Mixing
Numerical methods for the matrix ODE $dot L(t)=[L, Pi_m(Fcirc L)]$
Suppose $L$ is a lower-triangular matrix whose row sum is $0$. I want to numerically solve the following matrix ODE
$$dot L(t) = [L, Pi_m (F circ L)], qquad L(0)=L_0$$
where
$[cdot ,,cdot]$ is the Lie bracket, i.e., $[A,B]=A B - B A$.
$L_0$ is lower-triangular and its row sum is also $0$.
$circ$ is the entry-wise product, i.e., $(A circ B)_{ij} = a_{ij} b_{ij}$.
$F_{ij}=-frac{f(u_i)-f(u_j)}{u_i-u_j} $ for $i ne j.$ $F_{ii}=-u(i)$.
$f(x)=frac{x^2}{2}$ and $u=(u_1, u_2, dots, u_n)$ is a sequence of some numbers, where $u_i<u_j.$ So, $F$ is a symmetric constant matrix.
$Pi_m(Fcirc L)=Fcirc L- mbox{diag}((Fcirc L)*e),$ where $e=(1 ~1~...1)^T, $ a column vector. So, the row sum of $Pi_m(Fcirc L)$ is also zero.
I want to solve $L(t)$ numerically such that $L(t)$ preserves the property that the row sum is $0$.
For convenience, let $Pi_m (F circ L)=B(L(t))$.
By the forward Euler method, we have the following iteration equation:
$L_{n+1}=L_n+ hdot L=L_n +h (L_nB_n-B_nL_n).$
It is easy to prove that the product of two matrices that their row sums are $0$ preserve the property.
So, the row sums of $B_nL_n$ and $L_nB_n$ are $0$. Since the row sum of $L_0$ is $0$, then we have the row sum of $L_n$ is $0$ for any $n ge 0$.
So, we can conclude that the forward Euler method preserves the property (row sum is zero).
However, I used the Matlab to implement the forward method, and the numerical result blew up for some initial condition. However, my professor told me that the analytic solution would not blow up because he proved it analytically.
My questions:
Do we have the convergence of $L_n$, i.e. $L_n$ converges to the exact solution $L(t)$?
Do we have other methods that converge to the exact solution and preserve the property if the forward Euler does not converge?
differential-equations numerical-methods matlab
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
asked Nov 1 '18 at 17:45
learner
14117
add a comment |
Suppose $L$ is a lower-triangular matrix whose row sum is $0$. I want to numerically solve the following matrix ODE
$$dot L(t) = [L, Pi_m (F circ L)], qquad L(0)=L_0$$
where
$[cdot ,,cdot]$ is the Lie bracket, i.e., $[A,B]=A B - B A$.
$L_0$ is lower-triangular and its row sum is also $0$.
$circ$ is the entry-wise product, i.e., $(A circ B)_{ij} = a_{ij} b_{ij}$.
$F_{ij}=-frac{f(u_i)-f(u_j)}{u_i-u_j} $ for $i ne j.$ $F_{ii}=-u(i)$.
$f(x)=frac{x^2}{2}$ and $u=(u_1, u_2, dots, u_n)$ is a sequence of some numbers, where $u_i<u_j.$ So, $F$ is a symmetric constant matrix.
$Pi_m(Fcirc L)=Fcirc L- mbox{diag}((Fcirc L)*e),$ where $e=(1 ~1~...1)^T, $ a column vector. So, the row sum of $Pi_m(Fcirc L)$ is also zero.
I want to solve $L(t)$ numerically such that $L(t)$ preserves the property that the row sum is $0$.
For convenience, let $Pi_m (F circ L)=B(L(t))$.
By the forward Euler method, we have the following iteration equation:
$L_{n+1}=L_n+ hdot L=L_n +h (L_nB_n-B_nL_n).$
It is easy to prove that the product of two matrices that their row sums are $0$ preserve the property.
So, the row sums of $B_nL_n$ and $L_nB_n$ are $0$. Since the row sum of $L_0$ is $0$, then we have the row sum of $L_n$ is $0$ for any $n ge 0$.
So, we can conclude that the forward Euler method preserves the property (row sum is zero).
However, I used the Matlab to implement the forward method, and the numerical result blew up for some initial condition. However, my professor told me that the analytic solution would not blow up because he proved it analytically.
My questions:
Do we have the convergence of $L_n$, i.e. $L_n$ converges to the exact solution $L(t)$?
