Mixing
Show that the solution of an IVP eventually leaves a compact set
$begingroup$
I am reading a proof which states that a solution $(t,x(t))$ to an IVP ($x' = f(t,x), x(t_0) = x_0$), where $f$ is a function from an open set $U$ to $mathbb{R^n}$ that is locally Lipschitz continuous in its second argument, will leave any compact set $C$ (contained in $U$) as $t$ approaches $beta$. Here $(alpha, beta)$ is the maximal interval where a solution exists.
There are a few things in the proof that confuses me. I understand (or at least I think I do) that it relates to proving that there exists a solution $ x(beta)=zeta$, so that by Picard-Lindelof there will exist a neighborhood of $beta$ where a solution exists, contradicting the assumption of $(alpha, beta)$ being the maximal interval of existence.
So the proof is done by contradiction. We assume that there is a sequence $(t_k, x(t_k)) in C, forall k,$ where $t_k rightarrow beta$. And then there is also a convergent subsequence, which we will also denote by $(t_k, x(t_k))$, where $$(t_k, x(t_k)) rightarrow (beta, zeta) in C $$ as $k rightarrow infty$. Since $U$ is open, there exists an $epsilon > 0$ such that $R = [beta, beta+epsilon] times [zeta-epsilon, zeta+epsilon]$ in $U$. We let $M$ denote the maximum value of $|f|$ in $R$.
So far I believe that I have been able to follow the proof. But then it says that by Picard-Lindelof, that the solution passing through $(t_k, x(t_k))$ exists as long as the cone $$|x-x(t_k)| leq M|t-t_k|, tgeq t_k $$ remains in $R$. And that this is true for some $t > beta$ as long as we take $(t_k, x(t_k))$ close enough to $(beta,zeta)$. And that this gives the contradiction.
I don't quite understand what is meant by the cone remaining in $R$. And since we already know that $x(t_k) rightarrow zeta$ as $k rightarrow infty$, isn't that enough to conclude that $x(t) rightarrow zeta$?
ordinary-differential-equations initial-value-problems
asked Jan 31 at 20:31
j.eeej.eee
787
$endgroup$
add a comment |
$begingroup$
I am reading a proof which states that a solution $(t,x(t))$ to an IVP ($x' = f(t,x), x(t_0) = x_0$), where $f$ is a function from an open set $U$ to $mathbb{R^n}$ that is locally Lipschitz continuous in its second argument, will leave any compact set $C$ (contained in $U$) as $t$ approaches $beta$. Here $(alpha, beta)$ is the maximal interval where a solution exists.
There are a few things in the proof that confuses me. I understand (or at least I think I do) that it relates to proving that there exists a solution $ x(beta)=zeta$, so that by Picard-Lindelof there will exist a neighborhood of $beta$ where a solution exists, contradicting the assumption of $(alpha, beta)$ being the maximal interval of existence.
So the proof is done by contradiction. We assume that there is a sequence $(t_k, x(t_k)) in C, forall k,$ where $t_k rightarrow beta$. And then there is also a convergent subsequence, which we will also denote by $(t_k, x(t_k))$, where $$(t_k, x(t_k)) rightarrow (beta, zeta) in C $$ as $k rightarrow infty$. Since $U$ is open, there exists an $epsilon > 0$ such that $R = [beta, beta+epsilon] times [zeta-epsilon, zeta+epsilon]$ in $U$. We let $M$ denote the maximum value of $|f|$ in $R$.
So far I believe that I have been able to follow the proof. But then it says that by Picard-Lindelof, that the solution passing through $(t_k, x(t_k))$ exists as long as the cone $$|x-x(t_k)| leq M|t-t_k|, tgeq t_k $$ remains in $R$. And that this is true for some $t > beta$ as long as we take $(t_k, x(t_k))$ close enough to $(beta,zeta)$. And that this gives the contradiction.
