Uniqueness of solutions of an IVP
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I have a misunderstanding regarding a very common reasoning:
Let's i.e. look at the IVP $dot x=f(x), x(t_0)=x_0$ with $f(x)=(x-1)(x-2)$. Now, for $x_0in ]1,2[$ there can be made an argument that the solution always stays in between $]1,2[$.
The usual way to reason this is since otherwise there would be a point, let's say $t_1$, such that $lambda(t_1)=1$ or $lambda(t_1)=2$ and that conflicts with the uniqueness of the solution since it would have an intersecting point with one of the constant solutions.
But I don't understand this: The constant solutions don't even solve the IVP if $x_0in ]1,2[$, so why is it a problem that those solutions intersect?
Is the point here that a solution $lambda$ for the IVP $x(t_0)=x_0$ would then also solve the IVP with $x(t_1)=1$ which is soled by the constant solution and therefor they must coincide?
ordinary-differential-equations initial-value-problems
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
$endgroup$
add a comment |
$begingroup$
I have a misunderstanding regarding a very common reasoning:
Let's i.e. look at the IVP $dot x=f(x), x(t_0)=x_0$ with $f(x)=(x-1)(x-2)$. Now, for $x_0in ]1,2[$ there can be made an argument that the solution always stays in between $]1,2[$.
The usual way to reason this is since otherwise there would be a point, let's say $t_1$, such that $lambda(t_1)=1$ or $lambda(t_1)=2$ and that conflicts with the uniqueness of the solution since it would have an intersecting point with one of the constant solutions.
But I don't understand this: The constant solutions don't even solve the IVP if $x_0in ]1,2[$, so why is it a problem that those solutions intersect?
Is the point here that a solution $lambda$ for the IVP $x(t_0)=x_0$ would then also solve the IVP with $x(t_1)=1$ which is soled by the constant solution and therefor they must coincide?
ordinary-differential-equations initial-value-problems
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
$endgroup$
$begingroup$
Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness.
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– LutzL
Dec 19 '18 at 16:18
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Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$
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– Song
Dec 19 '18 at 16:36
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Dec 19 '18 at 16:43
add a comment |
$begingroup$
I have a misunderstanding regarding a very common reasoning:
Let's i.e. look at the IVP $dot x=f(x), x(t_0)=x_0$ with $f(x)=(x-1)(x-2)$. Now, for $x_0in ]1,2[$ there can be made an argument that the solution always stays in between $]1,2[$.
The usual way to reason this is since otherwise there would be a point, let's say $t_1$, such that $lambda(t_1)=1$ or $lambda(t_1)=2$ and that conflicts with the uniqueness of the solution since it would have an intersecting point with one of the constant solutions.
But I don't understand this: The constant solutions don't even solve the IVP if $x_0in ]1,2[$, so why is it a problem that those solutions intersect?
Is the point here that a solution $lambda$ for the IVP $x(t_0)=x_0$ would then also solve the IVP with $x(t_1)=1$ which is soled by the constant solution and therefor they must coincide?
ordinary-differential-equations initial-value-problems
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
$endgroup$
I have a misunderstanding regarding a very common reasoning:
Let's i.e. look at the IVP $dot x=f(x), x(t_0)=x_0$ with $f(x)=(x-1)(x-2)$. Now, for $x_0in ]1,2[$ there can be made an argument that the solution always stays in between $]1,2[$.
The usual way to reason this is since otherwise there would be a point, let's say $t_1$, such that $lambda(t_1)=1$ or $lambda(t_1)=2$ and that conflicts with the uniqueness of the solution since it would have an intersecting point with one of the constant solutions.
But I don't understand this: The constant solutions don't even solve the IVP if $x_0in ]1,2[$, so why is it a problem that those solutions intersect?
Is the point here that a solution $lambda$ for the IVP $x(t_0)=x_0$ would then also solve the IVP with $x(t_1)=1$ which is soled by the constant solution and therefor they must coincide?
ordinary-differential-equations initial-value-problems
ordinary-differential-equations initial-value-problems
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
asked Dec 19 '18 at 15:55
RedLanternRedLantern
415
415
$begingroup$
Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness.