Do we have other methods that converge to the exact solution and preserve the property if the forward Euler does not converge?
differential-equations numerical-methods matlab
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
asked Nov 1 '18 at 17:45
learner
14117
Are you maintaining all the conserved quantities? Like trace of L, other aspects of the characteristic polynomial?
– AHusain
Nov 16 '18 at 1:15
@AHusain No. I just need to preserve the property that the row sum is zero. Actually, the diagonal of $L$ is constant which is equal to the diagonal of the initial condition(matrix) because $L$ is the isospectral flow and the eigenvalues of isospectral flow remains unchanged.
– learner
Nov 16 '18 at 1:27
I was just hinting at making sure that your numeric method is isospectral as well. But it looks like you've already done that.
– AHusain
Nov 16 '18 at 1:34
@AHusain But I cannot show the convergence of Euler method for my problem.
– learner
Nov 16 '18 at 14:22
add a comment |
Suppose $L$ is a lower-triangular matrix whose row sum is $0$. I want to numerically solve the following matrix ODE
$$dot L(t) = [L, Pi_m (F circ L)], qquad L(0)=L_0$$
where
$[cdot ,,cdot]$ is the Lie bracket, i.e., $[A,B]=A B - B A$.
$L_0$ is lower-triangular and its row sum is also $0$.
$circ$ is the entry-wise product, i.e., $(A circ B)_{ij} = a_{ij} b_{ij}$.
$F_{ij}=-frac{f(u_i)-f(u_j)}{u_i-u_j} $ for $i ne j.$ $F_{ii}=-u(i)$.
$f(x)=frac{x^2}{2}$ and $u=(u_1, u_2, dots, u_n)$ is a sequence of some numbers, where $u_i<u_j.$ So, $F$ is a symmetric constant matrix.
$Pi_m(Fcirc L)=Fcirc L- mbox{diag}((Fcirc L)*e),$ where $e=(1 ~1~...1)^T, $ a column vector. So, the row sum of $Pi_m(Fcirc L)$ is also zero.
I want to solve $L(t)$ numerically such that $L(t)$ preserves the property that the row sum is $0$.
For convenience, let $Pi_m (F circ L)=B(L(t))$.
By the forward Euler method, we have the following iteration equation:
$L_{n+1}=L_n+ hdot L=L_n +h (L_nB_n-B_nL_n).$
It is easy to prove that the product of two matrices that their row sums are $0$ preserve the property.
So, the row sums of $B_nL_n$ and $L_nB_n$ are $0$. Since the row sum of $L_0$ is $0$, then we have the row sum of $L_n$ is $0$ for any $n ge 0$.
So, we can conclude that the forward Euler method preserves the property (row sum is zero).
However, I used the Matlab to implement the forward method, and the numerical result blew up for some initial condition. However, my professor told me that the analytic solution would not blow up because he proved it analytically.
My questions:
Do we have the convergence of $L_n$, i.e. $L_n$ converges to the exact solution $L(t)$?
Do we have other methods that converge to the exact solution and preserve the property if the forward Euler does not converge?
differential-equations numerical-methods matlab
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
asked Nov 1 '18 at 17:45
learner
14117
Suppose $L$ is a lower-triangular matrix whose row sum is $0$. I want to numerically solve the following matrix ODE
$$dot L(t) = [L, Pi_m (F circ L)], qquad L(0)=L_0$$
where
$[cdot ,,cdot]$ is the Lie bracket, i.e., $[A,B]=A B - B A$.
$L_0$ is lower-triangular and its row sum is also $0$.
$circ$ is the entry-wise product, i.e., $(A circ B)_{ij} = a_{ij} b_{ij}$.
$F_{ij}=-frac{f(u_i)-f(u_j)}{u_i-u_j} $ for $i ne j.$ $F_{ii}=-u(i)$.
$f(x)=frac{x^2}{2}$ and $u=(u_1, u_2, dots, u_n)$ is a sequence of some numbers, where $u_i<u_j.$ So, $F$ is a symmetric constant matrix.
$Pi_m(Fcirc L)=Fcirc L- mbox{diag}((Fcirc L)*e),$ where $e=(1 ~1~...1)^T, $ a column vector. So, the row sum of $Pi_m(Fcirc L)$ is also zero.
I want to solve $L(t)$ numerically such that $L(t)$ preserves the property that the row sum is $0$.
For convenience, let $Pi_m (F circ L)=B(L(t))$.