I don't quite understand what is meant by the cone remaining in $R$. And since we already know that $x(t_k) rightarrow zeta$ as $k rightarrow infty$, isn't that enough to conclude that $x(t) rightarrow zeta$?
ordinary-differential-equations initial-value-problems
asked Jan 31 at 20:31
j.eeej.eee
787
$endgroup$
add a comment |
$begingroup$
I am reading a proof which states that a solution $(t,x(t))$ to an IVP ($x' = f(t,x), x(t_0) = x_0$), where $f$ is a function from an open set $U$ to $mathbb{R^n}$ that is locally Lipschitz continuous in its second argument, will leave any compact set $C$ (contained in $U$) as $t$ approaches $beta$. Here $(alpha, beta)$ is the maximal interval where a solution exists.
There are a few things in the proof that confuses me. I understand (or at least I think I do) that it relates to proving that there exists a solution $ x(beta)=zeta$, so that by Picard-Lindelof there will exist a neighborhood of $beta$ where a solution exists, contradicting the assumption of $(alpha, beta)$ being the maximal interval of existence.
So the proof is done by contradiction. We assume that there is a sequence $(t_k, x(t_k)) in C, forall k,$ where $t_k rightarrow beta$. And then there is also a convergent subsequence, which we will also denote by $(t_k, x(t_k))$, where $$(t_k, x(t_k)) rightarrow (beta, zeta) in C $$ as $k rightarrow infty$. Since $U$ is open, there exists an $epsilon > 0$ such that $R = [beta, beta+epsilon] times [zeta-epsilon, zeta+epsilon]$ in $U$. We let $M$ denote the maximum value of $|f|$ in $R$.
So far I believe that I have been able to follow the proof. But then it says that by Picard-Lindelof, that the solution passing through $(t_k, x(t_k))$ exists as long as the cone $$|x-x(t_k)| leq M|t-t_k|, tgeq t_k $$ remains in $R$. And that this is true for some $t > beta$ as long as we take $(t_k, x(t_k))$ close enough to $(beta,zeta)$. And that this gives the contradiction.
I don't quite understand what is meant by the cone remaining in $R$. And since we already know that $x(t_k) rightarrow zeta$ as $k rightarrow infty$, isn't that enough to conclude that $x(t) rightarrow zeta$?
ordinary-differential-equations initial-value-problems
asked Jan 31 at 20:31
j.eeej.eee
787
$endgroup$
I am reading a proof which states that a solution $(t,x(t))$ to an IVP ($x' = f(t,x), x(t_0) = x_0$), where $f$ is a function from an open set $U$ to $mathbb{R^n}$ that is locally Lipschitz continuous in its second argument, will leave any compact set $C$ (contained in $U$) as $t$ approaches $beta$. Here $(alpha, beta)$ is the maximal interval where a solution exists.
There are a few things in the proof that confuses me. I understand (or at least I think I do) that it relates to proving that there exists a solution $ x(beta)=zeta$, so that by Picard-Lindelof there will exist a neighborhood of $beta$ where a solution exists, contradicting the assumption of $(alpha, beta)$ being the maximal interval of existence.
So the proof is done by contradiction. We assume that there is a sequence $(t_k, x(t_k)) in C, forall k,$ where $t_k rightarrow beta$. And then there is also a convergent subsequence, which we will also denote by $(t_k, x(t_k))$, where $$(t_k, x(t_k)) rightarrow (beta, zeta) in C $$ as $k rightarrow infty$. Since $U$ is open, there exists an $epsilon > 0$ such that $R = [beta, beta+epsilon] times [zeta-epsilon, zeta+epsilon]$ in $U$. We let $M$ denote the maximum value of $|f|$ in $R$.
So far I believe that I have been able to follow the proof. But then it says that by Picard-Lindelof, that the solution passing through $(t_k, x(t_k))$ exists as long as the cone $$|x-x(t_k)| leq M|t-t_k|, tgeq t_k $$ remains in $R$. And that this is true for some $t > beta$ as long as we take $(t_k, x(t_k))$ close enough to $(beta,zeta)$. And that this gives the contradiction.
