Relation of vector process to Heat Equation?
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The user Dinisaur conjectured that the following vector process could maybe be analyzed via a discrete version of the heat equation.
I am personally not aware of the heat equation and maybe someone here has a feeling how the result could look like.
Intuitively my process can be interpreted as follows. We have an $n$ node line graph and the temperature at time step $t$ can be interpreted for every inner node as the average over its two neighbors.
At the boundaries of the graph, we increase the temperature a little bit at every time step.
More formally:
I have n vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
It turns out that this procedure converges and in the end all vectors are the same. However, I cannot predict the vector to which the vectors converge.
Does anyone have an idea?
linear-algebra differential-equations convergence vectors heat-equation
asked Nov 26 at 18:35
Jannik
587
add a comment |
up vote
1
down vote
favorite
The user Dinisaur conjectured that the following vector process could maybe be analyzed via a discrete version of the heat equation.
I am personally not aware of the heat equation and maybe someone here has a feeling how the result could look like.
Intuitively my process can be interpreted as follows. We have an $n$ node line graph and the temperature at time step $t$ can be interpreted for every inner node as the average over its two neighbors.
At the boundaries of the graph, we increase the temperature a little bit at every time step.
More formally:
I have n vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
It turns out that this procedure converges and in the end all vectors are the same. However, I cannot predict the vector to which the vectors converge.
Does anyone have an idea?
linear-algebra differential-equations convergence vectors heat-equation
asked Nov 26 at 18:35
Jannik
587
I can partially explain why you can't predict the vector to which the procedure converges. If you consider this as a discrete dynamical system (very roughly speaking, something that takes $x$ as an input and produces output $bar{x}$ by some formula $bar{x} = F(x)$ with $F$ better be differentiable), for any unit vector $v$ your system has $p_v = (v_1, v_2 , dots, v_n) = (v, v, dots, v)$ as a fixed point: the point for which $p_v = F(p_v)$. There are points that arbitrarily close to each other on this set, hence none of them can be an attracting fixed point.
– Evgeny
Nov 27 at 8:51
Do you think with you intuition that it is impossible to predict the limit vector or only very difficult?
– Jannik
Nov 27 at 21:32
Let me say that I think it's difficult at this moment, but I wouldn't say impossible. Interesting fact: if you write $(bar{v}_1, dots, bar{v}_n) = f(v_1, dots, v_n)$, where $f$ is your mapping, you can see that for any rotation $R in O(n)$ holds $f(Rv_1, dots, Rv_n ) = (Rbar{v}_1, dots, Rbar{v}_n)$, i.e. $O(n)$ with action $R colon (x_1, dots, x_n) mapsto (Rx_1, dots, Rx_n)$ is a symmetry of this system. It might shed some light on this system, but I haven't found any interesting consequences of this yet.
– Evgeny
Nov 27 at 22:01
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The user Dinisaur conjectured that the following vector process could maybe be analyzed via a discrete version of the heat equation.
I am personally not aware of the heat equation and maybe someone here has a feeling how the result could look like.
Intuitively my process can be interpreted as follows. We have an $n$ node line graph and the temperature at time step $t$ can be interpreted for every inner node as the average over its two neighbors.
At the boundaries of the graph, we increase the temperature a little bit at every time step.
More formally:
I have n vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
It turns out that this procedure converges and in the end all vectors are the same. However, I cannot predict the vector to which the vectors converge.
Does anyone have an idea?
linear-algebra differential-equations convergence vectors heat-equation
asked Nov 26 at 18:35
Jannik
587
The user Dinisaur conjectured that the following vector process could maybe be analyzed via a discrete version of the heat equation.
I am personally not aware of the heat equation and maybe someone here has a feeling how the result could look like.
Intuitively my process can be interpreted as follows. We have an $n$ node line graph and the temperature at time step $t$ can be interpreted for every inner node as the average over its two neighbors.
At the boundaries of the graph, we increase the temperature a little bit at every time step.