By the forward Euler method, we have the following iteration equation:
$L_{n+1}=L_n+ hdot L=L_n +h (L_nB_n-B_nL_n).$
It is easy to prove that the product of two matrices that their row sums are $0$ preserve the property.
So, the row sums of $B_nL_n$ and $L_nB_n$ are $0$. Since the row sum of $L_0$ is $0$, then we have the row sum of $L_n$ is $0$ for any $n ge 0$.
So, we can conclude that the forward Euler method preserves the property (row sum is zero).
However, I used the Matlab to implement the forward method, and the numerical result blew up for some initial condition. However, my professor told me that the analytic solution would not blow up because he proved it analytically.
My questions:
Do we have the convergence of $L_n$, i.e. $L_n$ converges to the exact solution $L(t)$?
Do we have other methods that converge to the exact solution and preserve the property if the forward Euler does not converge?
differential-equations numerical-methods matlab
differential-equations numerical-methods matlab
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
asked Nov 1 '18 at 17:45
learner
14117
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
asked Nov 1 '18 at 17:45
learner
14117
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
edited Nov 7 '18 at 9:55
Rodrigo de Azevedo
12.8k41855
12.8k41855
asked Nov 1 '18 at 17:45
learner
14117
asked Nov 1 '18 at 17:45
learner
14117
asked Nov 1 '18 at 17:45
learner
14117
14117
Are you maintaining all the conserved quantities? Like trace of L, other aspects of the characteristic polynomial?
– AHusain
Nov 16 '18 at 1:15
@AHusain No. I just need to preserve the property that the row sum is zero. Actually, the diagonal of $L$ is constant which is equal to the diagonal of the initial condition(matrix) because $L$ is the isospectral flow and the eigenvalues of isospectral flow remains unchanged.
– learner
Nov 16 '18 at 1:27
I was just hinting at making sure that your numeric method is isospectral as well. But it looks like you've already done that.
– AHusain
Nov 16 '18 at 1:34
@AHusain But I cannot show the convergence of Euler method for my problem.
– learner
Nov 16 '18 at 14:22
add a comment |
Are you maintaining all the conserved quantities? Like trace of L, other aspects of the characteristic polynomial?
– AHusain
Nov 16 '18 at 1:15
@AHusain No. I just need to preserve the property that the row sum is zero. Actually, the diagonal of $L$ is constant which is equal to the diagonal of the initial condition(matrix) because $L$ is the isospectral flow and the eigenvalues of isospectral flow remains unchanged.
– learner
Nov 16 '18 at 1:27
I was just hinting at making sure that your numeric method is isospectral as well. But it looks like you've already done that.
– AHusain
Nov 16 '18 at 1:34
@AHusain But I cannot show the convergence of Euler method for my problem.
– learner
Nov 16 '18 at 14:22
Are you maintaining all the conserved quantities? Like trace of L, other aspects of the characteristic polynomial?
– AHusain
Nov 16 '18 at 1:15
Are you maintaining all the conserved quantities? Like trace of L, other aspects of the characteristic polynomial?
– AHusain
Nov 16 '18 at 1:15
@AHusain No. I just need to preserve the property that the row sum is zero. Actually, the diagonal of $L$ is constant which is equal to the diagonal of the initial condition(matrix) because $L$ is the isospectral flow and the eigenvalues of isospectral flow remains unchanged.
– learner
Nov 16 '18 at 1:27
@AHusain No. I just need to preserve the property that the row sum is zero. Actually, the diagonal of $L$ is constant which is equal to the diagonal of the initial condition(matrix) because $L$ is the isospectral flow and the eigenvalues of isospectral flow remains unchanged.
– learner
Nov 16 '18 at 1:27
I was just hinting at making sure that your numeric method is isospectral as well. But it looks like you've already done that.
– AHusain
Nov 16 '18 at 1:34
I was just hinting at making sure that your numeric method is isospectral as well. But it looks like you've already done that.
– AHusain
Nov 16 '18 at 1:34
@AHusain But I cannot show the convergence of Euler method for my problem.
– learner
Nov 16 '18 at 14:22
@AHusain But I cannot show the convergence of Euler method for my problem.
