Uniqueness of IVP using Picard-Lindelöf

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Let $f$ be lipschitz and consider the IVP $dot x=f(x), x(t_0)=x_0$ with a solution $varphi$. If $f(x_0)>0$, then $varphi$ is monotonically increasing.

I have a question about the following reasoning for this fact:

Let's assume $dotvarphi (t_1)=0$ for some $t_1inmathbb R$. Set $y:=varphi(t_1)$, then we have $f(y)=0$ and thus $x_1(t)equiv y$ solves the DE $dot x=f(x)$. Because $x_1$ and $varphi$ intersect in $y$, they must be equal, $varphi =x_1$, which is a contradiction to $dotvarphi(x_0)=f(x_0)neq 0$.

Why must $varphi$ and $x_1$ be equal? Picard-Lindelöf guarantees a unique solution since $f$ is lipschitz, but this is regarding the IVP. And, as I see it, $x_1 (t)equiv y$ is a solution for the IVP if and only if $y=x_0$, but that would be a contradiction since $f(x_0)>0$, so we don't have the trivial solution $x(t)equiv x_0$. Or, in other words, the constructed $x_1$ in the reasoning doesn't solve the IVP. Why should it be equal then with a solution that does(!) solve it?

differential-equations

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asked Jul 16 at 18:18

Buh

59427

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up vote
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Let $f$ be lipschitz and consider the IVP $dot x=f(x), x(t_0)=x_0$ with a solution $varphi$. If $f(x_0)>0$, then $varphi$ is monotonically increasing.

I have a question about the following reasoning for this fact:

Let's assume $dotvarphi (t_1)=0$ for some $t_1inmathbb R$. Set $y:=varphi(t_1)$, then we have $f(y)=0$ and thus $x_1(t)equiv y$ solves the DE $dot x=f(x)$. Because $x_1$ and $varphi$ intersect in $y$, they must be equal, $varphi =x_1$, which is a contradiction to $dotvarphi(x_0)=f(x_0)neq 0$.

Why must $varphi$ and $x_1$ be equal? Picard-Lindelöf guarantees a unique solution since $f$ is lipschitz, but this is regarding the IVP. And, as I see it, $x_1 (t)equiv y$ is a solution for the IVP if and only if $y=x_0$, but that would be a contradiction since $f(x_0)>0$, so we don't have the trivial solution $x(t)equiv x_0$. Or, in other words, the constructed $x_1$ in the reasoning doesn't solve the IVP. Why should it be equal then with a solution that does(!) solve it?

differential-equations

share|cite|improve this question

asked Jul 16 at 18:18

Buh

59427

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $f$ be lipschitz and consider the IVP $dot x=f(x), x(t_0)=x_0$ with a solution $varphi$. If $f(x_0)>0$, then $varphi$ is monotonically increasing.

I have a question about the following reasoning for this fact:

Let's assume $dotvarphi (t_1)=0$ for some $t_1inmathbb R$. Set $y:=varphi(t_1)$, then we have $f(y)=0$ and thus $x_1(t)equiv y$ solves the DE $dot x=f(x)$. Because $x_1$ and $varphi$ intersect in $y$, they must be equal, $varphi =x_1$, which is a contradiction to $dotvarphi(x_0)=f(x_0)neq 0$.

Why must $varphi$ and $x_1$ be equal? Picard-Lindelöf guarantees a unique solution since $f$ is lipschitz, but this is regarding the IVP. And, as I see it, $x_1 (t)equiv y$ is a solution for the IVP if and only if $y=x_0$, but that would be a contradiction since $f(x_0)>0$, so we don't have the trivial solution $x(t)equiv x_0$. Or, in other words, the constructed $x_1$ in the reasoning doesn't solve the IVP. Why should it be equal then with a solution that does(!) solve it?

differential-equations

share|cite|improve this question

asked Jul 16 at 18:18

Buh

59427

Let $f$ be lipschitz and consider the IVP $dot x=f(x), x(t_0)=x_0$ with a solution $varphi$. If $f(x_0)>0$, then $varphi$ is monotonically increasing.

I have a question about the following reasoning for this fact:

Let's assume $dotvarphi (t_1)=0$ for some $t_1inmathbb R$. Set $y:=varphi(t_1)$, then we have $f(y)=0$ and thus $x_1(t)equiv y$ solves the DE $dot x=f(x)$. Because $x_1$ and $varphi$ intersect in $y$, they must be equal, $varphi =x_1$, which is a contradiction to $dotvarphi(x_0)=f(x_0)neq 0$.

