what is the significance of $kappa$ in the proof of inconditional stability of the Crank-Nicolson scheme
$begingroup$
When using a centered difference approximation
$$
frac{partial}{partial t}u(t,x) = frac{u(t + Delta t/2,x) - u(t - Delta t/2, x)}{Delta t} + O((Delta t)^2)
$$
It is an approximation of the differential equation $dot{u} = ku''$ at the midpoint $(frac{t_i + t_{i+1}}{2}, x_j)$. The scheme is consistent with the error $ O((Delta t)^2) + O((Delta x)^2)$.
$$
frac{u_{i+1, j}-u_{i,j}}{kappa Delta t} = frac{u_{i,j-1} - 2u_{i,j} + u_{i,j + 1}}{2(Delta x)^2} + frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j + 1}}{2(Delta x)^2}
$$
The matrix notation leads to
$$
vec{u}_{i+1} - vec{u}_i = -frac{kappa Delta t}{2}(textbf{A}_n cdot vec{u}_{i+1} + textbf{A}_n cdot vec{u}_i)
$$
or
$$
Big(mathbb{I}_n + frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_{i+1} = Big(mathbb{I}_n - frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_i
$$
If the values $vec{u}_i$ at a given time are known we have to multiply the vector with a matrix and then solve a system of linear equations to determine the values $vec{u}_{i+1}$ at the next time level. We have an implicit method. We can use the eigenvalues and vectors of $textbf{A}_n$ to examine stability of the scheme. We are lead to examine the inequality
$$Bigg|frac{2 - kappa Delta t lambda_k}{2 + kappa Delta t lambda_k}Bigg|^2 < 1
$$
Since $lambda_k > 0$ we find that this scheme is $textbf{unconditionally stable.}$
My question
what is the significance of the $kappa$ in that proof? is it a condition number of a matrix? if yes which one?
pde finite-differences condition-number
edited Dec 3 '18 at 1:35
Mattos
2,73921321
asked Dec 2 '18 at 18:22
ecjbecjb
1618
$endgroup$
add a comment |
$begingroup$
When using a centered difference approximation
$$
frac{partial}{partial t}u(t,x) = frac{u(t + Delta t/2,x) - u(t - Delta t/2, x)}{Delta t} + O((Delta t)^2)
$$
It is an approximation of the differential equation $dot{u} = ku''$ at the midpoint $(frac{t_i + t_{i+1}}{2}, x_j)$. The scheme is consistent with the error $ O((Delta t)^2) + O((Delta x)^2)$.
$$
frac{u_{i+1, j}-u_{i,j}}{kappa Delta t} = frac{u_{i,j-1} - 2u_{i,j} + u_{i,j + 1}}{2(Delta x)^2} + frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j + 1}}{2(Delta x)^2}
$$
The matrix notation leads to
$$
vec{u}_{i+1} - vec{u}_i = -frac{kappa Delta t}{2}(textbf{A}_n cdot vec{u}_{i+1} + textbf{A}_n cdot vec{u}_i)
$$
or
$$
Big(mathbb{I}_n + frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_{i+1} = Big(mathbb{I}_n - frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_i
$$
If the values $vec{u}_i$ at a given time are known we have to multiply the vector with a matrix and then solve a system of linear equations to determine the values $vec{u}_{i+1}$ at the next time level. We have an implicit method. We can use the eigenvalues and vectors of $textbf{A}_n$ to examine stability of the scheme. We are lead to examine the inequality
$$Bigg|frac{2 - kappa Delta t lambda_k}{2 + kappa Delta t lambda_k}Bigg|^2 < 1
$$
Since $lambda_k > 0$ we find that this scheme is $textbf{unconditionally stable.}$
My question
what is the significance of the $kappa$ in that proof? is it a condition number of a matrix? if yes which one?
pde finite-differences condition-number
edited Dec 3 '18 at 1:35
Mattos
2,73921321
asked Dec 2 '18 at 18:22
ecjbecjb
1618
$endgroup$
1
$begingroup$
It is most probably that $kappa = k$ in the PDE $$dot{u} = color{red}{kappa}u''$$ and that the $1/(Delta x)^{2}$ term has been absorbed into the A matrix.
$endgroup$
– Mattos
Dec 3 '18 at 1:41
add a comment |
$begingroup$
When using a centered difference approximation
$$
frac{partial}{partial t}u(t,x) = frac{u(t + Delta t/2,x) - u(t - Delta t/2, x)}{Delta t} + O((Delta t)^2)
$$
It is an approximation of the differential equation $dot{u} = ku''$ at the midpoint $(frac{t_i + t_{i+1}}{2}, x_j)$. The scheme is consistent with the error $ O((Delta t)^2) + O((Delta x)^2)$.
