Iteration step of the Crank–Nicolson scheme












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












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