0$ and $y_0 > 0$, there is a unique $C$ such that $y_0 = sqrtfrac5 t_02+C t_0^5$, and the unique solution with this initial condition is $y = sqrtfrac5 t2+C t^5$ with that $C$. Similarly for $y_0 < 0$ with
$y_0 = -sqrtfrac5 t_02+C t_0^5$. And for $y_0 = 0$, the unique solution is $y = 0$.
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
It often happens that there are particular solutions that are not covered by the general solution, but may be a limit of the general solution as some parameter goes to $+infty$ or $-infty$. In this case taking $C to infty$ gives you $y=0$.
There is no violation of the Existence and Uniqueness Theorem here. If $t_0 > 0$ and $y_0 > 0$, there is a unique $C$ such that $y_0 = sqrtfrac5 t_02+C t_0^5$, and the unique solution with this initial condition is $y = sqrtfrac5 t2+C t^5$ with that $C$. Similarly for $y_0 < 0$ with
$y_0 = -sqrtfrac5 t_02+C t_0^5$. And for $y_0 = 0$, the unique solution is $y = 0$.
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
It often happens that there are particular solutions that are not covered by the general solution, but may be a limit of the general solution as some parameter goes to $+infty$ or $-infty$. In this case taking $C to infty$ gives you $y=0$.
There is no violation of the Existence and Uniqueness Theorem here. If $t_0 > 0$ and $y_0 > 0$, there is a unique $C$ such that $y_0 = sqrtfrac5 t_02+C t_0^5$, and the unique solution with this initial condition is $y = sqrtfrac5 t2+C t^5$ with that $C$. Similarly for $y_0 < 0$ with
$y_0 = -sqrtfrac5 t_02+C t_0^5$. And for $y_0 = 0$, the unique solution is $y = 0$.
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
$endgroup$
It often happens that there are particular solutions that are not covered by the general solution, but may be a limit of the general solution as some parameter goes to $+infty$ or $-infty$. In this case taking $C to infty$ gives you $y=0$.
There is no violation of the Existence and Uniqueness Theorem here. If $t_0 > 0$ and $y_0 > 0$, there is a unique $C$ such that $y_0 = sqrtfrac5 t_02+C t_0^5$, and the unique solution with this initial condition is $y = sqrtfrac5 t2+C t^5$ with that $C$. Similarly for $y_0 < 0$ with
$y_0 = -sqrtfrac5 t_02+C t_0^5$. And for $y_0 = 0$, the unique solution is $y = 0$.
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
answered Sep 12 '17 at 19:56
Robert IsraelRobert Israel
331k23220475
331k23220475
add a comment |
add a comment |
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$begingroup$
@Robert Israel Thank you for your answer. In general, is there any way we can find such particular solutions (which are not covered by the general solution) without guessing? If they exist, are they always asymptotes? Or are they always the result of some parameters going to infinity?
$endgroup$
– John Lei
Sep 13 '17 at 3:14