Iteration step of the Crank–Nicolson scheme
$begingroup$
For many dynamic problems a masse matrix $M$ has to be taken into account too. Consider a discretized systems of the form
$$
textbf{M}frac{d}{dt}vec{u}(t) = -textbf{A} cdot vec{u}(t) + vec{f}(t)
$$
Often linear systesm of equation with the matrix $M$ are easily solved, e.g., $M$ might be a diagonal matrix with positive entries. The generalized eigenvalues $lambda$ and eigenvectors $vec{v}$ are nonzero solutions of
$$
textbf{A} cdot vec{v} = lambda textbf{M} cdot vec{v}
$$
The Crank-Nicolson scheme will lead to
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) +frac{1}{2}(f(t) + f(t + Delta t)) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t) + frac{Delta t}{2}(f(t) + f(t + Delta t))
end{align}
Question:
Examine the discretization of a general parabolic problem, i.e. examine
$$
frac{d}{dt}textbf{M} cdot vec{u}(t) = -textbf{A}cdot vec{u}(t)
$$
Assume that both matrices M and A are symmetric and positive definite. The generalized eigenvalues $lambda_k$ are assumed to be positive. Use $textbf{A}vec{v}_k = lambda_k textbf{M}vec{v}_k$
Verify that the Crank-Nicolson scheme leads to an iteration step
$$
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
$$
the answer is:
Use a centered time derivative at $t + frac{Delta t}{2}$
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
end{align}
Here is what I don't understand:
It seems that we use the Crank Nicolson scheme and simply drop the expression $+frac{1}{2}(f(t) + f(t + Delta t)) $. What is the reasoning behind it?
pde eigenvalues-eigenvectors finite-element-method
$endgroup$
add a comment |
$begingroup$
For many dynamic problems a masse matrix $M$ has to be taken into account too. Consider a discretized systems of the form
$$
textbf{M}frac{d}{dt}vec{u}(t) = -textbf{A} cdot vec{u}(t) + vec{f}(t)
$$
Often linear systesm of equation with the matrix $M$ are easily solved, e.g., $M$ might be a diagonal matrix with positive entries. The generalized eigenvalues $lambda$ and eigenvectors $vec{v}$ are nonzero solutions of
$$
textbf{A} cdot vec{v} = lambda textbf{M} cdot vec{v}
$$
The Crank-Nicolson scheme will lead to
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) +frac{1}{2}(f(t) + f(t + Delta t)) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t) + frac{Delta t}{2}(f(t) + f(t + Delta t))
end{align}
Question:
Examine the discretization of a general parabolic problem, i.e. examine
$$
frac{d}{dt}textbf{M} cdot vec{u}(t) = -textbf{A}cdot vec{u}(t)
$$
Assume that both matrices M and A are symmetric and positive definite. The generalized eigenvalues $lambda_k$ are assumed to be positive. Use $textbf{A}vec{v}_k = lambda_k textbf{M}vec{v}_k$
Verify that the Crank-Nicolson scheme leads to an iteration step
$$
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
$$
the answer is:
Use a centered time derivative at $t + frac{Delta t}{2}$
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
end{align}
Here is what I don't understand:
It seems that we use the Crank Nicolson scheme and simply drop the expression $+frac{1}{2}(f(t) + f(t + Delta t)) $. What is the reasoning behind it?
