Mixing
Proof of Peano's existence theorem for ODE using Picard iterations
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I am curious why one does not use the standard Picard iteration scheme in order to prove the Cauchy--Peano existence theorem for ODE. This seems to give a relatively easy proof, as one can easily show by induction that the sequence of Picard iterates is uniformly bounded and equicontinuous on some compact set.
All other proofs seem to use the Tonelli sequence, which is more complicated. So I am wondering whether I have missed something...
thanks!
ordinary-differential-equations
asked Jan 26 at 21:30
user638579user638579
31
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add a comment |
$begingroup$
I am curious why one does not use the standard Picard iteration scheme in order to prove the Cauchy--Peano existence theorem for ODE. This seems to give a relatively easy proof, as one can easily show by induction that the sequence of Picard iterates is uniformly bounded and equicontinuous on some compact set.
All other proofs seem to use the Tonelli sequence, which is more complicated. So I am wondering whether I have missed something...
thanks!
ordinary-differential-equations
asked Jan 26 at 21:30
user638579user638579
31
$endgroup$
add a comment |
$begingroup$
I am curious why one does not use the standard Picard iteration scheme in order to prove the Cauchy--Peano existence theorem for ODE. This seems to give a relatively easy proof, as one can easily show by induction that the sequence of Picard iterates is uniformly bounded and equicontinuous on some compact set.
All other proofs seem to use the Tonelli sequence, which is more complicated. So I am wondering whether I have missed something...
thanks!
ordinary-differential-equations
asked Jan 26 at 21:30
user638579user638579
31
$endgroup$
I am curious why one does not use the standard Picard iteration scheme in order to prove the Cauchy--Peano existence theorem for ODE. This seems to give a relatively easy proof, as one can easily show by induction that the sequence of Picard iterates is uniformly bounded and equicontinuous on some compact set.
All other proofs seem to use the Tonelli sequence, which is more complicated. So I am wondering whether I have missed something...
thanks!
ordinary-differential-equations
ordinary-differential-equations
asked Jan 26 at 21:30
user638579user638579
31
asked Jan 26 at 21:30
user638579user638579
31
asked Jan 26 at 21:30
user638579user638579
31
asked Jan 26 at 21:30
user638579user638579
31
asked Jan 26 at 21:30
user638579user638579
31
31
add a comment |
add a comment |
1 Answer
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Because the sequence of Picard iterations may have no subsequence converging to a solution. There are converging subsequences, but the limits may not be solutions. Let
$$
f(t,x)=begin{cases}2,t &text{if }xle0\
-2,t & text{if }xge t^2\
lambda(2,t)+(1-lambda)(-2,t)&text{if }0<x<t^2
end{cases}
$$
where $lambdain(0,1)$ is such that $(t,x)=lambda(t,0)+(1-lambda)(t,t^2)$; $f$ is continuous. Consider the problem $x'=f(t,x)$, $x(0)=0$. It has a unique solution, since $f$ is decreasing in $x$ for each fixed $t$. The Picard iterations starting with $x(t)equiv0$ are $t^2,-t^2,t^2,-t^2,dots$ and neither $t^2$ nor $-t^2$ are solutions.
This example is due to M. Müller, Uber das Fundamentaltheorem in der Theorie der gewöhnlichen Differentialgleichungen, Math. Zeit, 26, 1927. It appears in the book Ecuaciones diferenzciales ordinarias by Miguel de Guzmán, Alhambra 1975. It was widely used as a textbook in Spain when I was an undergraduate student.
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Because the sequence of Picard iterations may have no subsequence converging to a solution. There are converging subsequences, but the limits may not be solutions. Let
$$
f(t,x)=begin{cases}2,t &text{if }xle0\
-2,t & text{if }xge t^2\
lambda(2,t)+(1-lambda)(-2,t)&text{if }0<x<t^2
end{cases}
$$
where $lambdain(0,1)$ is such that $(t,x)=lambda(t,0)+(1-lambda)(t,t^2)$; $f$ is continuous. Consider the problem $x'=f(t,x)$, $x(0)=0$. It has a unique solution, since $f$ is decreasing in $x$ for each fixed $t$. The Picard iterations starting with $x(t)equiv0$ are $t^2,-t^2,t^2,-t^2,dots$ and neither $t^2$ nor $-t^2$ are solutions.
