Prove finiteness ODE dynamics resulting from convex cost function and concave benefit function
$begingroup$
I have developed an ODE describing the dynamics of a state variable $x$.
However, the function is quite cumbersome. It results from a convex cost function and a concave benefit function and I want to say something about the dynamics. Stationary solutions are cumbersome as well but it should be possible to prove somehow that the function does not blow up to infinity or am I wrong?
I am happy about general or also more specific remarks.
It is hard to explain, maybe it gets clearer with some equations.
In equations, the problem results from the following:
begin{align}frac{text{d}y}{text{d}t}&=f(x,y,t)=y-frac{y}{1+alpha(1+frac{x}{1+x})y}-(1+x^2)beta yend{align}
Hence, I got a state variable $y$ and the dynamics change with respect to time depending on the value of $x$. The concave benefit of $x$ is in the denominator while the cost of $x$ is in the last term which is convex (blows up to infinity).
Now, the dynamics of $x$ change as well depending on wether $x$ has a beneficial impact or not. In particular, if $x$ is already large, the costs should have a higher impact than the benefit and $x$ should become smaller (so it is adaptive).
Hence, for the change of $x$ over time, I consider the partial derivative of $f$ with respect to $x$:
begin{align}frac{text{d}x}{text{d}t}&=frac{partial f/y}{partial x
}=frac{alpha y-2beta x(1+x+alpha y+2alpha xy)^2}{(1+x+alpha y+2alpha x y)^2}end{align}
Now, given this differential equation and knowing where it comes from, how can I prove that it does not blow up to infinity? It seems to be obvious but is there a nice theorem?
Note that I simplified the problem a bit. The real equations are more messy.
ordinary-differential-equations
asked Jan 15 at 9:55
MerlinMerlin
184
$endgroup$
|
show 1 more comment
$begingroup$
I have developed an ODE describing the dynamics of a state variable $x$.
However, the function is quite cumbersome. It results from a convex cost function and a concave benefit function and I want to say something about the dynamics. Stationary solutions are cumbersome as well but it should be possible to prove somehow that the function does not blow up to infinity or am I wrong?
I am happy about general or also more specific remarks.
It is hard to explain, maybe it gets clearer with some equations.
In equations, the problem results from the following:
begin{align}frac{text{d}y}{text{d}t}&=f(x,y,t)=y-frac{y}{1+alpha(1+frac{x}{1+x})y}-(1+x^2)beta yend{align}
Hence, I got a state variable $y$ and the dynamics change with respect to time depending on the value of $x$. The concave benefit of $x$ is in the denominator while the cost of $x$ is in the last term which is convex (blows up to infinity).
Now, the dynamics of $x$ change as well depending on wether $x$ has a beneficial impact or not. In particular, if $x$ is already large, the costs should have a higher impact than the benefit and $x$ should become smaller (so it is adaptive).
Hence, for the change of $x$ over time, I consider the partial derivative of $f$ with respect to $x$:
begin{align}frac{text{d}x}{text{d}t}&=frac{partial f/y}{partial x
}=frac{alpha y-2beta x(1+x+alpha y+2alpha xy)^2}{(1+x+alpha y+2alpha x y)^2}end{align}
Now, given this differential equation and knowing where it comes from, how can I prove that it does not blow up to infinity? It seems to be obvious but is there a nice theorem?
Note that I simplified the problem a bit. The real equations are more messy.
ordinary-differential-equations
asked Jan 15 at 9:55
MerlinMerlin
184
$endgroup$
$begingroup$
Can you please write in a one place a full system of equation that you are interested in? Also, do you have any numerical simulations?
$endgroup$
– Evgeny
Jan 15 at 10:11
$begingroup$
I am interested in the two equations which are both explicitly shown now.
$endgroup$
– Merlin
Jan 15 at 10:20
$begingroup$
Try simulations first. Pick many initial conditions for different $alpha$ and $beta$ and see whether trajectories go to infinity. It's pretty easy to automate, although it requires some time for checking. Meanwhile, try to check how vector field behaves on lines $x = mathrm{const}$ and $y = mathrm{const}$. If you are able to show that there exist some $M^ast$ and $K^ast$ such that vector field points to the inside of a rectangle $0 leqslant x leqslant M$, $0 leqslant y leqslant K$ for any $M geqslant M^ast$ and $K geqslant K^ast$ - you are done, there is no blow-up.