– learner
Nov 16 '18 at 14:22
add a comment |
1 Answer
1
active
oldest
votes
In general, suppose that the function $f(t)$ be well approximated by its truncated Taylor expansion
in some interval, that wlog we can assume to be around $t=0$
$$
f(t) approx f(0) + f'(0),t + cdots + {1 over {q!}}f^{,left( q right)} (0),t^{,q} quad left| {;0 leftarrow t le T} right.
$$
We know that the approximation error can be expressed as
$$
R_{,q} (t) = {1 over {left( {q + 1} right)!}}f^{,left( {q + 1} right)} (tau ),t^{,q + 1}
< varepsilon _{,q} (T),t^{,q + 1} quad left| {;0 le tau le T} right.
$$
Your ODE is of the form
$$
f'(t), = gleft( {f(t)} right),
$$
where you know $g(x)$ and are going to estimate $f(t)$ by a 1st order approximation
$$
f(t_{,n} + Delta t) = f(t_{,n} ) + f'(t_{,n} )Delta t = f(t_{,n} ) + gleft( {f(t_{,n} )} right)Delta t
$$
which is a feed-back mechanism in cumulating the errors.
It is clear that the critical point is in particular the function $g(x)$, which is the feed-back transfer function.
No wonder that it might cause your attempt to blow away from the right solution.
This function varies on the domain of the expected values for $f(t)$ and since it depends on the sequence of the $u$ values
which are not specified in your post, I cannot be of further help.
Thus take care to analyze the expected influence of $g(x)$ on the errors, put bounds on it and determine
the necessary degree in the quadrature approximation, and of course in the number of significant digits used.
answered Nov 20 '18 at 23:20
G Cab
18k31237
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
votes
In general, suppose that the function $f(t)$ be well approximated by its truncated Taylor expansion
in some interval, that wlog we can assume to be around $t=0$
$$
f(t) approx f(0) + f'(0),t + cdots + {1 over {q!}}f^{,left( q right)} (0),t^{,q} quad left| {;0 leftarrow t le T} right.
$$
We know that the approximation error can be expressed as
$$
R_{,q} (t) = {1 over {left( {q + 1} right)!}}f^{,left( {q + 1} right)} (tau ),t^{,q + 1}
< varepsilon _{,q} (T),t^{,q + 1} quad left| {;0 le tau le T} right.
$$
Your ODE is of the form
$$
f'(t), = gleft( {f(t)} right),
$$
where you know $g(x)$ and are going to estimate $f(t)$ by a 1st order approximation
$$
f(t_{,n} + Delta t) = f(t_{,n} ) + f'(t_{,n} )Delta t = f(t_{,n} ) + gleft( {f(t_{,n} )} right)Delta t
$$
which is a feed-back mechanism in cumulating the errors.
It is clear that the critical point is in particular the function $g(x)$, which is the feed-back transfer function.
No wonder that it might cause your attempt to blow away from the right solution.
This function varies on the domain of the expected values for $f(t)$ and since it depends on the sequence of the $u$ values
which are not specified in your post, I cannot be of further help.
Thus take care to analyze the expected influence of $g(x)$ on the errors, put bounds on it and determine
the necessary degree in the quadrature approximation, and of course in the number of significant digits used.
answered Nov 20 '18 at 23:20
G Cab
18k31237
add a comment |
In general, suppose that the function $f(t)$ be well approximated by its truncated Taylor expansion
in some interval, that wlog we can assume to be around $t=0$
$$
f(t) approx f(0) + f'(0),t + cdots + {1 over {q!}}f^{,left( q right)} (0),t^{,q} quad left| {;0 leftarrow t le T} right.
$$
We know that the approximation error can be expressed as
$$
R_{,q} (t) = {1 over {left( {q + 1} right)!}}f^{,left( {q + 1} right)} (tau ),t^{,q + 1}
< varepsilon _{,q} (T),t^{,q + 1} quad left| {;0 le tau le T} right.
$$
Your ODE is of the form
$$
f'(t), = gleft( {f(t)} right),
$$
where you know $g(x)$ and are going to estimate $f(t)$ by a 1st order approximation
$$
f(t_{,n} + Delta t) = f(t_{,n} ) + f'(t_{,n} )Delta t = f(t_{,n} ) + gleft( {f(t_{,n} )} right)Delta t
$$
which is a feed-back mechanism in cumulating the errors.
It is clear that the critical point is in particular the function $g(x)$, which is the feed-back transfer function.
No wonder that it might cause your attempt to blow away from the right solution.
This function varies on the domain of the expected values for $f(t)$ and since it depends on the sequence of the $u$ values
which are not specified in your post, I cannot be of further help.
