GMRES vs Newton-GMRES for Solving nonlinear PDE'sNewton-Raphson method for nonlinear partial differential equationsSolving a nonlinear algebraic system that includes a linear termIs it possible to solve nonlinear PDEs without using Newton-Raphson iteration?Solving a system of nonlinear equations with an ODE solver is faster than with the Newton method?Newton iteration applied to nonlinear PDESolving this nonlinear system of equationsNumerically solving a system of stiff nonlinear PDEsNewton method for a nonlinear system of time-independent PDEsNonlinear system with diagonal nonlinearityResidual value goes to NaN while solving a system of nonlinear equations
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GMRES vs Newton-GMRES for Solving nonlinear PDE's
Newton-Raphson method for nonlinear partial differential equationsSolving a nonlinear algebraic system that includes a linear termIs it possible to solve nonlinear PDEs without using Newton-Raphson iteration?Solving a system of nonlinear equations with an ODE solver is faster than with the Newton method?Newton iteration applied to nonlinear PDESolving this nonlinear system of equationsNumerically solving a system of stiff nonlinear PDEsNewton method for a nonlinear system of time-independent PDEsNonlinear system with diagonal nonlinearityResidual value goes to NaN while solving a system of nonlinear equations
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Often when numerically solving nonlinear PDE's using method of lines approach with an implicit integrator a system of nonlinear equations have to be solved.
To be more specific, let's say we have simple backward Euler method:
beginalign
y^n=y^n+1-h f(y^n+1,t)= G(y^n+1,t)
endalign
From the available literature, it seems the most common approach is using an Newton-GMRES method to solve the nonlinear system, especially if it is stiff.
As far as I can understand GMRES is used because it is a matrix-free method and so also works with vector functions which is good when approximating the Jacobian with finite differences.
My question is then, why not just use the GMRES to solve the initial problem?
I see no reason why GMRES couldn't solve this for $y^n+1$ when for the Newton-GMRES we solve the system:
beginalign
H(y^n+1_k) &= y^n+1_k-y^n-h f(y^n+1_k,t)rightarrow 0 text for krightarrow infty\
Jv &simeq fracH(y^n+1_k+epsilon v)-H(y^n+1_k))epsilon simeq -H(y^n+1_k)\
y^n+1_k+1&=y^n+1_k+v
endalign
where the GMRES is used for the second part to find $v$. I don't see why solving for $v$ with GMRES followed by the Newton step should be easier/better than solving the general system in the top for $y^n+1$ using GMRES? Does the finite difference approximation to the Jacobian have better properties compared the the general system $y^n=G(y^n+1,t)$?
pde nonlinear-equations newton-method implicit-methods gmres
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
asked Jun 3 at 20:29
RasmusRasmus
503
$endgroup$
add a comment |
$begingroup$
Often when numerically solving nonlinear PDE's using method of lines approach with an implicit integrator a system of nonlinear equations have to be solved.
To be more specific, let's say we have simple backward Euler method:
beginalign
y^n=y^n+1-h f(y^n+1,t)= G(y^n+1,t)
endalign
From the available literature, it seems the most common approach is using an Newton-GMRES method to solve the nonlinear system, especially if it is stiff.
As far as I can understand GMRES is used because it is a matrix-free method and so also works with vector functions which is good when approximating the Jacobian with finite differences.
My question is then, why not just use the GMRES to solve the initial problem?
I see no reason why GMRES couldn't solve this for $y^n+1$ when for the Newton-GMRES we solve the system:
beginalign
H(y^n+1_k) &= y^n+1_k-y^n-h f(y^n+1_k,t)rightarrow 0 text for krightarrow infty\
Jv &simeq fracH(y^n+1_k+epsilon v)-H(y^n+1_k))epsilon simeq -H(y^n+1_k)\
y^n+1_k+1&=y^n+1_k+v
endalign
where the GMRES is used for the second part to find $v$. I don't see why solving for $v$ with GMRES followed by the Newton step should be easier/better than solving the general system in the top for $y^n+1$ using GMRES? Does the finite difference approximation to the Jacobian have better properties compared the the general system $y^n=G(y^n+1,t)$?
pde nonlinear-equations newton-method implicit-methods gmres
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
asked Jun 3 at 20:29
RasmusRasmus
503
$endgroup$
add a comment |
$begingroup$
Often when numerically solving nonlinear PDE's using method of lines approach with an implicit integrator a system of nonlinear equations have to be solved.