This example is due to M. Müller, Uber das Fundamentaltheorem in der Theorie der gewöhnlichen Differentialgleichungen, Math. Zeit, 26, 1927. It appears in the book Ecuaciones diferenzciales ordinarias by Miguel de Guzmán, Alhambra 1975. It was widely used as a textbook in Spain when I was an undergraduate student.
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
add a comment |
$begingroup$
Because the sequence of Picard iterations may have no subsequence converging to a solution. There are converging subsequences, but the limits may not be solutions. Let
$$
f(t,x)=begin{cases}2,t &text{if }xle0\
-2,t & text{if }xge t^2\
lambda(2,t)+(1-lambda)(-2,t)&text{if }0<x<t^2
end{cases}
$$
where $lambdain(0,1)$ is such that $(t,x)=lambda(t,0)+(1-lambda)(t,t^2)$; $f$ is continuous. Consider the problem $x'=f(t,x)$, $x(0)=0$. It has a unique solution, since $f$ is decreasing in $x$ for each fixed $t$. The Picard iterations starting with $x(t)equiv0$ are $t^2,-t^2,t^2,-t^2,dots$ and neither $t^2$ nor $-t^2$ are solutions.
This example is due to M. Müller, Uber das Fundamentaltheorem in der Theorie der gewöhnlichen Differentialgleichungen, Math. Zeit, 26, 1927. It appears in the book Ecuaciones diferenzciales ordinarias by Miguel de Guzmán, Alhambra 1975. It was widely used as a textbook in Spain when I was an undergraduate student.
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
add a comment |
$begingroup$
Because the sequence of Picard iterations may have no subsequence converging to a solution. There are converging subsequences, but the limits may not be solutions. Let
$$
f(t,x)=begin{cases}2,t &text{if }xle0\
-2,t & text{if }xge t^2\
lambda(2,t)+(1-lambda)(-2,t)&text{if }0<x<t^2
end{cases}
$$
where $lambdain(0,1)$ is such that $(t,x)=lambda(t,0)+(1-lambda)(t,t^2)$; $f$ is continuous. Consider the problem $x'=f(t,x)$, $x(0)=0$. It has a unique solution, since $f$ is decreasing in $x$ for each fixed $t$. The Picard iterations starting with $x(t)equiv0$ are $t^2,-t^2,t^2,-t^2,dots$ and neither $t^2$ nor $-t^2$ are solutions.
This example is due to M. Müller, Uber das Fundamentaltheorem in der Theorie der gewöhnlichen Differentialgleichungen, Math. Zeit, 26, 1927. It appears in the book Ecuaciones diferenzciales ordinarias by Miguel de Guzmán, Alhambra 1975. It was widely used as a textbook in Spain when I was an undergraduate student.
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
Because the sequence of Picard iterations may have no subsequence converging to a solution. There are converging subsequences, but the limits may not be solutions. Let
$$
f(t,x)=begin{cases}2,t &text{if }xle0\
-2,t & text{if }xge t^2\
lambda(2,t)+(1-lambda)(-2,t)&text{if }0<x<t^2
end{cases}
$$
where $lambdain(0,1)$ is such that $(t,x)=lambda(t,0)+(1-lambda)(t,t^2)$; $f$ is continuous. Consider the problem $x'=f(t,x)$, $x(0)=0$. It has a unique solution, since $f$ is decreasing in $x$ for each fixed $t$. The Picard iterations starting with $x(t)equiv0$ are $t^2,-t^2,t^2,-t^2,dots$ and neither $t^2$ nor $-t^2$ are solutions.
This example is due to M. Müller, Uber das Fundamentaltheorem in der Theorie der gewöhnlichen Differentialgleichungen, Math. Zeit, 26, 1927. It appears in the book Ecuaciones diferenzciales ordinarias by Miguel de Guzmán, Alhambra 1975. It was widely used as a textbook in Spain when I was an undergraduate student.
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
answered Jan 27 at 16:55
Julián AguirreJulián Aguirre
69.4k24197
69.4k24197
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
add a comment |
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
many thanks for your response! I was initially confused as to what I must have done wrong in my proof, but I realise now that it is the presence of the $u_{n_k-1}$ in the iteration scheme which cause the limit not to solve the equation, as your great example shows. A follow up question: do you know of weaker conditions than locally Lipshitz which ensures convergence of the Picard iterates?
$endgroup$
– user638579
Jan 28 at 16:36
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
$begingroup$
No, I am not aware of weaker conditions. This does not mean that they do not exist.
$endgroup$
– Julián Aguirre
Jan 28 at 17:50
add a comment |
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