$endgroup$
– Evgeny
Jan 15 at 14:56
$begingroup$
OK, thank you for this approach! I will try to do this. However, I would still be thankful for any advice regarding a theorem dealing which such a problem :)
$endgroup$
– Merlin
Jan 17 at 8:47
$begingroup$
I don't know such theorem and I am slightly doubtful that there is one theorem that encompasses all cases. There are bunch of cases though. For example, you might had a Lyapunov-like function: a function $mathcal{F}(x, y)$ which is known to be non-increasing along trajectories and which has compact level-sets. Depending on level sets structure this would mean no trajectory can escape to infinity. My suggestion in previous comment is a bit similar to this idea (non-increasing of Lyapunov-like function geometrically means that vector field points in one direction w.r.t. level sets). ...
$endgroup$
– Evgeny
Jan 17 at 8:56
|
show 1 more comment
$begingroup$
I have developed an ODE describing the dynamics of a state variable $x$.
However, the function is quite cumbersome. It results from a convex cost function and a concave benefit function and I want to say something about the dynamics. Stationary solutions are cumbersome as well but it should be possible to prove somehow that the function does not blow up to infinity or am I wrong?
I am happy about general or also more specific remarks.
It is hard to explain, maybe it gets clearer with some equations.
In equations, the problem results from the following:
begin{align}frac{text{d}y}{text{d}t}&=f(x,y,t)=y-frac{y}{1+alpha(1+frac{x}{1+x})y}-(1+x^2)beta yend{align}
Hence, I got a state variable $y$ and the dynamics change with respect to time depending on the value of $x$. The concave benefit of $x$ is in the denominator while the cost of $x$ is in the last term which is convex (blows up to infinity).
Now, the dynamics of $x$ change as well depending on wether $x$ has a beneficial impact or not. In particular, if $x$ is already large, the costs should have a higher impact than the benefit and $x$ should become smaller (so it is adaptive).
Hence, for the change of $x$ over time, I consider the partial derivative of $f$ with respect to $x$:
begin{align}frac{text{d}x}{text{d}t}&=frac{partial f/y}{partial x
}=frac{alpha y-2beta x(1+x+alpha y+2alpha xy)^2}{(1+x+alpha y+2alpha x y)^2}end{align}
Now, given this differential equation and knowing where it comes from, how can I prove that it does not blow up to infinity? It seems to be obvious but is there a nice theorem?
Note that I simplified the problem a bit. The real equations are more messy.
ordinary-differential-equations
asked Jan 15 at 9:55
MerlinMerlin
184
$endgroup$
I have developed an ODE describing the dynamics of a state variable $x$.
However, the function is quite cumbersome. It results from a convex cost function and a concave benefit function and I want to say something about the dynamics. Stationary solutions are cumbersome as well but it should be possible to prove somehow that the function does not blow up to infinity or am I wrong?
I am happy about general or also more specific remarks.
It is hard to explain, maybe it gets clearer with some equations.
In equations, the problem results from the following:
begin{align}frac{text{d}y}{text{d}t}&=f(x,y,t)=y-frac{y}{1+alpha(1+frac{x}{1+x})y}-(1+x^2)beta yend{align}
Hence, I got a state variable $y$ and the dynamics change with respect to time depending on the value of $x$. The concave benefit of $x$ is in the denominator while the cost of $x$ is in the last term which is convex (blows up to infinity).
Now, the dynamics of $x$ change as well depending on wether $x$ has a beneficial impact or not. In particular, if $x$ is already large, the costs should have a higher impact than the benefit and $x$ should become smaller (so it is adaptive).
Hence, for the change of $x$ over time, I consider the partial derivative of $f$ with respect to $x$:
begin{align}frac{text{d}x}{text{d}t}&=frac{partial f/y}{partial x
}=frac{alpha y-2beta x(1+x+alpha y+2alpha xy)^2}{(1+x+alpha y+2alpha x y)^2}end{align}
Now, given this differential equation and knowing where it comes from, how can I prove that it does not blow up to infinity? It seems to be obvious but is there a nice theorem?
Note that I simplified the problem a bit. The real equations are more messy.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 15 at 9:55
MerlinMerlin
184
asked Jan 15 at 9:55
MerlinMerlin
184
edited Jan 15 at 10:20
Merlin
asked Jan 15 at 9:55
MerlinMerlin
184
asked Jan 15 at 9:55
MerlinMerlin
184
asked Jan 15 at 9:55
MerlinMerlin
184
184
$begingroup$
Can you please write in a one place a full system of equation that you are interested in? Also, do you have any numerical simulations?