$endgroup$
– LutzL
Dec 19 '18 at 16:18
$begingroup$
Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$
$endgroup$
– Song
Dec 19 '18 at 16:36
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Dec 19 '18 at 16:43
add a comment |
$begingroup$
Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness.
$endgroup$
– LutzL
Dec 19 '18 at 16:18
$begingroup$
Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$
$endgroup$
– Song
Dec 19 '18 at 16:36
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Dec 19 '18 at 16:43
$begingroup$
Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness.
$endgroup$
– LutzL
Dec 19 '18 at 16:18
$begingroup$
Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness.
$endgroup$
– LutzL
Dec 19 '18 at 16:18
$begingroup$
Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$
$endgroup$
– Song
Dec 19 '18 at 16:36
$begingroup$
Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$
$endgroup$
– Song
Dec 19 '18 at 16:36
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Dec 19 '18 at 16:43
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Dec 19 '18 at 16:43
add a comment |
1 Answer
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Firstly, the IVP
$$ dot{x}=f(x),x(t_0)=x_0.tag{1} $$
has a unique solution for $x_0in]1,2[$. Clearly $x=underline{x}(t)equiv1$ is a subsolution of (1) and $x=bar{x}(t)equiv2$ is a supersolution of (1). Since $underline{x}(t)le bar{x}(t)$. Therefore there is a unique solution $x=x(t)$ such that
$$ underline{x}(t)le x(t)le bar{x}(t). $$
Namely the solution always stays between $]1,2[$ if $x_0in]1,2[$.
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
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$begingroup$
Firstly, the IVP
$$ dot{x}=f(x),x(t_0)=x_0.tag{1} $$
has a unique solution for $x_0in]1,2[$. Clearly $x=underline{x}(t)equiv1$ is a subsolution of (1) and $x=bar{x}(t)equiv2$ is a supersolution of (1). Since $underline{x}(t)le bar{x}(t)$. Therefore there is a unique solution $x=x(t)$ such that
$$ underline{x}(t)le x(t)le bar{x}(t). $$
Namely the solution always stays between $]1,2[$ if $x_0in]1,2[$.
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
$endgroup$
add a comment |
$begingroup$
Firstly, the IVP
$$ dot{x}=f(x),x(t_0)=x_0.tag{1} $$
has a unique solution for $x_0in]1,2[$. Clearly $x=underline{x}(t)equiv1$ is a subsolution of (1) and $x=bar{x}(t)equiv2$ is a supersolution of (1). Since $underline{x}(t)le bar{x}(t)$. Therefore there is a unique solution $x=x(t)$ such that
$$ underline{x}(t)le x(t)le bar{x}(t). $$
Namely the solution always stays between $]1,2[$ if $x_0in]1,2[$.
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
$endgroup$
add a comment |
$begingroup$
Firstly, the IVP
$$ dot{x}=f(x),x(t_0)=x_0.tag{1} $$
has a unique solution for $x_0in]1,2[$. Clearly $x=underline{x}(t)equiv1$ is a subsolution of (1) and $x=bar{x}(t)equiv2$ is a supersolution of (1). Since $underline{x}(t)le bar{x}(t)$. Therefore there is a unique solution $x=x(t)$ such that
$$ underline{x}(t)le x(t)le bar{x}(t). $$
Namely the solution always stays between $]1,2[$ if $x_0in]1,2[$.
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
$endgroup$
Firstly, the IVP
$$ dot{x}=f(x),x(t_0)=x_0.tag{1} $$
has a unique solution for $x_0in]1,2[$. Clearly $x=underline{x}(t)equiv1$ is a subsolution of (1) and $x=bar{x}(t)equiv2$ is a supersolution of (1). Since $underline{x}(t)le bar{x}(t)$. Therefore there is a unique solution $x=x(t)$ such that
$$ underline{x}(t)le x(t)le bar{x}(t). $$
Namely the solution always stays between $]1,2[$ if $x_0in]1,2[$.
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
answered Dec 20 '18 at 14:38
xpaulxpaul
22.9k24455
22.9k24455
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$begingroup$
Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness.
$endgroup$
– LutzL
Dec 19 '18 at 16:18
$begingroup$
Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$
$endgroup$
– Song
Dec 19 '18 at 16:36
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Dec 19 '18 at 16:43