I don't quite understand what is meant by the cone remaining in $R$. And since we already know that $x(t_k) rightarrow zeta$ as $k rightarrow infty$, isn't that enough to conclude that $x(t) rightarrow zeta$?
ordinary-differential-equations initial-value-problems
ordinary-differential-equations initial-value-problems
asked Jan 31 at 20:31
j.eeej.eee
787
asked Jan 31 at 20:31
j.eeej.eee
787
asked Jan 31 at 20:31
j.eeej.eee
787
asked Jan 31 at 20:31
j.eeej.eee
787
asked Jan 31 at 20:31
j.eeej.eee
787
787
add a comment |
add a comment |
1 Answer
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oldest
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No, $x(t_k)toxi$ does not imply the convergence of the solution, as that could oscillate like for example $sin(1/(β-t))$.
The proof of Picard-Lindelöf has 3 stages: In the first, some compact cylinder is fixed and on this the quantities $M=max|f|$ and the Lipschitz constant $L$ "computed". In the second stage the "butterfly" set centered at the initial condition is constructed, in the third this is further constricted using the Lipschitz constant to gain contractivity of the Picard operator.
Here we consider simultaneously initial value problems for data $(t_k,x(t_k))$ for $k$ large enough. As first stage set take the rectangle $R=[β-δ,β+δ]×[ζ−ϵ,ζ+ϵ]$ inside the domain, note that the time interval is two-sided and in its size decoupled from the space extension. For the simultaneous second stage shrink $δ$ so that $Mδleϵ$. Then choose $N$ so that $t_k>β-δ/3$ and $|x(t_k)-ζ|<ϵ/3$ for $kge N$. Now every box $[t_k-δ/2,t_k+δ/2]×[x(t_k)−ϵ/2,x(t_k)+ϵ/2]$ for the $k$-th IVP satisfies the "butterfly" condition for the second stage. This means that the third stage gives an interval radius $h$ that is independent of $kge N$ so that a local solution of the $k$-th IVP on $[t_k-h,t_k+h]$ exists. Now let $kto infty$, $t_ktoβ$ to get a contradiction to the non-extensibility of the maximum domain.
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
$endgroup$
$begingroup$
Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
$endgroup$
– j.eee
Feb 6 at 15:59
$begingroup$
Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
$endgroup$
– LutzL
Feb 6 at 16:49
$begingroup$
So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
$endgroup$
– j.eee
Feb 9 at 23:44
1
$begingroup$
Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
$endgroup$
– LutzL
Feb 10 at 13:51
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
No, $x(t_k)toxi$ does not imply the convergence of the solution, as that could oscillate like for example $sin(1/(β-t))$.
The proof of Picard-Lindelöf has 3 stages: In the first, some compact cylinder is fixed and on this the quantities $M=max|f|$ and the Lipschitz constant $L$ "computed". In the second stage the "butterfly" set centered at the initial condition is constructed, in the third this is further constricted using the Lipschitz constant to gain contractivity of the Picard operator.
Here we consider simultaneously initial value problems for data $(t_k,x(t_k))$ for $k$ large enough. As first stage set take the rectangle $R=[β-δ,β+δ]×[ζ−ϵ,ζ+ϵ]$ inside the domain, note that the time interval is two-sided and in its size decoupled from the space extension. For the simultaneous second stage shrink $δ$ so that $Mδleϵ$. Then choose $N$ so that $t_k>β-δ/3$ and $|x(t_k)-ζ|<ϵ/3$ for $kge N$. Now every box $[t_k-δ/2,t_k+δ/2]×[x(t_k)−ϵ/2,x(t_k)+ϵ/2]$ for the $k$-th IVP satisfies the "butterfly" condition for the second stage. This means that the third stage gives an interval radius $h$ that is independent of $kge N$ so that a local solution of the $k$-th IVP on $[t_k-h,t_k+h]$ exists. Now let $kto infty$, $t_ktoβ$ to get a contradiction to the non-extensibility of the maximum domain.
