Mixing
Existence of Solution for Initial value Problem $y'(t)=cos(ty), t>0$ and $y(0)=frac{1}{n}, ninmathbb{N}$...
$begingroup$
Consider the cauchy problem
(1)begin{cases}
y'(t)=frac{1}{1+ty}, & t>0 \
y(0)=1+frac{1}{n} & ninmathbb{N}
end{cases}
(2)begin{cases}
y'(t)=cos(ty), & t>0 \
y(0)=frac{1}{n} & ninmathbb{N}
end{cases}
(a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.
(b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.
My attempt
(1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
begin{equation}
0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
end{equation}
which is the required solution.
(1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.
(2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$
functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
Consider the cauchy problem
(1)begin{cases}
y'(t)=frac{1}{1+ty}, & t>0 \
y(0)=1+frac{1}{n} & ninmathbb{N}
end{cases}
(2)begin{cases}
y'(t)=cos(ty), & t>0 \
y(0)=frac{1}{n} & ninmathbb{N}
end{cases}
(a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.
(b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.
My attempt
(1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
begin{equation}
0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
end{equation}
which is the required solution.
(1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.
(2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$
functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
Consider the cauchy problem
(1)begin{cases}
y'(t)=frac{1}{1+ty}, & t>0 \
y(0)=1+frac{1}{n} & ninmathbb{N}
end{cases}
(2)begin{cases}
y'(t)=cos(ty), & t>0 \
y(0)=frac{1}{n} & ninmathbb{N}
end{cases}
(a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.
(b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.
My attempt
(1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
begin{equation}
0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
end{equation}
which is the required solution.
(1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.
(2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$
functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
Consider the cauchy problem
(1)begin{cases}
y'(t)=frac{1}{1+ty}, & t>0 \
y(0)=1+frac{1}{n} & ninmathbb{N}
end{cases}
(2)begin{cases}
y'(t)=cos(ty), & t>0 \
y(0)=frac{1}{n} & ninmathbb{N}
end{cases}
(a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $mathbb{R}^+$.
(b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=cos(ty), t>0$ and $y(0)=0$.
My attempt
(1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, forall ninmathbb{N}$. Thus
begin{equation}
0<y_n(t)=y_n(0)+int_0^ty_n'(s)dsleq 2+t, t>0
end{equation}
which is the required solution.
(1b) Let $I=[0,1]$. Since $0leq y_n'(t)leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.
(2) My problem is here. Since $-1leq y'(t)leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$
functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli
functional-analysis ordinary-differential-equations initial-value-problems equicontinuity arzela-ascoli
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
asked Jan 29 at 12:16
Muhammad MubarakMuhammad Mubarak
9810
9810
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.
For two solutions $y,z$ you get for the difference the upper bounds
$$
|dot z(t)-dot y(t)|le t|z(t)-y(t)|
implies
|z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
$$
This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
$$
|y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
$$
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
(a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
which implies
begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
Thus $(y_n)_n$ is the required solution.
(b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).
Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.
For two solutions $y,z$ you get for the difference the upper bounds
$$
|dot z(t)-dot y(t)|le t|z(t)-y(t)|
implies
|z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
$$
This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
$$
|y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
$$
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.
For two solutions $y,z$ you get for the difference the upper bounds
$$
|dot z(t)-dot y(t)|le t|z(t)-y(t)|
implies
|z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
$$
This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
$$
|y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
$$
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.
For two solutions $y,z$ you get for the difference the upper bounds
$$
|dot z(t)-dot y(t)|le t|z(t)-y(t)|
implies
|z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
$$
This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
$$
|y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
$$
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
$endgroup$
The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.
For two solutions $y,z$ you get for the difference the upper bounds
$$
|dot z(t)-dot y(t)|le t|z(t)-y(t)|
implies
|z(t)-y(t)|le e^{t^2/2},|z(0)-y(0)|.
$$
This implies uniform convergence on any finite intervals, as for $tin[0,T]$ one has
$$
|y_n(t)-y_infty(t)|le frac1n, e^{T^2/2}
$$
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
edited Jan 29 at 13:46
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
answered Jan 29 at 13:09
LutzLLutzL
60.1k42057
60.1k42057
add a comment |
add a comment |
$begingroup$
(a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
which implies
begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
Thus $(y_n)_n$ is the required solution.
(b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).
Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
(a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
which implies
begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
Thus $(y_n)_n$ is the required solution.
(b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).
Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
add a comment |
$begingroup$
(a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
which implies
begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
Thus $(y_n)_n$ is the required solution.
(b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).
Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
$endgroup$
(a) Since $-1leq y'(t)leq 1, t>0$, we have that $-1leq y_n'(t)leq 1, t>0, ninmathbb{N}$. Thus
begin{equation} y_n(t)=frac{1}{n}+int_0^t y_n'(s)dsend{equation}
which implies
begin{equation}-tleq y_n(t)leq frac{1}{n}+tleq 1+t.end{equation}
Thus $(y_n)_n$ is the required solution.
(b) Let $I=[0,1]$. This implies $-1leq y_n(t)leq 2$ equivalently $|y_n(t)|leq 2$ (uniformly bounded).
Since $y_n'$ is continuous for each $ninmathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=lim frac{1}{n_k}=0$. Hence, the result.
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
edited Jan 30 at 10:37
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
answered Jan 29 at 13:19
Muhammad MubarakMuhammad Mubarak
9810
9810
add a comment |
add a comment |
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