Why must $varphi$ and $x_1$ be equal? Picard-Lindelöf guarantees a unique solution since $f$ is lipschitz, but this is regarding the IVP. And, as I see it, $x_1 (t)equiv y$ is a solution for the IVP if and only if $y=x_0$, but that would be a contradiction since $f(x_0)>0$, so we don't have the trivial solution $x(t)equiv x_0$. Or, in other words, the constructed $x_1$ in the reasoning doesn't solve the IVP. Why should it be equal then with a solution that does(!) solve it?

differential-equations

share|cite|improve this question

asked Jul 16 at 18:18

Buh

59427

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 18:18

Buh

59427

asked Jul 16 at 18:18

Buh

59427

asked Jul 16 at 18:18

Buh

59427

59427

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

I'm very glad I ran across this statement, since I've been unfamiliar with it!

Keep in mind though that here, Picard‒Lindelöf is not applied to the original initial value problem, but to the IVP

$$
begincases
x_1'(t) = f(x_1(t)) \
x_1(t_1) = y;
endcases
$$

the endpoint in the argument above for equality is $t_1$, not $t_0$.

share|cite|improve this answer

edited Jul 17 at 6:07

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
  – Buh
  Jul 16 at 19:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do indeed @Buh
  – AlgebraicsAnonymous
  Jul 17 at 6:08
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

I'm very glad I ran across this statement, since I've been unfamiliar with it!

Keep in mind though that here, Picard‒Lindelöf is not applied to the original initial value problem, but to the IVP

$$
begincases
x_1'(t) = f(x_1(t)) \
x_1(t_1) = y;
endcases
$$

the endpoint in the argument above for equality is $t_1$, not $t_0$.

share|cite|improve this answer

edited Jul 17 at 6:07

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
  – Buh
  Jul 16 at 19:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do indeed @Buh
  – AlgebraicsAnonymous
  Jul 17 at 6:08
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

I'm very glad I ran across this statement, since I've been unfamiliar with it!

Keep in mind though that here, Picard‒Lindelöf is not applied to the original initial value problem, but to the IVP

$$
begincases
x_1'(t) = f(x_1(t)) \
x_1(t_1) = y;
endcases
$$

the endpoint in the argument above for equality is $t_1$, not $t_0$.

share|cite|improve this answer

edited Jul 17 at 6:07

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
  – Buh
  Jul 16 at 19:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do indeed @Buh
  – AlgebraicsAnonymous
  Jul 17 at 6:08
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

I'm very glad I ran across this statement, since I've been unfamiliar with it!

Keep in mind though that here, Picard‒Lindelöf is not applied to the original initial value problem, but to the IVP

$$
begincases
x_1'(t) = f(x_1(t)) \
x_1(t_1) = y;
endcases
$$

the endpoint in the argument above for equality is $t_1$, not $t_0$.

share|cite|improve this answer

edited Jul 17 at 6:07

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

I'm very glad I ran across this statement, since I've been unfamiliar with it!

Keep in mind though that here, Picard‒Lindelöf is not applied to the original initial value problem, but to the IVP

$$
begincases
x_1'(t) = f(x_1(t)) \
x_1(t_1) = y;
endcases
$$

the endpoint in the argument above for equality is $t_1$, not $t_0$.

share|cite|improve this answer

edited Jul 17 at 6:07

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 6:07

edited Jul 17 at 6:07

edited Jul 17 at 6:07

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

answered Jul 16 at 18:38

AlgebraicsAnonymous

69111

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
  – Buh
  Jul 16 at 19:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do indeed @Buh
  – AlgebraicsAnonymous
  Jul 17 at 6:08
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
  – Buh
  Jul 16 at 19:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do indeed @Buh
  – AlgebraicsAnonymous
  Jul 17 at 6:08
  

  
  
  
  

1

1

Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
– Buh
Jul 16 at 19:54

Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $varphi$ isn't a solution.
– Buh
Jul 16 at 19:54

I do indeed @Buh
– AlgebraicsAnonymous
Jul 17 at 6:08

I do indeed @Buh
– AlgebraicsAnonymous
Jul 17 at 6:08

add a comment | 

 

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