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Solving a stiff PDE system with operator splitting
The Next CEO of Stack OverflowSolving a system of differential equation including non-linearityTimestepping PDE with positive eigenvaluesViscous Burgers' equation using Lax-Wendroff schemeCoupled linear PDE equations (2nd and 1st order) - Numerical MethodSolving system of differential equations using matlabA little PDE solvingDifferential Equation System With The Euler MethodSolving an inhomogenous system PDENumerical Solution of PDE with SingularityApproximation of partial derivatives in Backward Euler Scheme
$begingroup$
I have a particular system of PDE that have given me a lot of grief and I wanted to try to solve it using operator splitting however I don't fully understand how its done. I'll start by explaining what I've done and my issues with it
The equations of interest are
(1) $ u_t=iu_xx+f(epsilon)u$
(2) $epsilon_t=q(x)-(1+g(u))epsilon$
(3) $u(0,t)=u(L,t)=0, u(x,0)=p(x), epsilon(x,0)=0$
Where the function q(x) is positive over the domain.
My basic attempt is to apply finite difference on the derivative and solve the system of equations directly. I used backwards Euler and treated the nonlinear part of (1) explicitly. Equation (2) can be solved exactly in backwards Euler (and in crank-nicholson as well).
(4) $epsilon^n+1=fracepsilon^n+Delta tq(x)1+Delta t(1+g(u^n+1))$
So my general plan, was to solve equation (1), feed those results into equation (4), and repeat. While it does work it has a major flaw in that when I change the time step size the resultant solutions distort.
I want to move on to operator splitting but I don't understand how to apply it.
Splitting equation (1) is simple
(5) $u_t=(A+B)u$
(6) $A=fracpartial^2partial x^2, B=f(epsilon)$
Equation (2) I'm not sure how to split since it's inhomogeneous. Would it go like this?
(7) $epsilon_t=q(x)-(C+D)epsilon$
(8) $C=1, D=g(u)$
Or could I say the operator C is
(9) $Cepsilon=q(x)-epsilon$
Theoretically the solutions for operator splitting would be
(10) $u(t+h)=e^(A+B)hu(t)$
(11) $epsilon(t+h)=e^(C+D)hepsilon(t)$
If I did strang splitting I would have for instance
(12) $u(t+h)=e^0.5hAe^hBe^0.5hAu(t)$
But how do I actually use this? I know you can expand the exponential operators
(13) $e^hAapprox1+hA+frac(hA)^22+...$
But I don't see how this makes the new problem any more solvable.
I've seen you can break the operators into sub problems which would be very useful because the equations $u_t=Au$ and $epsilon_t=Cepsilon$ are both exactly solvable if the boundary/initial conditions are unchanged. But again I don't see how use them.
What's the procedure here?
pde numerical-methods
asked Mar 28 at 1:50
DenisDenis
394
$endgroup$
add a comment |
$begingroup$
I have a particular system of PDE that have given me a lot of grief and I wanted to try to solve it using operator splitting however I don't fully understand how its done. I'll start by explaining what I've done and my issues with it
The equations of interest are
(1) $ u_t=iu_xx+f(epsilon)u$
(2) $epsilon_t=q(x)-(1+g(u))epsilon$
(3) $u(0,t)=u(L,t)=0, u(x,0)=p(x), epsilon(x,0)=0$
Where the function q(x) is positive over the domain.
My basic attempt is to apply finite difference on the derivative and solve the system of equations directly. I used backwards Euler and treated the nonlinear part of (1) explicitly. Equation (2) can be solved exactly in backwards Euler (and in crank-nicholson as well).
(4) $epsilon^n+1=fracepsilon^n+Delta tq(x)1+Delta t(1+g(u^n+1))$
So my general plan, was to solve equation (1), feed those results into equation (4), and repeat. While it does work it has a major flaw in that when I change the time step size the resultant solutions distort.
I want to move on to operator splitting but I don't understand how to apply it.
