Mixing
Existence of solution of ordinary differential equation
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I am reading a proof of the existence of solutions for ordinary differential equations and I have some basic doubt. I'll copy the statement, the part of the proof I don't understand and my question:
Statement
Let $I$ be an interval of the real line. Let $f(t,x)$ be a Lipschitz function on the variable $x$ on $I times mathbb R$. Let $tau in I, xi in mathbb R$. If $tau in I^{circ}$, then there is $lambda >0$ and a continuous differentiable function $x:[tau-lambda,tau+lambda] subset I to mathbb R$ such that $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
Sketch of the proof
It is easy to show that finding a solution to the system $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
is equivalent to find a solution to the integral equation $$x(t)=xi+int_tau^t f(s,x(s))ds$$
So we define inductively define $$x_0=xi,$$ $$x_1=xi+int_tau^t f(s,x_0(s))ds,$$ $$x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$$
and if we show that this sequence of functions converges uniformly on the interval
$[tau-lambda,tau+lambda]$ to a function $x(t)$, then we have that $x(t)$ is continuous and that $int_tau^t f(s,x_k(s))ds to int_tau^t f(s,x(s))ds$. This means $x(t)$ is a solution of the integral equation and therefore a solution to the original system.
Question
Now, in my textbook it says that the functions $x_k(t)$ are defined on the interval $I$, it may be a silly little detail but I don't see why these functions are defined on that interval just from the fact $x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$, I would appreciate if someone could explaine me how to check this.
analysis ordinary-differential-equations
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
$endgroup$
add a comment |
$begingroup$
I am reading a proof of the existence of solutions for ordinary differential equations and I have some basic doubt. I'll copy the statement, the part of the proof I don't understand and my question:
Statement
Let $I$ be an interval of the real line. Let $f(t,x)$ be a Lipschitz function on the variable $x$ on $I times mathbb R$. Let $tau in I, xi in mathbb R$. If $tau in I^{circ}$, then there is $lambda >0$ and a continuous differentiable function $x:[tau-lambda,tau+lambda] subset I to mathbb R$ such that $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
Sketch of the proof
It is easy to show that finding a solution to the system $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
is equivalent to find a solution to the integral equation $$x(t)=xi+int_tau^t f(s,x(s))ds$$
So we define inductively define $$x_0=xi,$$ $$x_1=xi+int_tau^t f(s,x_0(s))ds,$$ $$x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$$
and if we show that this sequence of functions converges uniformly on the interval
$[tau-lambda,tau+lambda]$ to a function $x(t)$, then we have that $x(t)$ is continuous and that $int_tau^t f(s,x_k(s))ds to int_tau^t f(s,x(s))ds$. This means $x(t)$ is a solution of the integral equation and therefore a solution to the original system.
Question
Now, in my textbook it says that the functions $x_k(t)$ are defined on the interval $I$, it may be a silly little detail but I don't see why these functions are defined on that interval just from the fact $x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$, I would appreciate if someone could explaine me how to check this.
analysis ordinary-differential-equations
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
$endgroup$
add a comment |
$begingroup$
I am reading a proof of the existence of solutions for ordinary differential equations and I have some basic doubt. I'll copy the statement, the part of the proof I don't understand and my question:
Statement
Let $I$ be an interval of the real line. Let $f(t,x)$ be a Lipschitz function on the variable $x$ on $I times mathbb R$. Let $tau in I, xi in mathbb R$. If $tau in I^{circ}$, then there is $lambda >0$ and a continuous differentiable function $x:[tau-lambda,tau+lambda] subset I to mathbb R$ such that $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
Sketch of the proof
It is easy to show that finding a solution to the system $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
is equivalent to find a solution to the integral equation $$x(t)=xi+int_tau^t f(s,x(s))ds$$
So we define inductively define $$x_0=xi,$$ $$x_1=xi+int_tau^t f(s,x_0(s))ds,$$ $$x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$$
and if we show that this sequence of functions converges uniformly on the interval
$[tau-lambda,tau+lambda]$ to a function $x(t)$, then we have that $x(t)$ is continuous and that $int_tau^t f(s,x_k(s))ds to int_tau^t f(s,x(s))ds$. This means $x(t)$ is a solution of the integral equation and therefore a solution to the original system.
