Stability region for two step Nyström method
$begingroup$
I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.
Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.
For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.
So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.
numerical-methods stability-theory
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
$endgroup$
add a comment |
$begingroup$
I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.
Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.
For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.
So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.
numerical-methods stability-theory
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
$endgroup$
add a comment |
$begingroup$
I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.
Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.
For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.
So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.
numerical-methods stability-theory
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
$endgroup$
I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2hcdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = Acdot w_k$ with $A=left(begin{matrix} 2hlambda & 1 \ 1 & 0end{matrix}right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=lambdacdot u = f(u)$ with $lambdainmathbb{C}$. To find the stability region I must set some limitations on the matrix A.
Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2hlambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.
For (Forward) Euler we have $u_{k+1}=u_k + hcdot f(u_k) = (1 + hlambda)u_k$ and so we require that $hlambda$ does not have a positive real part.
So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$.
numerical-methods stability-theory
numerical-methods stability-theory
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
edited Jan 15 at 21:17
The Coding Wombat
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
asked Jan 15 at 20:54
The Coding WombatThe Coding Wombat
280110
280110
add a comment |
add a comment |
1 Answer
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The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.
In conclusion, the method is nowhere stable.
Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.
Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
1
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
1
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
|
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$begingroup$
The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.
In conclusion, the method is nowhere stable.
Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.
Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
1
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
1
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
|
show 5 more comments
$begingroup$
The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.
In conclusion, the method is nowhere stable.
Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.
Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
1
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
1
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
|
show 5 more comments
$begingroup$
The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.
In conclusion, the method is nowhere stable.
Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.
Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
$endgroup$
The eigenvalues of $A$ are $hλpmsqrt{1+(hλ)^2}approx pmexp(pm hλ)$. The method is stable if for $Re(λ)<0$ the iteration converges to zero. As the product of the eigenvalues is always $-1$, you will always have one eigenvalue greater than $1$ in absolute value, and thus the component of the solution corresponding to it growing, not falling to $0$.
In conclusion, the method is nowhere stable.
Following Hairer/Wanner: "Solving ODE II", ch. V.1 "Stability of multi-step methods", the characteristic polynomials here are $rho(zeta)=zeta^2-1$ and $sigma(zeta)=2zeta$ with characteristic equation $rho(zeta)-musigma(zeta)=0$, where $mu=hλ$, and $zeta=e^{hλ}$ is the step factor of the exact solution.
Then the stability domain is defined as the set of all $mu$ where the characteristic equation only has solutions in the closed unit disk. But again, here the solutions have product $-1$, so they can not all be inside the unit disk, and both roots are on the boundary for $mu$ on the segment between $-i$ and $i$. This is called "weak instability" and causes the oscillating behavior (slowly increasing from floating point noise) shown in my previous answer resp. vol. I, chapter III.9 of the cited book.
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
edited Jan 15 at 21:33
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
answered Jan 15 at 21:02
LutzLLutzL
60.1k42057
60.1k42057
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
1
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
1
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
|
show 5 more comments
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
1
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
1
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
Are you assuming $lambda$ has to be a real number? It can be complex, I will update my question.
$endgroup$
– The Coding Wombat
Jan 15 at 21:16
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
If I impose conditions on the real part of $λ$, it obviously can be complex.
$endgroup$
– LutzL
Jan 15 at 21:21
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
$begingroup$
Okay, so you're saying one component of the solution isn't falling to 0, but the second component can stay at 1, because that would just mean that the second component equals 1 times the same term because of the 1 in the bottom left?
$endgroup$
– The Coding Wombat
Jan 15 at 21:40
1
1
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
$begingroup$
For $h$ small enough, $q=hλ+sqrt{1+(hλ)^2}=hλ+(1+frac12(hλ)^2-frac18(hλ)^4+...)=e^{hλ}+O((hλ)^3)$. For stability you need $|A^k|le 1$ for large $k$. The eigenvalues of $A$ tell you if that is the case.
$endgroup$
– LutzL
Jan 15 at 22:39
1
1
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
$begingroup$
Yes it can, for the differential equation $y'=0$, that is, for $λ=0$. Hopefully any implementation of a solver will return the constant solution in this case. Also for $y''=-y$ you get $λ=pm i$ so that for $h<1$ the points $pm ihpmsqrt{1-h^2}$ are on the unit circle. But that is the best you get.
$endgroup$
– LutzL
Jan 16 at 21:51
|
show 5 more comments
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