Mixing
Is it useful to convert a higher order PDE into a 1st order system?
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I just learned how a higher order PDE can be converted into a system of PDEs.I am just wondering whether this is a standard way to solve some higher order PDEs which are immune to other methods or is it more efficient?
pde linear-pde parabolic-pde
asked Jan 23 at 13:28
ben tenysonben tenyson
414
$endgroup$
add a comment |
$begingroup$
I just learned how a higher order PDE can be converted into a system of PDEs.I am just wondering whether this is a standard way to solve some higher order PDEs which are immune to other methods or is it more efficient?
pde linear-pde parabolic-pde
asked Jan 23 at 13:28
ben tenysonben tenyson
414
$endgroup$
add a comment |
$begingroup$
I just learned how a higher order PDE can be converted into a system of PDEs.I am just wondering whether this is a standard way to solve some higher order PDEs which are immune to other methods or is it more efficient?
pde linear-pde parabolic-pde
asked Jan 23 at 13:28
ben tenysonben tenyson
414
$endgroup$
I just learned how a higher order PDE can be converted into a system of PDEs.I am just wondering whether this is a standard way to solve some higher order PDEs which are immune to other methods or is it more efficient?
pde linear-pde parabolic-pde
pde linear-pde parabolic-pde
asked Jan 23 at 13:28
ben tenysonben tenyson
414
asked Jan 23 at 13:28
ben tenysonben tenyson
414
edited Jan 23 at 13:53
ben tenyson
asked Jan 23 at 13:28
ben tenysonben tenyson
414
asked Jan 23 at 13:28
ben tenysonben tenyson
414
asked Jan 23 at 13:28
ben tenysonben tenyson
414
414
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In addition to @GiuseppeNegro's answer, it has several advantages to convert a high-order PDE into a first-order system, e.g. a system of the form $boldsymbol{u}_t + boldsymbol{f}(boldsymbol{u})_x = boldsymbol{r}(boldsymbol{u})$.
- For some particular linear hyperbolic systems, we may be able to decouple the equations, so that the method of characteristics can be used to solve initial- and boundary-value problems.
- For some particular hyperbolic systems, we may be able to derive an analytical solution to the Riemann problem (a particular initial-value problem).
- We may be able to exploit theoretical results available for this kind of system (existence, uniqueness, smoothness).
- We can make use of well-suited numerical methods, such as finite-volume methods of Godunov type.
However, not every higher-order PDE can be written in the previous form.
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
$endgroup$
add a comment |
$begingroup$
This is connected to the Hamiltonian formalism, and is also done in PDEs sometimes. I am far from being an expert, so let me just present the case of the wave equation $u_{tt} = Delta u$. Introduce the two-component vector (position-momentum):
$$
boldsymbol{u}:=begin{bmatrix} u \ u_tend{bmatrix}.$$
Then the wave equation is rewritten as
$$
partial_t boldsymbol{u} = J H boldsymbol{u},qquad J=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, qquad H=begin{bmatrix} -Delta & 0 \ 0 & 1 end{bmatrix}.$$
Is this useful? Well, to begin with, it lets the abstract theory of semigroups come into play. (That theory is the generalization to the infinite-dimensional case of the equation $dot{x}= A x$, where $A$ is replaced with a linear operator on some Banach space). It also reveals a conservation law; if $boldsymbol{u}$ solves the wave equation then
$$
begin{split}
partial_t langle Hboldsymbol{u}(t), boldsymbol{u}(t)rangle &= langle HJHboldsymbol{u}| boldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle \ &= -langle Hboldsymbol{u}|JHboldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle =0,
end{split}
$$
where
$$
langle (f_1, g_1) |(f_2, g_2)rangle:= int f_1f_2 + int g_1g_2,$$
and where we used the fact that $H$ is a symmetric operator, while $J$ is an anti-symmetric one. We have thus shown that
$$
langle H boldsymbol{u}|boldsymbol{u}rangle=int -Delta u, u + int u_t^2
$$
is constant along the wave flow, which is a fundamental fact known as conservation of energy. (The first summand equals $int |nabla u|^2$, which is how it usually appears).
This is all very basic, but already shows that, sometimes, this kind of manipulation can be useful.