Splitting equation (1) is simple
(5) $u_t=(A+B)u$
(6) $A=fracpartial^2partial x^2, B=f(epsilon)$
Equation (2) I'm not sure how to split since it's inhomogeneous. Would it go like this?
(7) $epsilon_t=q(x)-(C+D)epsilon$
(8) $C=1, D=g(u)$
Or could I say the operator C is
(9) $Cepsilon=q(x)-epsilon$
Theoretically the solutions for operator splitting would be
(10) $u(t+h)=e^(A+B)hu(t)$
(11) $epsilon(t+h)=e^(C+D)hepsilon(t)$
If I did strang splitting I would have for instance
(12) $u(t+h)=e^0.5hAe^hBe^0.5hAu(t)$
But how do I actually use this? I know you can expand the exponential operators
(13) $e^hAapprox1+hA+frac(hA)^22+...$
But I don't see how this makes the new problem any more solvable.
I've seen you can break the operators into sub problems which would be very useful because the equations $u_t=Au$ and $epsilon_t=Cepsilon$ are both exactly solvable if the boundary/initial conditions are unchanged. But again I don't see how use them.
What's the procedure here?
pde numerical-methods
asked Mar 28 at 1:50
DenisDenis
394
$endgroup$
$begingroup$
When splitting, we usually split the linear and nonlinear terms in the derivatives (hence the term operator splitting). Because your PDEs aren't nonlinear in the derivatives, I don't think you need to split the operators, you can probably just go with solving the system using an IMEX scheme.
$endgroup$
– Mattos
yesterday
add a comment |
$begingroup$
I have a particular system of PDE that have given me a lot of grief and I wanted to try to solve it using operator splitting however I don't fully understand how its done. I'll start by explaining what I've done and my issues with it
The equations of interest are
(1) $ u_t=iu_xx+f(epsilon)u$
(2) $epsilon_t=q(x)-(1+g(u))epsilon$
(3) $u(0,t)=u(L,t)=0, u(x,0)=p(x), epsilon(x,0)=0$
Where the function q(x) is positive over the domain.
My basic attempt is to apply finite difference on the derivative and solve the system of equations directly. I used backwards Euler and treated the nonlinear part of (1) explicitly. Equation (2) can be solved exactly in backwards Euler (and in crank-nicholson as well).
(4) $epsilon^n+1=fracepsilon^n+Delta tq(x)1+Delta t(1+g(u^n+1))$
So my general plan, was to solve equation (1), feed those results into equation (4), and repeat. While it does work it has a major flaw in that when I change the time step size the resultant solutions distort.
I want to move on to operator splitting but I don't understand how to apply it.
Splitting equation (1) is simple
(5) $u_t=(A+B)u$
(6) $A=fracpartial^2partial x^2, B=f(epsilon)$
Equation (2) I'm not sure how to split since it's inhomogeneous. Would it go like this?
(7) $epsilon_t=q(x)-(C+D)epsilon$
(8) $C=1, D=g(u)$
Or could I say the operator C is
(9) $Cepsilon=q(x)-epsilon$
Theoretically the solutions for operator splitting would be
(10) $u(t+h)=e^(A+B)hu(t)$
(11) $epsilon(t+h)=e^(C+D)hepsilon(t)$
If I did strang splitting I would have for instance
(12) $u(t+h)=e^0.5hAe^hBe^0.5hAu(t)$
But how do I actually use this? I know you can expand the exponential operators
(13) $e^hAapprox1+hA+frac(hA)^22+...$
But I don't see how this makes the new problem any more solvable.
I've seen you can break the operators into sub problems which would be very useful because the equations $u_t=Au$ and $epsilon_t=Cepsilon$ are both exactly solvable if the boundary/initial conditions are unchanged. But again I don't see how use them.