Thus take care to analyze the expected influence of $g(x)$ on the errors, put bounds on it and determine
the necessary degree in the quadrature approximation, and of course in the number of significant digits used.
answered Nov 20 '18 at 23:20
G Cab
18k31237
add a comment |
In general, suppose that the function $f(t)$ be well approximated by its truncated Taylor expansion
in some interval, that wlog we can assume to be around $t=0$
$$
f(t) approx f(0) + f'(0),t + cdots + {1 over {q!}}f^{,left( q right)} (0),t^{,q} quad left| {;0 leftarrow t le T} right.
$$
We know that the approximation error can be expressed as
$$
R_{,q} (t) = {1 over {left( {q + 1} right)!}}f^{,left( {q + 1} right)} (tau ),t^{,q + 1}
< varepsilon _{,q} (T),t^{,q + 1} quad left| {;0 le tau le T} right.
$$
Your ODE is of the form
$$
f'(t), = gleft( {f(t)} right),
$$
where you know $g(x)$ and are going to estimate $f(t)$ by a 1st order approximation
$$
f(t_{,n} + Delta t) = f(t_{,n} ) + f'(t_{,n} )Delta t = f(t_{,n} ) + gleft( {f(t_{,n} )} right)Delta t
$$
which is a feed-back mechanism in cumulating the errors.
It is clear that the critical point is in particular the function $g(x)$, which is the feed-back transfer function.
No wonder that it might cause your attempt to blow away from the right solution.
This function varies on the domain of the expected values for $f(t)$ and since it depends on the sequence of the $u$ values
which are not specified in your post, I cannot be of further help.
Thus take care to analyze the expected influence of $g(x)$ on the errors, put bounds on it and determine
the necessary degree in the quadrature approximation, and of course in the number of significant digits used.
answered Nov 20 '18 at 23:20
G Cab
18k31237
In general, suppose that the function $f(t)$ be well approximated by its truncated Taylor expansion
in some interval, that wlog we can assume to be around $t=0$
$$
f(t) approx f(0) + f'(0),t + cdots + {1 over {q!}}f^{,left( q right)} (0),t^{,q} quad left| {;0 leftarrow t le T} right.
$$
We know that the approximation error can be expressed as
$$
R_{,q} (t) = {1 over {left( {q + 1} right)!}}f^{,left( {q + 1} right)} (tau ),t^{,q + 1}
< varepsilon _{,q} (T),t^{,q + 1} quad left| {;0 le tau le T} right.
$$
Your ODE is of the form
$$
f'(t), = gleft( {f(t)} right),
$$
where you know $g(x)$ and are going to estimate $f(t)$ by a 1st order approximation
$$
f(t_{,n} + Delta t) = f(t_{,n} ) + f'(t_{,n} )Delta t = f(t_{,n} ) + gleft( {f(t_{,n} )} right)Delta t
$$
which is a feed-back mechanism in cumulating the errors.
It is clear that the critical point is in particular the function $g(x)$, which is the feed-back transfer function.
No wonder that it might cause your attempt to blow away from the right solution.
This function varies on the domain of the expected values for $f(t)$ and since it depends on the sequence of the $u$ values
which are not specified in your post, I cannot be of further help.
Thus take care to analyze the expected influence of $g(x)$ on the errors, put bounds on it and determine
the necessary degree in the quadrature approximation, and of course in the number of significant digits used.
answered Nov 20 '18 at 23:20
G Cab
18k31237
answered Nov 20 '18 at 23:20
G Cab
18k31237
answered Nov 20 '18 at 23:20
G Cab
18k31237
answered Nov 20 '18 at 23:20
G Cab
18k31237
18k31237
add a comment |
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
Are you maintaining all the conserved quantities? Like trace of L, other aspects of the characteristic polynomial?
– AHusain
Nov 16 '18 at 1:15
@AHusain No. I just need to preserve the property that the row sum is zero. Actually, the diagonal of $L$ is constant which is equal to the diagonal of the initial condition(matrix) because $L$ is the isospectral flow and the eigenvalues of isospectral flow remains unchanged.
– learner
Nov 16 '18 at 1:27
I was just hinting at making sure that your numeric method is isospectral as well. But it looks like you've already done that.
– AHusain
Nov 16 '18 at 1:34
@AHusain But I cannot show the convergence of Euler method for my problem.
– learner
Nov 16 '18 at 14:22