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
add a comment |
$begingroup$
It's not clear to me how this would work for PDE's. For ODE's:
Yes, any higher-order ODE is equivalent to a system of first-order ODE's. For example $$y''=f(y,y',t)$$is equivalent to the system $$y_1'=y_2, y_2'=f(y_1,y_2,t).$$I don't know that this is important for solving ODE's, but it's certainly useful in proving things about them. Because that system of two ODE"s is just $$Y'=F(Y,t)$$for a suitable $F$, where $Y=(y_1,y_2)$. Proofs of things for ODE's often work just as well for "vector-valued" ODE's, which then have corollaries for higher-order (scalar-valued) ODE's. Hmm, I suppose the same holds for numerical methods for approximate solutions.
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
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$begingroup$
In addition to @GiuseppeNegro's answer, it has several advantages to convert a high-order PDE into a first-order system, e.g. a system of the form $boldsymbol{u}_t + boldsymbol{f}(boldsymbol{u})_x = boldsymbol{r}(boldsymbol{u})$.
- For some particular linear hyperbolic systems, we may be able to decouple the equations, so that the method of characteristics can be used to solve initial- and boundary-value problems.
- For some particular hyperbolic systems, we may be able to derive an analytical solution to the Riemann problem (a particular initial-value problem).
- We may be able to exploit theoretical results available for this kind of system (existence, uniqueness, smoothness).
- We can make use of well-suited numerical methods, such as finite-volume methods of Godunov type.
However, not every higher-order PDE can be written in the previous form.
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
$endgroup$
add a comment |
$begingroup$
In addition to @GiuseppeNegro's answer, it has several advantages to convert a high-order PDE into a first-order system, e.g. a system of the form $boldsymbol{u}_t + boldsymbol{f}(boldsymbol{u})_x = boldsymbol{r}(boldsymbol{u})$.
- For some particular linear hyperbolic systems, we may be able to decouple the equations, so that the method of characteristics can be used to solve initial- and boundary-value problems.
- For some particular hyperbolic systems, we may be able to derive an analytical solution to the Riemann problem (a particular initial-value problem).
- We may be able to exploit theoretical results available for this kind of system (existence, uniqueness, smoothness).
- We can make use of well-suited numerical methods, such as finite-volume methods of Godunov type.
However, not every higher-order PDE can be written in the previous form.
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
$endgroup$
add a comment |
$begingroup$
In addition to @GiuseppeNegro's answer, it has several advantages to convert a high-order PDE into a first-order system, e.g. a system of the form $boldsymbol{u}_t + boldsymbol{f}(boldsymbol{u})_x = boldsymbol{r}(boldsymbol{u})$.
- For some particular linear hyperbolic systems, we may be able to decouple the equations, so that the method of characteristics can be used to solve initial- and boundary-value problems.
- For some particular hyperbolic systems, we may be able to derive an analytical solution to the Riemann problem (a particular initial-value problem).
- We may be able to exploit theoretical results available for this kind of system (existence, uniqueness, smoothness).
- We can make use of well-suited numerical methods, such as finite-volume methods of Godunov type.
However, not every higher-order PDE can be written in the previous form.
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
$endgroup$
In addition to @GiuseppeNegro's answer, it has several advantages to convert a high-order PDE into a first-order system, e.g. a system of the form $boldsymbol{u}_t + boldsymbol{f}(boldsymbol{u})_x = boldsymbol{r}(boldsymbol{u})$.
- For some particular linear hyperbolic systems, we may be able to decouple the equations, so that the method of characteristics can be used to solve initial- and boundary-value problems.
- For some particular hyperbolic systems, we may be able to derive an analytical solution to the Riemann problem (a particular initial-value problem).
- We may be able to exploit theoretical results available for this kind of system (existence, uniqueness, smoothness).
- We can make use of well-suited numerical methods, such as finite-volume methods of Godunov type.
However, not every higher-order PDE can be written in the previous form.
