Stability of solutions to $y''+4=0$
$begingroup$
Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.
Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.
Here is my attempt at showing the $phi(t)$ is a stable solution.
Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.
This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.
A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$
ordinary-differential-equations
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
$endgroup$
add a comment |
$begingroup$
Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.
Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.
Here is my attempt at showing the $phi(t)$ is a stable solution.
Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.
This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.
A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$
ordinary-differential-equations
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
$endgroup$
2
$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55
add a comment |
$begingroup$
Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.
Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.
Here is my attempt at showing the $phi(t)$ is a stable solution.
Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.
This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.
A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$
ordinary-differential-equations
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
$endgroup$
Consider the system $y''+4y=0$. Here, the solutions are of the form $y(t)=C_1cos(2t)+C_2sin(2t)$. The solution in question is $phi(t)=cos(2t)$.
Now the definition of stability I work with states that given $epsilon>0$, $exists delta>0$ such that if $|y(o)-phi(0)|<delta$, then $|y(t)-phi(t)|<epsilon$ for all $tgeq 0$ iff the solution $phi(t)$ is stable.
Here is my attempt at showing the $phi(t)$ is a stable solution.
Now, the problem for me here is that there are two unknown constants in the general solution, $C_1$ and $C_2$, and when I take the difference $|y(0)-phi(0)|=|(C_1-1)cos(2t)|$, which is indeed $leq|(C_1-1)|$ which I can make small. But $C_2$ could still be large as far as I know.
This problem can be overcome by converting into a first order system. Given initial data $phi(o)=begin{bmatrix}eta_1 \ eta_2end{bmatrix}$, I have computed a solution $phi(t)=begin{bmatrix}(-2eta_1)cos(2t)+i(2eta_2)sin(2t) \ (-2eta_2)cos(2t)+i(2eta_1)sin(2t)end{bmatrix}$. I have run into trouble now converting the solution in question, $phi(t)=cos(2t)$ into vector form.
A similar question I have is showing that the solution $y(t)=0$ is a stable solution to the equation $y''+4y'+4=0$, where I compute a general solution to be $phi(t)=C_1e^{-2t}+C_2te^{-2t}$
ordinary-differential-equations
ordinary-differential-equations
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
edited Dec 8 '18 at 0:59
Jungleshrimp
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
asked Dec 8 '18 at 0:53
JungleshrimpJungleshrimp
314111
314111
2
$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55
add a comment |
2
$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55
2
2
$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55
$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55
add a comment |
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$begingroup$
In second order ODEs there are two initial conditions, for the value and the first derivative at an initial time. In your case, $phi(0)=1$ and $phi'(0)=0$. So, you should consider $y(t)$ such that $y(0)$ is close to $1$ and $y'(0)$ is close to $0$.
$endgroup$
– user539887
Dec 8 '18 at 12:55