$$
frac{u_{i+1, j}-u_{i,j}}{kappa Delta t} = frac{u_{i,j-1} - 2u_{i,j} + u_{i,j + 1}}{2(Delta x)^2} + frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j + 1}}{2(Delta x)^2}
$$
The matrix notation leads to
$$
vec{u}_{i+1} - vec{u}_i = -frac{kappa Delta t}{2}(textbf{A}_n cdot vec{u}_{i+1} + textbf{A}_n cdot vec{u}_i)
$$
or
$$
Big(mathbb{I}_n + frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_{i+1} = Big(mathbb{I}_n - frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_i
$$
If the values $vec{u}_i$ at a given time are known we have to multiply the vector with a matrix and then solve a system of linear equations to determine the values $vec{u}_{i+1}$ at the next time level. We have an implicit method. We can use the eigenvalues and vectors of $textbf{A}_n$ to examine stability of the scheme. We are lead to examine the inequality
$$Bigg|frac{2 - kappa Delta t lambda_k}{2 + kappa Delta t lambda_k}Bigg|^2 < 1
$$
Since $lambda_k > 0$ we find that this scheme is $textbf{unconditionally stable.}$
My question
what is the significance of the $kappa$ in that proof? is it a condition number of a matrix? if yes which one?
pde finite-differences condition-number
edited Dec 3 '18 at 1:35
Mattos
2,73921321
asked Dec 2 '18 at 18:22
ecjbecjb
1618
$endgroup$
When using a centered difference approximation
$$
frac{partial}{partial t}u(t,x) = frac{u(t + Delta t/2,x) - u(t - Delta t/2, x)}{Delta t} + O((Delta t)^2)
$$
It is an approximation of the differential equation $dot{u} = ku''$ at the midpoint $(frac{t_i + t_{i+1}}{2}, x_j)$. The scheme is consistent with the error $ O((Delta t)^2) + O((Delta x)^2)$.
$$
frac{u_{i+1, j}-u_{i,j}}{kappa Delta t} = frac{u_{i,j-1} - 2u_{i,j} + u_{i,j + 1}}{2(Delta x)^2} + frac{u_{i+1,j-1} - 2u_{i+1,j} + u_{i+1,j + 1}}{2(Delta x)^2}
$$
The matrix notation leads to
$$
vec{u}_{i+1} - vec{u}_i = -frac{kappa Delta t}{2}(textbf{A}_n cdot vec{u}_{i+1} + textbf{A}_n cdot vec{u}_i)
$$
or
$$
Big(mathbb{I}_n + frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_{i+1} = Big(mathbb{I}_n - frac{kappa Delta t}{2}textbf{A}_nBig) cdot vec{u}_i
$$
If the values $vec{u}_i$ at a given time are known we have to multiply the vector with a matrix and then solve a system of linear equations to determine the values $vec{u}_{i+1}$ at the next time level. We have an implicit method. We can use the eigenvalues and vectors of $textbf{A}_n$ to examine stability of the scheme. We are lead to examine the inequality
$$Bigg|frac{2 - kappa Delta t lambda_k}{2 + kappa Delta t lambda_k}Bigg|^2 < 1
$$
Since $lambda_k > 0$ we find that this scheme is $textbf{unconditionally stable.}$
My question
what is the significance of the $kappa$ in that proof? is it a condition number of a matrix? if yes which one?
pde finite-differences condition-number
pde finite-differences condition-number
edited Dec 3 '18 at 1:35
Mattos
2,73921321
asked Dec 2 '18 at 18:22
ecjbecjb
1618
edited Dec 3 '18 at 1:35
Mattos
2,73921321
asked Dec 2 '18 at 18:22
ecjbecjb
1618
edited Dec 3 '18 at 1:35
Mattos
2,73921321
edited Dec 3 '18 at 1:35
Mattos
2,73921321
edited Dec 3 '18 at 1:35
Mattos
2,73921321
2,73921321
asked Dec 2 '18 at 18:22
ecjbecjb
1618
asked Dec 2 '18 at 18:22
ecjbecjb
1618
asked Dec 2 '18 at 18:22
ecjbecjb
1618
1618
1
$begingroup$
It is most probably that $kappa = k$ in the PDE $$dot{u} = color{red}{kappa}u''$$ and that the $1/(Delta x)^{2}$ term has been absorbed into the A matrix.
$endgroup$
– Mattos
Dec 3 '18 at 1:41
add a comment |
1
$begingroup$
It is most probably that $kappa = k$ in the PDE $$dot{u} = color{red}{kappa}u''$$ and that the $1/(Delta x)^{2}$ term has been absorbed into the A matrix.
$endgroup$
– Mattos
Dec 3 '18 at 1:41
1
1
$begingroup$
It is most probably that $kappa = k$ in the PDE $$dot{u} = color{red}{kappa}u''$$ and that the $1/(Delta x)^{2}$ term has been absorbed into the A matrix.
$endgroup$
– Mattos
Dec 3 '18 at 1:41
$begingroup$
It is most probably that $kappa = k$ in the PDE $$dot{u} = color{red}{kappa}u''$$ and that the $1/(Delta x)^{2}$ term has been absorbed into the A matrix.
$endgroup$
– Mattos
Dec 3 '18 at 1:41
add a comment |
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$begingroup$
It is most probably that $kappa = k$ in the PDE $$dot{u} = color{red}{kappa}u''$$ and that the $1/(Delta x)^{2}$ term has been absorbed into the A matrix.
$endgroup$
– Mattos
Dec 3 '18 at 1:41