Mixing
Numerical scheme for coupled PDEs
$begingroup$
I am trying to solve the three coupled PDEs;
$frac{partial{Q}}{partial{t}} = -RaPra^2theta - Pra^2Q + Prfrac{partial^2{Q}}{partial{z}^2}, (1)$
$frac{partial{theta}}{partial{t}} = w - a^2theta + frac{partial^2{theta}}{partial{z}^2}, (2)$
$ Q = -a^2w + frac{partial^2{w}}{partial{z}^2}, (3)$
where $Ra, Pr, a$ are constants. I want a method which is second order accurate in both time and space. I tried using the Crank-Nicolson scheme for the first two equations;
$frac{Q^{n+1} - Q^n}{Delta{t}} = frac{1}{2}(F^{n+1} + F^n)$
$frac{theta^{n+1} - theta^n}{Delta{t}} = frac{1}{2}(G^{n+1} + G^n)$,
and a centered space finite difference scheme for the third equaiton - $F$ and $G$ are the right hand sides of Eq.1 and Eq.2 respectively. My problem is that when using the Crank-Nicolson scheme I do not know $theta^{n+1}$ or $Q^{n+1}$ and therefore $w^{n+1}$. So far, I have just used;
$frac{Q^{n+1} - Q^n}{Delta{t}}= -RaPra^2theta^n + frac{1}{2}(H^{n+1} + H^n),$
$frac{theta^{n+1} - theta^n}{Delta{t}} = w^n + frac{1}{2}(K^{n+1} + K^n)$,
which isn't fully second order in time. I had an idea that was to solve these equations using the scheme above. Then use RK2 or a similar predictor-corrector method where the above scheme is my predictor. Does this make sense?
What scheme can I use to solve these equations which will be accurate to second order in space and time?
I am similar with finite difference methods and Runge-Kutta methods so anything involving these would be the best.
pde numerical-methods systems-of-equations finite-differences runge-kutta-methods
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
$endgroup$
add a comment |
$begingroup$
I am trying to solve the three coupled PDEs;
$frac{partial{Q}}{partial{t}} = -RaPra^2theta - Pra^2Q + Prfrac{partial^2{Q}}{partial{z}^2}, (1)$
$frac{partial{theta}}{partial{t}} = w - a^2theta + frac{partial^2{theta}}{partial{z}^2}, (2)$
$ Q = -a^2w + frac{partial^2{w}}{partial{z}^2}, (3)$
where $Ra, Pr, a$ are constants. I want a method which is second order accurate in both time and space. I tried using the Crank-Nicolson scheme for the first two equations;
$frac{Q^{n+1} - Q^n}{Delta{t}} = frac{1}{2}(F^{n+1} + F^n)$
$frac{theta^{n+1} - theta^n}{Delta{t}} = frac{1}{2}(G^{n+1} + G^n)$,
and a centered space finite difference scheme for the third equaiton - $F$ and $G$ are the right hand sides of Eq.1 and Eq.2 respectively. My problem is that when using the Crank-Nicolson scheme I do not know $theta^{n+1}$ or $Q^{n+1}$ and therefore $w^{n+1}$. So far, I have just used;
$frac{Q^{n+1} - Q^n}{Delta{t}}= -RaPra^2theta^n + frac{1}{2}(H^{n+1} + H^n),$
$frac{theta^{n+1} - theta^n}{Delta{t}} = w^n + frac{1}{2}(K^{n+1} + K^n)$,
which isn't fully second order in time. I had an idea that was to solve these equations using the scheme above. Then use RK2 or a similar predictor-corrector method where the above scheme is my predictor. Does this make sense?
What scheme can I use to solve these equations which will be accurate to second order in space and time?
I am similar with finite difference methods and Runge-Kutta methods so anything involving these would be the best.