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
$endgroup$
$begingroup$
Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
$endgroup$
– j.eee
Feb 6 at 15:59
$begingroup$
Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
$endgroup$
– LutzL
Feb 6 at 16:49
$begingroup$
So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
$endgroup$
– j.eee
Feb 9 at 23:44
1
$begingroup$
Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
$endgroup$
– LutzL
Feb 10 at 13:51
add a comment |
$begingroup$
No, $x(t_k)toxi$ does not imply the convergence of the solution, as that could oscillate like for example $sin(1/(β-t))$.
The proof of Picard-Lindelöf has 3 stages: In the first, some compact cylinder is fixed and on this the quantities $M=max|f|$ and the Lipschitz constant $L$ "computed". In the second stage the "butterfly" set centered at the initial condition is constructed, in the third this is further constricted using the Lipschitz constant to gain contractivity of the Picard operator.
Here we consider simultaneously initial value problems for data $(t_k,x(t_k))$ for $k$ large enough. As first stage set take the rectangle $R=[β-δ,β+δ]×[ζ−ϵ,ζ+ϵ]$ inside the domain, note that the time interval is two-sided and in its size decoupled from the space extension. For the simultaneous second stage shrink $δ$ so that $Mδleϵ$. Then choose $N$ so that $t_k>β-δ/3$ and $|x(t_k)-ζ|<ϵ/3$ for $kge N$. Now every box $[t_k-δ/2,t_k+δ/2]×[x(t_k)−ϵ/2,x(t_k)+ϵ/2]$ for the $k$-th IVP satisfies the "butterfly" condition for the second stage. This means that the third stage gives an interval radius $h$ that is independent of $kge N$ so that a local solution of the $k$-th IVP on $[t_k-h,t_k+h]$ exists. Now let $kto infty$, $t_ktoβ$ to get a contradiction to the non-extensibility of the maximum domain.
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
$endgroup$
$begingroup$
Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
$endgroup$
– j.eee
Feb 6 at 15:59
$begingroup$
Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
$endgroup$
– LutzL
Feb 6 at 16:49
$begingroup$
So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
$endgroup$
– j.eee
Feb 9 at 23:44
1
$begingroup$
Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
$endgroup$
– LutzL
Feb 10 at 13:51
add a comment |
$begingroup$
No, $x(t_k)toxi$ does not imply the convergence of the solution, as that could oscillate like for example $sin(1/(β-t))$.
The proof of Picard-Lindelöf has 3 stages: In the first, some compact cylinder is fixed and on this the quantities $M=max|f|$ and the Lipschitz constant $L$ "computed". In the second stage the "butterfly" set centered at the initial condition is constructed, in the third this is further constricted using the Lipschitz constant to gain contractivity of the Picard operator.
Here we consider simultaneously initial value problems for data $(t_k,x(t_k))$ for $k$ large enough. As first stage set take the rectangle $R=[β-δ,β+δ]×[ζ−ϵ,ζ+ϵ]$ inside the domain, note that the time interval is two-sided and in its size decoupled from the space extension. For the simultaneous second stage shrink $δ$ so that $Mδleϵ$. Then choose $N$ so that $t_k>β-δ/3$ and $|x(t_k)-ζ|<ϵ/3$ for $kge N$. Now every box $[t_k-δ/2,t_k+δ/2]×[x(t_k)−ϵ/2,x(t_k)+ϵ/2]$ for the $k$-th IVP satisfies the "butterfly" condition for the second stage. This means that the third stage gives an interval radius $h$ that is independent of $kge N$ so that a local solution of the $k$-th IVP on $[t_k-h,t_k+h]$ exists. Now let $kto infty$, $t_ktoβ$ to get a contradiction to the non-extensibility of the maximum domain.
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
$endgroup$
No, $x(t_k)toxi$ does not imply the convergence of the solution, as that could oscillate like for example $sin(1/(β-t))$.