Question
Now, in my textbook it says that the functions $x_k(t)$ are defined on the interval $I$, it may be a silly little detail but I don't see why these functions are defined on that interval just from the fact $x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$, I would appreciate if someone could explaine me how to check this.
analysis ordinary-differential-equations
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
$endgroup$
I am reading a proof of the existence of solutions for ordinary differential equations and I have some basic doubt. I'll copy the statement, the part of the proof I don't understand and my question:
Statement
Let $I$ be an interval of the real line. Let $f(t,x)$ be a Lipschitz function on the variable $x$ on $I times mathbb R$. Let $tau in I, xi in mathbb R$. If $tau in I^{circ}$, then there is $lambda >0$ and a continuous differentiable function $x:[tau-lambda,tau+lambda] subset I to mathbb R$ such that $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
Sketch of the proof
It is easy to show that finding a solution to the system $$x'(t)=f(t,x(t)) forall t in [tau-lambda,tau+lambda]$$ $$x(tau)=xi$$
is equivalent to find a solution to the integral equation $$x(t)=xi+int_tau^t f(s,x(s))ds$$
So we define inductively define $$x_0=xi,$$ $$x_1=xi+int_tau^t f(s,x_0(s))ds,$$ $$x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$$
and if we show that this sequence of functions converges uniformly on the interval
$[tau-lambda,tau+lambda]$ to a function $x(t)$, then we have that $x(t)$ is continuous and that $int_tau^t f(s,x_k(s))ds to int_tau^t f(s,x(s))ds$. This means $x(t)$ is a solution of the integral equation and therefore a solution to the original system.
Question
Now, in my textbook it says that the functions $x_k(t)$ are defined on the interval $I$, it may be a silly little detail but I don't see why these functions are defined on that interval just from the fact $x_k(t)=xi+int_tau^t f(s,x_{k-1}(s))ds$, I would appreciate if someone could explaine me how to check this.
analysis ordinary-differential-equations
analysis ordinary-differential-equations
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
asked Jul 29 '14 at 20:16
user156441user156441
1,4561626
1,4561626
add a comment |
add a comment |
1 Answer
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You must add the condition that $f(t,x)$ is continuous on $Itimesmathbb{R}$.
$x_0(t)$ is constantt, so it is defined on $I$. The function $f(s,x_0(s))$ is defined and continuous on $I$, and hence, Riemann integrable on $I$. Then $x_1$ is defined and continuous on $I$. Induction shows that $x_k$ is defined on $I$ for all $k$.
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You must add the condition that $f(t,x)$ is continuous on $Itimesmathbb{R}$.
$x_0(t)$ is constantt, so it is defined on $I$. The function $f(s,x_0(s))$ is defined and continuous on $I$, and hence, Riemann integrable on $I$. Then $x_1$ is defined and continuous on $I$. Induction shows that $x_k$ is defined on $I$ for all $k$.
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
add a comment |
$begingroup$
You must add the condition that $f(t,x)$ is continuous on $Itimesmathbb{R}$.
$x_0(t)$ is constantt, so it is defined on $I$. The function $f(s,x_0(s))$ is defined and continuous on $I$, and hence, Riemann integrable on $I$. Then $x_1$ is defined and continuous on $I$. Induction shows that $x_k$ is defined on $I$ for all $k$.
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
add a comment |
$begingroup$
You must add the condition that $f(t,x)$ is continuous on $Itimesmathbb{R}$.
$x_0(t)$ is constantt, so it is defined on $I$. The function $f(s,x_0(s))$ is defined and continuous on $I$, and hence, Riemann integrable on $I$. Then $x_1$ is defined and continuous on $I$. Induction shows that $x_k$ is defined on $I$ for all $k$.
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
You must add the condition that $f(t,x)$ is continuous on $Itimesmathbb{R}$.
$x_0(t)$ is constantt, so it is defined on $I$. The function $f(s,x_0(s))$ is defined and continuous on $I$, and hence, Riemann integrable on $I$. Then $x_1$ is defined and continuous on $I$. Induction shows that $x_k$ is defined on $I$ for all $k$.
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
answered Jul 30 '14 at 9:51
Julián AguirreJulián Aguirre
69.2k24096
69.2k24096
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
add a comment |
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
I was wondering: are you sure you need continuity of $f$? In my textbook the hypothesis is that $f$ is Lipschitz on $x$ but not necessarily continuous. Using the Lipschitz condition, one can show that $f(s,x_0(s))=f(s,xi)$ is continuous, the rest follows exactly as you've said.
$endgroup$
– user156441
Aug 2 '14 at 1:16
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
$begingroup$
If $f$ is discontinuous how do you define the solution?
$endgroup$
– Julián Aguirre
Aug 2 '14 at 20:36
add a comment |
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