pde numerical-methods systems-of-equations finite-differences runge-kutta-methods
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
$endgroup$
add a comment |
$begingroup$
I am trying to solve the three coupled PDEs;
$frac{partial{Q}}{partial{t}} = -RaPra^2theta - Pra^2Q + Prfrac{partial^2{Q}}{partial{z}^2}, (1)$
$frac{partial{theta}}{partial{t}} = w - a^2theta + frac{partial^2{theta}}{partial{z}^2}, (2)$
$ Q = -a^2w + frac{partial^2{w}}{partial{z}^2}, (3)$
where $Ra, Pr, a$ are constants. I want a method which is second order accurate in both time and space. I tried using the Crank-Nicolson scheme for the first two equations;
$frac{Q^{n+1} - Q^n}{Delta{t}} = frac{1}{2}(F^{n+1} + F^n)$
$frac{theta^{n+1} - theta^n}{Delta{t}} = frac{1}{2}(G^{n+1} + G^n)$,
and a centered space finite difference scheme for the third equaiton - $F$ and $G$ are the right hand sides of Eq.1 and Eq.2 respectively. My problem is that when using the Crank-Nicolson scheme I do not know $theta^{n+1}$ or $Q^{n+1}$ and therefore $w^{n+1}$. So far, I have just used;
$frac{Q^{n+1} - Q^n}{Delta{t}}= -RaPra^2theta^n + frac{1}{2}(H^{n+1} + H^n),$
$frac{theta^{n+1} - theta^n}{Delta{t}} = w^n + frac{1}{2}(K^{n+1} + K^n)$,
which isn't fully second order in time. I had an idea that was to solve these equations using the scheme above. Then use RK2 or a similar predictor-corrector method where the above scheme is my predictor. Does this make sense?
What scheme can I use to solve these equations which will be accurate to second order in space and time?
I am similar with finite difference methods and Runge-Kutta methods so anything involving these would be the best.
pde numerical-methods systems-of-equations finite-differences runge-kutta-methods
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
$endgroup$
I am trying to solve the three coupled PDEs;
$frac{partial{Q}}{partial{t}} = -RaPra^2theta - Pra^2Q + Prfrac{partial^2{Q}}{partial{z}^2}, (1)$
$frac{partial{theta}}{partial{t}} = w - a^2theta + frac{partial^2{theta}}{partial{z}^2}, (2)$
$ Q = -a^2w + frac{partial^2{w}}{partial{z}^2}, (3)$
where $Ra, Pr, a$ are constants. I want a method which is second order accurate in both time and space. I tried using the Crank-Nicolson scheme for the first two equations;
$frac{Q^{n+1} - Q^n}{Delta{t}} = frac{1}{2}(F^{n+1} + F^n)$
$frac{theta^{n+1} - theta^n}{Delta{t}} = frac{1}{2}(G^{n+1} + G^n)$,
and a centered space finite difference scheme for the third equaiton - $F$ and $G$ are the right hand sides of Eq.1 and Eq.2 respectively. My problem is that when using the Crank-Nicolson scheme I do not know $theta^{n+1}$ or $Q^{n+1}$ and therefore $w^{n+1}$. So far, I have just used;
$frac{Q^{n+1} - Q^n}{Delta{t}}= -RaPra^2theta^n + frac{1}{2}(H^{n+1} + H^n),$
$frac{theta^{n+1} - theta^n}{Delta{t}} = w^n + frac{1}{2}(K^{n+1} + K^n)$,
which isn't fully second order in time. I had an idea that was to solve these equations using the scheme above. Then use RK2 or a similar predictor-corrector method where the above scheme is my predictor. Does this make sense?
What scheme can I use to solve these equations which will be accurate to second order in space and time?
I am similar with finite difference methods and Runge-Kutta methods so anything involving these would be the best.
pde numerical-methods systems-of-equations finite-differences runge-kutta-methods
pde numerical-methods systems-of-equations finite-differences runge-kutta-methods
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
edited Jan 6 at 20:08
Patrick Lewis
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
asked Jan 6 at 18:11
Patrick LewisPatrick Lewis
112
112
add a comment |
add a comment |
1 Answer
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$begingroup$
I think your problem arises from the fact that after space discretization, you obtain a differential-algebraic system of equations (DAEs) instead of a system of ordinary differential equations (ODEs).
In a standard method of lines approach we discretize in space first (using centered finite differences as you mentioned). We define new vector-valued time-dependent functions $boldsymbol{Q}, boldsymbol{theta}, boldsymbol{w}$ with $boldsymbol{Q}(t) = (Q_1(t),Q_2(t),dots,Q_n(t))^{top}$, where $Q_i(t) simeq Q(z_i,t)$, $i = 1,2,dots,n$, and analogous for $boldsymbol{theta}, boldsymbol{w}$.
For these new functions we now obtain the DAEs
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& boldsymbol{w} - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta},\
boldsymbol{Q} &=& - a^2 boldsymbol{w} + boldsymbol{underline{A}}_w boldsymbol{w} + boldsymbol{b}_w,
end{eqnarray}
with tridiagonal matrices $boldsymbol{underline{A}}_Q, boldsymbol{underline{A}}_{theta}, boldsymbol{underline{A}}_w$ and with vectors $boldsymbol{b}_Q, boldsymbol{b}_{theta}, boldsymbol{b}_w$ which arise from the centered finite differences and from the boundary conditions on $Q, theta, w$.