$endgroup$
– Evgeny
Jan 15 at 10:11
$begingroup$
I am interested in the two equations which are both explicitly shown now.
$endgroup$
– Merlin
Jan 15 at 10:20
$begingroup$
Try simulations first. Pick many initial conditions for different $alpha$ and $beta$ and see whether trajectories go to infinity. It's pretty easy to automate, although it requires some time for checking. Meanwhile, try to check how vector field behaves on lines $x = mathrm{const}$ and $y = mathrm{const}$. If you are able to show that there exist some $M^ast$ and $K^ast$ such that vector field points to the inside of a rectangle $0 leqslant x leqslant M$, $0 leqslant y leqslant K$ for any $M geqslant M^ast$ and $K geqslant K^ast$ - you are done, there is no blow-up.
$endgroup$
– Evgeny
Jan 15 at 14:56
$begingroup$
OK, thank you for this approach! I will try to do this. However, I would still be thankful for any advice regarding a theorem dealing which such a problem :)
$endgroup$
– Merlin
Jan 17 at 8:47
$begingroup$
I don't know such theorem and I am slightly doubtful that there is one theorem that encompasses all cases. There are bunch of cases though. For example, you might had a Lyapunov-like function: a function $mathcal{F}(x, y)$ which is known to be non-increasing along trajectories and which has compact level-sets. Depending on level sets structure this would mean no trajectory can escape to infinity. My suggestion in previous comment is a bit similar to this idea (non-increasing of Lyapunov-like function geometrically means that vector field points in one direction w.r.t. level sets). ...
$endgroup$
– Evgeny
Jan 17 at 8:56
|
show 1 more comment
$begingroup$
Can you please write in a one place a full system of equation that you are interested in? Also, do you have any numerical simulations?
$endgroup$
– Evgeny
Jan 15 at 10:11
$begingroup$
I am interested in the two equations which are both explicitly shown now.
$endgroup$
– Merlin
Jan 15 at 10:20
$begingroup$
Try simulations first. Pick many initial conditions for different $alpha$ and $beta$ and see whether trajectories go to infinity. It's pretty easy to automate, although it requires some time for checking. Meanwhile, try to check how vector field behaves on lines $x = mathrm{const}$ and $y = mathrm{const}$. If you are able to show that there exist some $M^ast$ and $K^ast$ such that vector field points to the inside of a rectangle $0 leqslant x leqslant M$, $0 leqslant y leqslant K$ for any $M geqslant M^ast$ and $K geqslant K^ast$ - you are done, there is no blow-up.
$endgroup$
– Evgeny
Jan 15 at 14:56
$begingroup$
OK, thank you for this approach! I will try to do this. However, I would still be thankful for any advice regarding a theorem dealing which such a problem :)
$endgroup$
– Merlin
Jan 17 at 8:47
$begingroup$
I don't know such theorem and I am slightly doubtful that there is one theorem that encompasses all cases. There are bunch of cases though. For example, you might had a Lyapunov-like function: a function $mathcal{F}(x, y)$ which is known to be non-increasing along trajectories and which has compact level-sets. Depending on level sets structure this would mean no trajectory can escape to infinity. My suggestion in previous comment is a bit similar to this idea (non-increasing of Lyapunov-like function geometrically means that vector field points in one direction w.r.t. level sets). ...
$endgroup$
– Evgeny
Jan 17 at 8:56
$begingroup$
Can you please write in a one place a full system of equation that you are interested in? Also, do you have any numerical simulations?
$endgroup$
– Evgeny
Jan 15 at 10:11
$begingroup$
Can you please write in a one place a full system of equation that you are interested in? Also, do you have any numerical simulations?
$endgroup$
– Evgeny
Jan 15 at 10:11
$begingroup$
I am interested in the two equations which are both explicitly shown now.
$endgroup$
– Merlin
Jan 15 at 10:20
$begingroup$
I am interested in the two equations which are both explicitly shown now.