More formally:
I have n vectors $v_1(t), dots, v_n(t)$.
Time is divided into discrete rounds.
Initially, all vectors have length $leq 1$.
The vectors at the next time step $t+1$ can be calculated as follows:
begin{align*}
v_1(t+1) &= frac{v_1(t)}{2|v_1(t)|} + frac{1}{2}v_2(t)\
v_i(t+1) &= frac{1}{2}v_{i-1}(t) + frac{1}{2}v_{i+1}(t) \
v_n(t+1) &= frac{1}{2}v_{n-1}(t) + frac{v_n(t)}{2|v_n(t)|}\
end{align*}
It turns out that this procedure converges and in the end all vectors are the same. However, I cannot predict the vector to which the vectors converge.
Does anyone have an idea?
linear-algebra differential-equations convergence vectors heat-equation
linear-algebra differential-equations convergence vectors heat-equation
asked Nov 26 at 18:35
Jannik
587
asked Nov 26 at 18:35
Jannik
587
asked Nov 26 at 18:35
Jannik
587
asked Nov 26 at 18:35
Jannik
587
asked Nov 26 at 18:35
Jannik
587
587
I can partially explain why you can't predict the vector to which the procedure converges. If you consider this as a discrete dynamical system (very roughly speaking, something that takes $x$ as an input and produces output $bar{x}$ by some formula $bar{x} = F(x)$ with $F$ better be differentiable), for any unit vector $v$ your system has $p_v = (v_1, v_2 , dots, v_n) = (v, v, dots, v)$ as a fixed point: the point for which $p_v = F(p_v)$. There are points that arbitrarily close to each other on this set, hence none of them can be an attracting fixed point.
– Evgeny
Nov 27 at 8:51
Do you think with you intuition that it is impossible to predict the limit vector or only very difficult?
– Jannik
Nov 27 at 21:32
Let me say that I think it's difficult at this moment, but I wouldn't say impossible. Interesting fact: if you write $(bar{v}_1, dots, bar{v}_n) = f(v_1, dots, v_n)$, where $f$ is your mapping, you can see that for any rotation $R in O(n)$ holds $f(Rv_1, dots, Rv_n ) = (Rbar{v}_1, dots, Rbar{v}_n)$, i.e. $O(n)$ with action $R colon (x_1, dots, x_n) mapsto (Rx_1, dots, Rx_n)$ is a symmetry of this system. It might shed some light on this system, but I haven't found any interesting consequences of this yet.
– Evgeny
Nov 27 at 22:01
add a comment |
I can partially explain why you can't predict the vector to which the procedure converges. If you consider this as a discrete dynamical system (very roughly speaking, something that takes $x$ as an input and produces output $bar{x}$ by some formula $bar{x} = F(x)$ with $F$ better be differentiable), for any unit vector $v$ your system has $p_v = (v_1, v_2 , dots, v_n) = (v, v, dots, v)$ as a fixed point: the point for which $p_v = F(p_v)$. There are points that arbitrarily close to each other on this set, hence none of them can be an attracting fixed point.
– Evgeny
Nov 27 at 8:51
Do you think with you intuition that it is impossible to predict the limit vector or only very difficult?
– Jannik
Nov 27 at 21:32
Let me say that I think it's difficult at this moment, but I wouldn't say impossible. Interesting fact: if you write $(bar{v}_1, dots, bar{v}_n) = f(v_1, dots, v_n)$, where $f$ is your mapping, you can see that for any rotation $R in O(n)$ holds $f(Rv_1, dots, Rv_n ) = (Rbar{v}_1, dots, Rbar{v}_n)$, i.e. $O(n)$ with action $R colon (x_1, dots, x_n) mapsto (Rx_1, dots, Rx_n)$ is a symmetry of this system. It might shed some light on this system, but I haven't found any interesting consequences of this yet.