To be more specific, let's say we have simple backward Euler method:
beginalign
y^n=y^n+1-h f(y^n+1,t)= G(y^n+1,t)
endalign
From the available literature, it seems the most common approach is using an Newton-GMRES method to solve the nonlinear system, especially if it is stiff.
As far as I can understand GMRES is used because it is a matrix-free method and so also works with vector functions which is good when approximating the Jacobian with finite differences.
My question is then, why not just use the GMRES to solve the initial problem?
I see no reason why GMRES couldn't solve this for $y^n+1$ when for the Newton-GMRES we solve the system:
beginalign
H(y^n+1_k) &= y^n+1_k-y^n-h f(y^n+1_k,t)rightarrow 0 text for krightarrow infty\
Jv &simeq fracH(y^n+1_k+epsilon v)-H(y^n+1_k))epsilon simeq -H(y^n+1_k)\
y^n+1_k+1&=y^n+1_k+v
endalign
where the GMRES is used for the second part to find $v$. I don't see why solving for $v$ with GMRES followed by the Newton step should be easier/better than solving the general system in the top for $y^n+1$ using GMRES? Does the finite difference approximation to the Jacobian have better properties compared the the general system $y^n=G(y^n+1,t)$?
pde nonlinear-equations newton-method implicit-methods gmres
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
asked Jun 3 at 20:29
RasmusRasmus
503
$endgroup$
Often when numerically solving nonlinear PDE's using method of lines approach with an implicit integrator a system of nonlinear equations have to be solved.
To be more specific, let's say we have simple backward Euler method:
beginalign
y^n=y^n+1-h f(y^n+1,t)= G(y^n+1,t)
endalign
From the available literature, it seems the most common approach is using an Newton-GMRES method to solve the nonlinear system, especially if it is stiff.
As far as I can understand GMRES is used because it is a matrix-free method and so also works with vector functions which is good when approximating the Jacobian with finite differences.
My question is then, why not just use the GMRES to solve the initial problem?
I see no reason why GMRES couldn't solve this for $y^n+1$ when for the Newton-GMRES we solve the system:
beginalign
H(y^n+1_k) &= y^n+1_k-y^n-h f(y^n+1_k,t)rightarrow 0 text for krightarrow infty\
Jv &simeq fracH(y^n+1_k+epsilon v)-H(y^n+1_k))epsilon simeq -H(y^n+1_k)\
y^n+1_k+1&=y^n+1_k+v
endalign
where the GMRES is used for the second part to find $v$. I don't see why solving for $v$ with GMRES followed by the Newton step should be easier/better than solving the general system in the top for $y^n+1$ using GMRES? Does the finite difference approximation to the Jacobian have better properties compared the the general system $y^n=G(y^n+1,t)$?
pde nonlinear-equations newton-method implicit-methods gmres
pde nonlinear-equations newton-method implicit-methods gmres
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
asked Jun 3 at 20:29
RasmusRasmus
503
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
asked Jun 3 at 20:29
RasmusRasmus
503
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
edited Jun 3 at 20:35
Anton Menshov♦
4,93322076
4,93322076
asked Jun 3 at 20:29
RasmusRasmus
503
asked Jun 3 at 20:29
RasmusRasmus
503
asked Jun 3 at 20:29
RasmusRasmus
503
503
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^-1$ using a matrix polynomial of $A$.
In this case (I assume) $f(y^n+1,t)$ is not necessarily linear in the vector $y^n+1$, and so $y^n=G(y^n+1,t)$ can't be written in the form $y^n=A(t)y^n+1$, where $A(t)$ is a matrix-valued function of time. So you can't use GMRES directly.
The "Newton step" is actually the formation of the linear system of equations with the Jacobian; the point of Newton's method is that it approximates the solution of a nonlinear equation with the solution of a sequence of linear equations. GMRES is just a tool that is used to implement Newton's method.
answered Jun 3 at 21:16
bgavbgav
862
$endgroup$
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
1
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^-1$ using a matrix polynomial of $A$.