$endgroup$
– Merlin
Jan 15 at 10:20
$begingroup$
Try simulations first. Pick many initial conditions for different $alpha$ and $beta$ and see whether trajectories go to infinity. It's pretty easy to automate, although it requires some time for checking. Meanwhile, try to check how vector field behaves on lines $x = mathrm{const}$ and $y = mathrm{const}$. If you are able to show that there exist some $M^ast$ and $K^ast$ such that vector field points to the inside of a rectangle $0 leqslant x leqslant M$, $0 leqslant y leqslant K$ for any $M geqslant M^ast$ and $K geqslant K^ast$ - you are done, there is no blow-up.
$endgroup$
– Evgeny
Jan 15 at 14:56
$begingroup$
Try simulations first. Pick many initial conditions for different $alpha$ and $beta$ and see whether trajectories go to infinity. It's pretty easy to automate, although it requires some time for checking. Meanwhile, try to check how vector field behaves on lines $x = mathrm{const}$ and $y = mathrm{const}$. If you are able to show that there exist some $M^ast$ and $K^ast$ such that vector field points to the inside of a rectangle $0 leqslant x leqslant M$, $0 leqslant y leqslant K$ for any $M geqslant M^ast$ and $K geqslant K^ast$ - you are done, there is no blow-up.
$endgroup$
– Evgeny
Jan 15 at 14:56
$begingroup$
OK, thank you for this approach! I will try to do this. However, I would still be thankful for any advice regarding a theorem dealing which such a problem :)
$endgroup$
– Merlin
Jan 17 at 8:47
$begingroup$
OK, thank you for this approach! I will try to do this. However, I would still be thankful for any advice regarding a theorem dealing which such a problem :)
$endgroup$
– Merlin
Jan 17 at 8:47
$begingroup$
I don't know such theorem and I am slightly doubtful that there is one theorem that encompasses all cases. There are bunch of cases though. For example, you might had a Lyapunov-like function: a function $mathcal{F}(x, y)$ which is known to be non-increasing along trajectories and which has compact level-sets. Depending on level sets structure this would mean no trajectory can escape to infinity. My suggestion in previous comment is a bit similar to this idea (non-increasing of Lyapunov-like function geometrically means that vector field points in one direction w.r.t. level sets). ...
$endgroup$
– Evgeny
Jan 17 at 8:56
$begingroup$
I don't know such theorem and I am slightly doubtful that there is one theorem that encompasses all cases. There are bunch of cases though. For example, you might had a Lyapunov-like function: a function $mathcal{F}(x, y)$ which is known to be non-increasing along trajectories and which has compact level-sets. Depending on level sets structure this would mean no trajectory can escape to infinity. My suggestion in previous comment is a bit similar to this idea (non-increasing of Lyapunov-like function geometrically means that vector field points in one direction w.r.t. level sets). ...
$endgroup$
– Evgeny
Jan 17 at 8:56
|
show 1 more comment
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$begingroup$
Can you please write in a one place a full system of equation that you are interested in? Also, do you have any numerical simulations?
$endgroup$
– Evgeny
Jan 15 at 10:11
$begingroup$
I am interested in the two equations which are both explicitly shown now.
$endgroup$
– Merlin
Jan 15 at 10:20
$begingroup$
Try simulations first. Pick many initial conditions for different $alpha$ and $beta$ and see whether trajectories go to infinity. It's pretty easy to automate, although it requires some time for checking. Meanwhile, try to check how vector field behaves on lines $x = mathrm{const}$ and $y = mathrm{const}$. If you are able to show that there exist some $M^ast$ and $K^ast$ such that vector field points to the inside of a rectangle $0 leqslant x leqslant M$, $0 leqslant y leqslant K$ for any $M geqslant M^ast$ and $K geqslant K^ast$ - you are done, there is no blow-up.
$endgroup$
– Evgeny
Jan 15 at 14:56
$begingroup$
OK, thank you for this approach! I will try to do this. However, I would still be thankful for any advice regarding a theorem dealing which such a problem :)
$endgroup$
– Merlin
Jan 17 at 8:47
$begingroup$
I don't know such theorem and I am slightly doubtful that there is one theorem that encompasses all cases. There are bunch of cases though. For example, you might had a Lyapunov-like function: a function $mathcal{F}(x, y)$ which is known to be non-increasing along trajectories and which has compact level-sets. Depending on level sets structure this would mean no trajectory can escape to infinity. My suggestion in previous comment is a bit similar to this idea (non-increasing of Lyapunov-like function geometrically means that vector field points in one direction w.r.t. level sets). ...
$endgroup$
– Evgeny
Jan 17 at 8:56