These are not ODEs because the time derivative of $boldsymbol{w}$ is missing. However, because the third equation is linear, I would suggest to eliminate $boldsymbol{w} = (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w)$ using the third equation ($boldsymbol{underline{I}}$ denoting the identity matrix) and plug into the second equation to obtain an actual (linear) system of ODEs for $boldsymbol{Q}$ and $boldsymbol{theta}$ only:
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w) - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta}.
end{eqnarray}
You can now use your favorite Runge-Kutta method to solve the system of ODEs for $boldsymbol{Q}, boldsymbol{theta}$.
answered Jan 7 at 4:38
ChristophChristoph
4616
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add a comment |
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$begingroup$
I think your problem arises from the fact that after space discretization, you obtain a differential-algebraic system of equations (DAEs) instead of a system of ordinary differential equations (ODEs).
In a standard method of lines approach we discretize in space first (using centered finite differences as you mentioned). We define new vector-valued time-dependent functions $boldsymbol{Q}, boldsymbol{theta}, boldsymbol{w}$ with $boldsymbol{Q}(t) = (Q_1(t),Q_2(t),dots,Q_n(t))^{top}$, where $Q_i(t) simeq Q(z_i,t)$, $i = 1,2,dots,n$, and analogous for $boldsymbol{theta}, boldsymbol{w}$.
For these new functions we now obtain the DAEs
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& boldsymbol{w} - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta},\
boldsymbol{Q} &=& - a^2 boldsymbol{w} + boldsymbol{underline{A}}_w boldsymbol{w} + boldsymbol{b}_w,
end{eqnarray}
with tridiagonal matrices $boldsymbol{underline{A}}_Q, boldsymbol{underline{A}}_{theta}, boldsymbol{underline{A}}_w$ and with vectors $boldsymbol{b}_Q, boldsymbol{b}_{theta}, boldsymbol{b}_w$ which arise from the centered finite differences and from the boundary conditions on $Q, theta, w$.
These are not ODEs because the time derivative of $boldsymbol{w}$ is missing. However, because the third equation is linear, I would suggest to eliminate $boldsymbol{w} = (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w)$ using the third equation ($boldsymbol{underline{I}}$ denoting the identity matrix) and plug into the second equation to obtain an actual (linear) system of ODEs for $boldsymbol{Q}$ and $boldsymbol{theta}$ only:
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w) - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta}.
end{eqnarray}
You can now use your favorite Runge-Kutta method to solve the system of ODEs for $boldsymbol{Q}, boldsymbol{theta}$.
answered Jan 7 at 4:38
ChristophChristoph
4616
$endgroup$
add a comment |
$begingroup$
I think your problem arises from the fact that after space discretization, you obtain a differential-algebraic system of equations (DAEs) instead of a system of ordinary differential equations (ODEs).
In a standard method of lines approach we discretize in space first (using centered finite differences as you mentioned). We define new vector-valued time-dependent functions $boldsymbol{Q}, boldsymbol{theta}, boldsymbol{w}$ with $boldsymbol{Q}(t) = (Q_1(t),Q_2(t),dots,Q_n(t))^{top}$, where $Q_i(t) simeq Q(z_i,t)$, $i = 1,2,dots,n$, and analogous for $boldsymbol{theta}, boldsymbol{w}$.
For these new functions we now obtain the DAEs
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& boldsymbol{w} - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta},\
boldsymbol{Q} &=& - a^2 boldsymbol{w} + boldsymbol{underline{A}}_w boldsymbol{w} + boldsymbol{b}_w,
end{eqnarray}
with tridiagonal matrices $boldsymbol{underline{A}}_Q, boldsymbol{underline{A}}_{theta}, boldsymbol{underline{A}}_w$ and with vectors $boldsymbol{b}_Q, boldsymbol{b}_{theta}, boldsymbol{b}_w$ which arise from the centered finite differences and from the boundary conditions on $Q, theta, w$.
These are not ODEs because the time derivative of $boldsymbol{w}$ is missing. However, because the third equation is linear, I would suggest to eliminate $boldsymbol{w} = (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w)$ using the third equation ($boldsymbol{underline{I}}$ denoting the identity matrix) and plug into the second equation to obtain an actual (linear) system of ODEs for $boldsymbol{Q}$ and $boldsymbol{theta}$ only:
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w) - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta}.
end{eqnarray}
You can now use your favorite Runge-Kutta method to solve the system of ODEs for $boldsymbol{Q}, boldsymbol{theta}$.
answered Jan 7 at 4:38
ChristophChristoph
4616
$endgroup$
add a comment |
$begingroup$
I think your problem arises from the fact that after space discretization, you obtain a differential-algebraic system of equations (DAEs) instead of a system of ordinary differential equations (ODEs).