What's the procedure here?
pde numerical-methods
asked Mar 28 at 1:50
DenisDenis
394
$endgroup$
I have a particular system of PDE that have given me a lot of grief and I wanted to try to solve it using operator splitting however I don't fully understand how its done. I'll start by explaining what I've done and my issues with it
The equations of interest are
(1) $ u_t=iu_xx+f(epsilon)u$
(2) $epsilon_t=q(x)-(1+g(u))epsilon$
(3) $u(0,t)=u(L,t)=0, u(x,0)=p(x), epsilon(x,0)=0$
Where the function q(x) is positive over the domain.
My basic attempt is to apply finite difference on the derivative and solve the system of equations directly. I used backwards Euler and treated the nonlinear part of (1) explicitly. Equation (2) can be solved exactly in backwards Euler (and in crank-nicholson as well).
(4) $epsilon^n+1=fracepsilon^n+Delta tq(x)1+Delta t(1+g(u^n+1))$
So my general plan, was to solve equation (1), feed those results into equation (4), and repeat. While it does work it has a major flaw in that when I change the time step size the resultant solutions distort.
I want to move on to operator splitting but I don't understand how to apply it.
Splitting equation (1) is simple
(5) $u_t=(A+B)u$
(6) $A=fracpartial^2partial x^2, B=f(epsilon)$
Equation (2) I'm not sure how to split since it's inhomogeneous. Would it go like this?
(7) $epsilon_t=q(x)-(C+D)epsilon$
(8) $C=1, D=g(u)$
Or could I say the operator C is
(9) $Cepsilon=q(x)-epsilon$
Theoretically the solutions for operator splitting would be
(10) $u(t+h)=e^(A+B)hu(t)$
(11) $epsilon(t+h)=e^(C+D)hepsilon(t)$
If I did strang splitting I would have for instance
(12) $u(t+h)=e^0.5hAe^hBe^0.5hAu(t)$
But how do I actually use this? I know you can expand the exponential operators
(13) $e^hAapprox1+hA+frac(hA)^22+...$
But I don't see how this makes the new problem any more solvable.
I've seen you can break the operators into sub problems which would be very useful because the equations $u_t=Au$ and $epsilon_t=Cepsilon$ are both exactly solvable if the boundary/initial conditions are unchanged. But again I don't see how use them.
What's the procedure here?
pde numerical-methods
pde numerical-methods
asked Mar 28 at 1:50
DenisDenis
394
asked Mar 28 at 1:50
DenisDenis
394
asked Mar 28 at 1:50
DenisDenis
394
asked Mar 28 at 1:50
DenisDenis
394
asked Mar 28 at 1:50
DenisDenis
394
394
$begingroup$
When splitting, we usually split the linear and nonlinear terms in the derivatives (hence the term operator splitting). Because your PDEs aren't nonlinear in the derivatives, I don't think you need to split the operators, you can probably just go with solving the system using an IMEX scheme.
$endgroup$
– Mattos
yesterday
add a comment |
$begingroup$
When splitting, we usually split the linear and nonlinear terms in the derivatives (hence the term operator splitting). Because your PDEs aren't nonlinear in the derivatives, I don't think you need to split the operators, you can probably just go with solving the system using an IMEX scheme.
$endgroup$
– Mattos
yesterday
$begingroup$
When splitting, we usually split the linear and nonlinear terms in the derivatives (hence the term operator splitting). Because your PDEs aren't nonlinear in the derivatives, I don't think you need to split the operators, you can probably just go with solving the system using an IMEX scheme.
$endgroup$
– Mattos
yesterday
$begingroup$
When splitting, we usually split the linear and nonlinear terms in the derivatives (hence the term operator splitting). Because your PDEs aren't nonlinear in the derivatives, I don't think you need to split the operators, you can probably just go with solving the system using an IMEX scheme.
$endgroup$
– Mattos
yesterday
add a comment |
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$begingroup$
When splitting, we usually split the linear and nonlinear terms in the derivatives (hence the term operator splitting). Because your PDEs aren't nonlinear in the derivatives, I don't think you need to split the operators, you can probably just go with solving the system using an IMEX scheme.
$endgroup$
– Mattos
yesterday