– Evgeny
Nov 27 at 22:01
I can partially explain why you can't predict the vector to which the procedure converges. If you consider this as a discrete dynamical system (very roughly speaking, something that takes $x$ as an input and produces output $bar{x}$ by some formula $bar{x} = F(x)$ with $F$ better be differentiable), for any unit vector $v$ your system has $p_v = (v_1, v_2 , dots, v_n) = (v, v, dots, v)$ as a fixed point: the point for which $p_v = F(p_v)$. There are points that arbitrarily close to each other on this set, hence none of them can be an attracting fixed point.
– Evgeny
Nov 27 at 8:51
I can partially explain why you can't predict the vector to which the procedure converges. If you consider this as a discrete dynamical system (very roughly speaking, something that takes $x$ as an input and produces output $bar{x}$ by some formula $bar{x} = F(x)$ with $F$ better be differentiable), for any unit vector $v$ your system has $p_v = (v_1, v_2 , dots, v_n) = (v, v, dots, v)$ as a fixed point: the point for which $p_v = F(p_v)$. There are points that arbitrarily close to each other on this set, hence none of them can be an attracting fixed point.
– Evgeny
Nov 27 at 8:51
Do you think with you intuition that it is impossible to predict the limit vector or only very difficult?
– Jannik
Nov 27 at 21:32
Do you think with you intuition that it is impossible to predict the limit vector or only very difficult?
– Jannik
Nov 27 at 21:32
Let me say that I think it's difficult at this moment, but I wouldn't say impossible. Interesting fact: if you write $(bar{v}_1, dots, bar{v}_n) = f(v_1, dots, v_n)$, where $f$ is your mapping, you can see that for any rotation $R in O(n)$ holds $f(Rv_1, dots, Rv_n ) = (Rbar{v}_1, dots, Rbar{v}_n)$, i.e. $O(n)$ with action $R colon (x_1, dots, x_n) mapsto (Rx_1, dots, Rx_n)$ is a symmetry of this system. It might shed some light on this system, but I haven't found any interesting consequences of this yet.
– Evgeny
Nov 27 at 22:01
Let me say that I think it's difficult at this moment, but I wouldn't say impossible. Interesting fact: if you write $(bar{v}_1, dots, bar{v}_n) = f(v_1, dots, v_n)$, where $f$ is your mapping, you can see that for any rotation $R in O(n)$ holds $f(Rv_1, dots, Rv_n ) = (Rbar{v}_1, dots, Rbar{v}_n)$, i.e. $O(n)$ with action $R colon (x_1, dots, x_n) mapsto (Rx_1, dots, Rx_n)$ is a symmetry of this system. It might shed some light on this system, but I haven't found any interesting consequences of this yet.
– Evgeny
Nov 27 at 22:01
add a comment |
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I can partially explain why you can't predict the vector to which the procedure converges. If you consider this as a discrete dynamical system (very roughly speaking, something that takes $x$ as an input and produces output $bar{x}$ by some formula $bar{x} = F(x)$ with $F$ better be differentiable), for any unit vector $v$ your system has $p_v = (v_1, v_2 , dots, v_n) = (v, v, dots, v)$ as a fixed point: the point for which $p_v = F(p_v)$. There are points that arbitrarily close to each other on this set, hence none of them can be an attracting fixed point.
– Evgeny
Nov 27 at 8:51
Do you think with you intuition that it is impossible to predict the limit vector or only very difficult?
– Jannik
Nov 27 at 21:32
Let me say that I think it's difficult at this moment, but I wouldn't say impossible. Interesting fact: if you write $(bar{v}_1, dots, bar{v}_n) = f(v_1, dots, v_n)$, where $f$ is your mapping, you can see that for any rotation $R in O(n)$ holds $f(Rv_1, dots, Rv_n ) = (Rbar{v}_1, dots, Rbar{v}_n)$, i.e. $O(n)$ with action $R colon (x_1, dots, x_n) mapsto (Rx_1, dots, Rx_n)$ is a symmetry of this system. It might shed some light on this system, but I haven't found any interesting consequences of this yet.
– Evgeny
Nov 27 at 22:01