pde eigenvalues-eigenvectors finite-element-method
$endgroup$
$begingroup$
The question is asking you to examine $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u$$ not $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u + f$$
$endgroup$
– Mattos
Dec 3 '18 at 1:17
add a comment |
$begingroup$
For many dynamic problems a masse matrix $M$ has to be taken into account too. Consider a discretized systems of the form
$$
textbf{M}frac{d}{dt}vec{u}(t) = -textbf{A} cdot vec{u}(t) + vec{f}(t)
$$
Often linear systesm of equation with the matrix $M$ are easily solved, e.g., $M$ might be a diagonal matrix with positive entries. The generalized eigenvalues $lambda$ and eigenvectors $vec{v}$ are nonzero solutions of
$$
textbf{A} cdot vec{v} = lambda textbf{M} cdot vec{v}
$$
The Crank-Nicolson scheme will lead to
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) +frac{1}{2}(f(t) + f(t + Delta t)) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t) + frac{Delta t}{2}(f(t) + f(t + Delta t))
end{align}
Question:
Examine the discretization of a general parabolic problem, i.e. examine
$$
frac{d}{dt}textbf{M} cdot vec{u}(t) = -textbf{A}cdot vec{u}(t)
$$
Assume that both matrices M and A are symmetric and positive definite. The generalized eigenvalues $lambda_k$ are assumed to be positive. Use $textbf{A}vec{v}_k = lambda_k textbf{M}vec{v}_k$
Verify that the Crank-Nicolson scheme leads to an iteration step
$$
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
$$
the answer is:
Use a centered time derivative at $t + frac{Delta t}{2}$
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
end{align}
Here is what I don't understand:
It seems that we use the Crank Nicolson scheme and simply drop the expression $+frac{1}{2}(f(t) + f(t + Delta t)) $. What is the reasoning behind it?
pde eigenvalues-eigenvectors finite-element-method
$endgroup$
For many dynamic problems a masse matrix $M$ has to be taken into account too. Consider a discretized systems of the form
$$
textbf{M}frac{d}{dt}vec{u}(t) = -textbf{A} cdot vec{u}(t) + vec{f}(t)
$$
Often linear systesm of equation with the matrix $M$ are easily solved, e.g., $M$ might be a diagonal matrix with positive entries. The generalized eigenvalues $lambda$ and eigenvectors $vec{v}$ are nonzero solutions of
$$
textbf{A} cdot vec{v} = lambda textbf{M} cdot vec{v}
$$
The Crank-Nicolson scheme will lead to
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) +frac{1}{2}(f(t) + f(t + Delta t)) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t) + frac{Delta t}{2}(f(t) + f(t + Delta t))
end{align}
Question:
Examine the discretization of a general parabolic problem, i.e. examine
$$
frac{d}{dt}textbf{M} cdot vec{u}(t) = -textbf{A}cdot vec{u}(t)
$$
Assume that both matrices M and A are symmetric and positive definite. The generalized eigenvalues $lambda_k$ are assumed to be positive. Use $textbf{A}vec{v}_k = lambda_k textbf{M}vec{v}_k$
Verify that the Crank-Nicolson scheme leads to an iteration step
$$
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
$$
the answer is:
Use a centered time derivative at $t + frac{Delta t}{2}$
begin{align}
frac{1}{Delta t} textbf{M} (vec{u}(t + Delta t) - vec{u}(t)) & = -frac{1}{2}(textbf{A} cdot vec{u}(t + Delta t) + textbf{A} cdot vec{u}(t) \
Bigg(textbf{M} + frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t + Delta t) & = Bigg(textbf{M} - frac{Delta t}{2}textbf{A}Bigg)cdot vec{u}(t)
end{align}
Here is what I don't understand:
It seems that we use the Crank Nicolson scheme and simply drop the expression $+frac{1}{2}(f(t) + f(t + Delta t)) $. What is the reasoning behind it?
pde eigenvalues-eigenvectors finite-element-method
pde eigenvalues-eigenvectors finite-element-method
asked Dec 2 '18 at 16:40
ecjbecjb
1618
1618
$begingroup$
The question is asking you to examine $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u$$ not $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u + f$$
$endgroup$
– Mattos
Dec 3 '18 at 1:17
add a comment |
$begingroup$
The question is asking you to examine $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u$$ not $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u + f$$
$endgroup$
– Mattos
Dec 3 '18 at 1:17
$begingroup$
The question is asking you to examine $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u$$ not $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u + f$$
$endgroup$
– Mattos
Dec 3 '18 at 1:17
$begingroup$
The question is asking you to examine $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u$$ not $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u + f$$
$endgroup$
– Mattos
Dec 3 '18 at 1:17
add a comment |
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$begingroup$
The question is asking you to examine $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u$$ not $$boldsymbol{M} cdot u' = boldsymbol{A} cdot u + f$$
$endgroup$
– Mattos
Dec 3 '18 at 1:17