The proof of Picard-Lindelöf has 3 stages: In the first, some compact cylinder is fixed and on this the quantities $M=max|f|$ and the Lipschitz constant $L$ "computed". In the second stage the "butterfly" set centered at the initial condition is constructed, in the third this is further constricted using the Lipschitz constant to gain contractivity of the Picard operator.
Here we consider simultaneously initial value problems for data $(t_k,x(t_k))$ for $k$ large enough. As first stage set take the rectangle $R=[β-δ,β+δ]×[ζ−ϵ,ζ+ϵ]$ inside the domain, note that the time interval is two-sided and in its size decoupled from the space extension. For the simultaneous second stage shrink $δ$ so that $Mδleϵ$. Then choose $N$ so that $t_k>β-δ/3$ and $|x(t_k)-ζ|<ϵ/3$ for $kge N$. Now every box $[t_k-δ/2,t_k+δ/2]×[x(t_k)−ϵ/2,x(t_k)+ϵ/2]$ for the $k$-th IVP satisfies the "butterfly" condition for the second stage. This means that the third stage gives an interval radius $h$ that is independent of $kge N$ so that a local solution of the $k$-th IVP on $[t_k-h,t_k+h]$ exists. Now let $kto infty$, $t_ktoβ$ to get a contradiction to the non-extensibility of the maximum domain.
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
answered Jan 31 at 21:32
LutzLLutzL
60.3k42057
60.3k42057
$begingroup$
Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
$endgroup$
– j.eee
Feb 6 at 15:59
$begingroup$
Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
$endgroup$
– LutzL
Feb 6 at 16:49
$begingroup$
So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
$endgroup$
– j.eee
Feb 9 at 23:44
1
$begingroup$
Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
$endgroup$
– LutzL
Feb 10 at 13:51
add a comment |
$begingroup$
Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
$endgroup$
– j.eee
Feb 6 at 15:59
$begingroup$
Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
$endgroup$
– LutzL
Feb 6 at 16:49
$begingroup$
So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
$endgroup$
– j.eee
Feb 9 at 23:44
1
$begingroup$
Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
$endgroup$
– LutzL
Feb 10 at 13:51
$begingroup$
Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
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– j.eee
Feb 6 at 15:59
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Sorry that it's taken so long for me to answer, I've had to think about this a lot and I'm still uncertain about a few things. So I can follow your proof just fine, but now I'm trying to "translate" it into the language of the proof that I was confused about (because I also wish to understand how they reasoned). So the "butterfly" I suppose is the cone mentioned in my post. And that corresponds to the boxes that you mentioned. As boxes I understand it, but as cones/butterflies I don't, because in the proof I've read of Picard-Lindelov, they also deal with boxes/rectangles. @LutzL
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– j.eee
Feb 6 at 15:59
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Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
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– LutzL
Feb 6 at 16:49
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Yes, I mean the two-sided cone $|x-x_k|le M,|t-t_k|$ cut off at the boundary of the box. Any solution going through $(t_k,x-k)$ has to stay inside that cone. In the second stage it is ensured that the cone exits through the sides of the (shrunken) box, so any local solution has to exist for the whole length of the box.
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– LutzL
Feb 6 at 16:49
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So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
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– j.eee
Feb 9 at 23:44
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So this is my current interpretation: Any solution going through $(t_k,x(t_k))$ must be continuous and therefore must stay in the cone. And then we ensure that this cone stays in the schrunken box, because then we can apply Picard-Lindelof to show that a local solution exists along that interval (the lenght of the box). Is this correct? @LutzL
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– j.eee
Feb 9 at 23:44
1
1
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Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
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– LutzL
Feb 10 at 13:51
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Yes, the aim is to show that there is some $h$ so that the solution exists up to $t_k+h$, for all $k$ with the same $h$. Which then contradicts the claim that the maximal solution ends at $β$.
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– LutzL
Feb 10 at 13:51
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