In this case (I assume) $f(y^n+1,t)$ is not necessarily linear in the vector $y^n+1$, and so $y^n=G(y^n+1,t)$ can't be written in the form $y^n=A(t)y^n+1$, where $A(t)$ is a matrix-valued function of time. So you can't use GMRES directly.
The "Newton step" is actually the formation of the linear system of equations with the Jacobian; the point of Newton's method is that it approximates the solution of a nonlinear equation with the solution of a sequence of linear equations. GMRES is just a tool that is used to implement Newton's method.
answered Jun 3 at 21:16
bgavbgav
862
$endgroup$
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
1
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
add a comment |
$begingroup$
The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^-1$ using a matrix polynomial of $A$.
In this case (I assume) $f(y^n+1,t)$ is not necessarily linear in the vector $y^n+1$, and so $y^n=G(y^n+1,t)$ can't be written in the form $y^n=A(t)y^n+1$, where $A(t)$ is a matrix-valued function of time. So you can't use GMRES directly.
The "Newton step" is actually the formation of the linear system of equations with the Jacobian; the point of Newton's method is that it approximates the solution of a nonlinear equation with the solution of a sequence of linear equations. GMRES is just a tool that is used to implement Newton's method.
answered Jun 3 at 21:16
bgavbgav
862
$endgroup$
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
1
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
add a comment |
$begingroup$
The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^-1$ using a matrix polynomial of $A$.
In this case (I assume) $f(y^n+1,t)$ is not necessarily linear in the vector $y^n+1$, and so $y^n=G(y^n+1,t)$ can't be written in the form $y^n=A(t)y^n+1$, where $A(t)$ is a matrix-valued function of time. So you can't use GMRES directly.
The "Newton step" is actually the formation of the linear system of equations with the Jacobian; the point of Newton's method is that it approximates the solution of a nonlinear equation with the solution of a sequence of linear equations. GMRES is just a tool that is used to implement Newton's method.
answered Jun 3 at 21:16
bgavbgav
862
$endgroup$
The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^-1$ using a matrix polynomial of $A$.
In this case (I assume) $f(y^n+1,t)$ is not necessarily linear in the vector $y^n+1$, and so $y^n=G(y^n+1,t)$ can't be written in the form $y^n=A(t)y^n+1$, where $A(t)$ is a matrix-valued function of time. So you can't use GMRES directly.
The "Newton step" is actually the formation of the linear system of equations with the Jacobian; the point of Newton's method is that it approximates the solution of a nonlinear equation with the solution of a sequence of linear equations. GMRES is just a tool that is used to implement Newton's method.
answered Jun 3 at 21:16
bgavbgav
862
answered Jun 3 at 21:16
bgavbgav
862
answered Jun 3 at 21:16
bgavbgav
862
answered Jun 3 at 21:16
bgavbgav
862
862
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
1
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
add a comment |
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
1
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
$begingroup$
But isn't the function evaluations $H(y^n+1_k(+epsilon v))$ used in the Jacobian approximation also non-linear? Or is this OK, since it approximates a linear operator J and if so couldn't you argue that there is some linear operator $L$ such that $Ly^n+1simeq G(y^n+1,t)$ such that you in effect solve that system instead?
$endgroup$
– Rasmus
Jun 4 at 7:54
1
1
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Approximating $G(y^n+1,1)$ with a linear operation $Ly^n+1$ is kind of what you're doing by using the Jacobian; what you're doing here is you're saying that $$ H(y^n+1_k)=G(y^n+1_k,t)-y^napprox H(y^n+1_k-1)+J(y^n+1_k-1)(y^n+1_k-y^n+1_k-1) $$ By taking the Jacobian of $H$ evaluated at $y^n+1_k-1$, you're linearizing the function $H$ around the point $y^n+1_k-1$. You can't do this just once, though; that linear approximation is only valid in the neighborhood around the point $y^n+1_k-1$, so you need to iterate in $k$ to solve the problem.
$endgroup$
– bgav
Jun 4 at 18:27
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
$begingroup$
Ahh, i see. That makes sense to me, thank you very much :)
$endgroup$
– Rasmus
Jun 5 at 6:44
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
