Picard-Lindelöf Theorem application
$begingroup$
I'm really lost here guys. I'd appreciate it if you could help.
Consider the differential equation
$$
frac{dy}{dt} = f(t, y) tag{1}
$$
with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?
ordinary-differential-equations initial-value-problems
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
$endgroup$
add a comment |
$begingroup$
I'm really lost here guys. I'd appreciate it if you could help.
Consider the differential equation
$$
frac{dy}{dt} = f(t, y) tag{1}
$$
with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?
ordinary-differential-equations initial-value-problems
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
$endgroup$
add a comment |
$begingroup$
I'm really lost here guys. I'd appreciate it if you could help.
Consider the differential equation
$$
frac{dy}{dt} = f(t, y) tag{1}
$$
with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?
ordinary-differential-equations initial-value-problems
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
$endgroup$
I'm really lost here guys. I'd appreciate it if you could help.
Consider the differential equation
$$
frac{dy}{dt} = f(t, y) tag{1}
$$
with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?
ordinary-differential-equations initial-value-problems
ordinary-differential-equations initial-value-problems
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
edited Dec 7 '18 at 19:43
user618968
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
asked Jul 6 '16 at 21:09
Da MikeDa Mike
527
527
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
So, for anyone interested, after a couple of days and with some indications, I solved my question.
Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.
Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem
$$
frac{dy}{dt} = f(t, y) , y(t_0) = 1
$$
has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
So, for anyone interested, after a couple of days and with some indications, I solved my question.
Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.
Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem
$$
frac{dy}{dt} = f(t, y) , y(t_0) = 1
$$
has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
$endgroup$
add a comment |
$begingroup$
So, for anyone interested, after a couple of days and with some indications, I solved my question.
Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.
Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem
$$
frac{dy}{dt} = f(t, y) , y(t_0) = 1
$$
has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
$endgroup$
add a comment |
$begingroup$
So, for anyone interested, after a couple of days and with some indications, I solved my question.
Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.
Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem
$$
frac{dy}{dt} = f(t, y) , y(t_0) = 1
$$
has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
$endgroup$
So, for anyone interested, after a couple of days and with some indications, I solved my question.
Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.
Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem
$$
frac{dy}{dt} = f(t, y) , y(t_0) = 1
$$
has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
answered Jul 9 '16 at 11:54
Da MikeDa Mike
527
527
add a comment |
add a comment |
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