In a standard method of lines approach we discretize in space first (using centered finite differences as you mentioned). We define new vector-valued time-dependent functions $boldsymbol{Q}, boldsymbol{theta}, boldsymbol{w}$ with $boldsymbol{Q}(t) = (Q_1(t),Q_2(t),dots,Q_n(t))^{top}$, where $Q_i(t) simeq Q(z_i,t)$, $i = 1,2,dots,n$, and analogous for $boldsymbol{theta}, boldsymbol{w}$.
For these new functions we now obtain the DAEs
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& boldsymbol{w} - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta},\
boldsymbol{Q} &=& - a^2 boldsymbol{w} + boldsymbol{underline{A}}_w boldsymbol{w} + boldsymbol{b}_w,
end{eqnarray}
with tridiagonal matrices $boldsymbol{underline{A}}_Q, boldsymbol{underline{A}}_{theta}, boldsymbol{underline{A}}_w$ and with vectors $boldsymbol{b}_Q, boldsymbol{b}_{theta}, boldsymbol{b}_w$ which arise from the centered finite differences and from the boundary conditions on $Q, theta, w$.
These are not ODEs because the time derivative of $boldsymbol{w}$ is missing. However, because the third equation is linear, I would suggest to eliminate $boldsymbol{w} = (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w)$ using the third equation ($boldsymbol{underline{I}}$ denoting the identity matrix) and plug into the second equation to obtain an actual (linear) system of ODEs for $boldsymbol{Q}$ and $boldsymbol{theta}$ only:
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w) - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta}.
end{eqnarray}
You can now use your favorite Runge-Kutta method to solve the system of ODEs for $boldsymbol{Q}, boldsymbol{theta}$.
answered Jan 7 at 4:38
ChristophChristoph
4616
$endgroup$
I think your problem arises from the fact that after space discretization, you obtain a differential-algebraic system of equations (DAEs) instead of a system of ordinary differential equations (ODEs).
In a standard method of lines approach we discretize in space first (using centered finite differences as you mentioned). We define new vector-valued time-dependent functions $boldsymbol{Q}, boldsymbol{theta}, boldsymbol{w}$ with $boldsymbol{Q}(t) = (Q_1(t),Q_2(t),dots,Q_n(t))^{top}$, where $Q_i(t) simeq Q(z_i,t)$, $i = 1,2,dots,n$, and analogous for $boldsymbol{theta}, boldsymbol{w}$.
For these new functions we now obtain the DAEs
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& boldsymbol{w} - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta},\
boldsymbol{Q} &=& - a^2 boldsymbol{w} + boldsymbol{underline{A}}_w boldsymbol{w} + boldsymbol{b}_w,
end{eqnarray}
with tridiagonal matrices $boldsymbol{underline{A}}_Q, boldsymbol{underline{A}}_{theta}, boldsymbol{underline{A}}_w$ and with vectors $boldsymbol{b}_Q, boldsymbol{b}_{theta}, boldsymbol{b}_w$ which arise from the centered finite differences and from the boundary conditions on $Q, theta, w$.
These are not ODEs because the time derivative of $boldsymbol{w}$ is missing. However, because the third equation is linear, I would suggest to eliminate $boldsymbol{w} = (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w)$ using the third equation ($boldsymbol{underline{I}}$ denoting the identity matrix) and plug into the second equation to obtain an actual (linear) system of ODEs for $boldsymbol{Q}$ and $boldsymbol{theta}$ only:
begin{eqnarray}
boldsymbol{dot{Q}} &=& -RaPra^2 boldsymbol{theta} - Pra^2 boldsymbol{Q} + Pr (boldsymbol{underline{A}}_Q boldsymbol{Q} + boldsymbol{b}_Q),\
boldsymbol{dot{theta}} &=& (boldsymbol{underline{A}}_w - a^2 boldsymbol{underline{I}})^{-1}(boldsymbol{Q} - boldsymbol{b}_w) - a^2 boldsymbol{theta} + boldsymbol{underline{A}}_{theta} boldsymbol{theta} + boldsymbol{b}_{theta}.
end{eqnarray}
You can now use your favorite Runge-Kutta method to solve the system of ODEs for $boldsymbol{Q}, boldsymbol{theta}$.
answered Jan 7 at 4:38
ChristophChristoph
4616
edited Jan 7 at 4:46
answered Jan 7 at 4:38
ChristophChristoph
4616
answered Jan 7 at 4:38
ChristophChristoph
4616
answered Jan 7 at 4:38
ChristophChristoph
4616
4616
add a comment |
add a comment |
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