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
edited Jan 23 at 16:30
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
answered Jan 23 at 16:24
Harry49Harry49
7,52431341
7,52431341
add a comment |
add a comment |
$begingroup$
This is connected to the Hamiltonian formalism, and is also done in PDEs sometimes. I am far from being an expert, so let me just present the case of the wave equation $u_{tt} = Delta u$. Introduce the two-component vector (position-momentum):
$$
boldsymbol{u}:=begin{bmatrix} u \ u_tend{bmatrix}.$$
Then the wave equation is rewritten as
$$
partial_t boldsymbol{u} = J H boldsymbol{u},qquad J=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, qquad H=begin{bmatrix} -Delta & 0 \ 0 & 1 end{bmatrix}.$$
Is this useful? Well, to begin with, it lets the abstract theory of semigroups come into play. (That theory is the generalization to the infinite-dimensional case of the equation $dot{x}= A x$, where $A$ is replaced with a linear operator on some Banach space). It also reveals a conservation law; if $boldsymbol{u}$ solves the wave equation then
$$
begin{split}
partial_t langle Hboldsymbol{u}(t), boldsymbol{u}(t)rangle &= langle HJHboldsymbol{u}| boldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle \ &= -langle Hboldsymbol{u}|JHboldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle =0,
end{split}
$$
where
$$
langle (f_1, g_1) |(f_2, g_2)rangle:= int f_1f_2 + int g_1g_2,$$
and where we used the fact that $H$ is a symmetric operator, while $J$ is an anti-symmetric one. We have thus shown that
$$
langle H boldsymbol{u}|boldsymbol{u}rangle=int -Delta u, u + int u_t^2
$$
is constant along the wave flow, which is a fundamental fact known as conservation of energy. (The first summand equals $int |nabla u|^2$, which is how it usually appears).
This is all very basic, but already shows that, sometimes, this kind of manipulation can be useful.
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
add a comment |
$begingroup$
This is connected to the Hamiltonian formalism, and is also done in PDEs sometimes. I am far from being an expert, so let me just present the case of the wave equation $u_{tt} = Delta u$. Introduce the two-component vector (position-momentum):
$$
boldsymbol{u}:=begin{bmatrix} u \ u_tend{bmatrix}.$$
Then the wave equation is rewritten as
$$
partial_t boldsymbol{u} = J H boldsymbol{u},qquad J=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, qquad H=begin{bmatrix} -Delta & 0 \ 0 & 1 end{bmatrix}.$$
Is this useful? Well, to begin with, it lets the abstract theory of semigroups come into play. (That theory is the generalization to the infinite-dimensional case of the equation $dot{x}= A x$, where $A$ is replaced with a linear operator on some Banach space). It also reveals a conservation law; if $boldsymbol{u}$ solves the wave equation then
$$
begin{split}
partial_t langle Hboldsymbol{u}(t), boldsymbol{u}(t)rangle &= langle HJHboldsymbol{u}| boldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle \ &= -langle Hboldsymbol{u}|JHboldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle =0,
end{split}
$$
where
$$
langle (f_1, g_1) |(f_2, g_2)rangle:= int f_1f_2 + int g_1g_2,$$
and where we used the fact that $H$ is a symmetric operator, while $J$ is an anti-symmetric one. We have thus shown that
$$
langle H boldsymbol{u}|boldsymbol{u}rangle=int -Delta u, u + int u_t^2
$$
is constant along the wave flow, which is a fundamental fact known as conservation of energy. (The first summand equals $int |nabla u|^2$, which is how it usually appears).
This is all very basic, but already shows that, sometimes, this kind of manipulation can be useful.
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
add a comment |
$begingroup$
This is connected to the Hamiltonian formalism, and is also done in PDEs sometimes. I am far from being an expert, so let me just present the case of the wave equation $u_{tt} = Delta u$. Introduce the two-component vector (position-momentum):
$$
boldsymbol{u}:=begin{bmatrix} u \ u_tend{bmatrix}.$$
Then the wave equation is rewritten as
$$
partial_t boldsymbol{u} = J H boldsymbol{u},qquad J=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, qquad H=begin{bmatrix} -Delta & 0 \ 0 & 1 end{bmatrix}.$$
Is this useful? Well, to begin with, it lets the abstract theory of semigroups come into play. (That theory is the generalization to the infinite-dimensional case of the equation $dot{x}= A x$, where $A$ is replaced with a linear operator on some Banach space). It also reveals a conservation law; if $boldsymbol{u}$ solves the wave equation then
$$
begin{split}
partial_t langle Hboldsymbol{u}(t), boldsymbol{u}(t)rangle &= langle HJHboldsymbol{u}| boldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle \ &= -langle Hboldsymbol{u}|JHboldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle =0,
end{split}
$$
where
$$
langle (f_1, g_1) |(f_2, g_2)rangle:= int f_1f_2 + int g_1g_2,$$
and where we used the fact that $H$ is a symmetric operator, while $J$ is an anti-symmetric one. We have thus shown that
$$
langle H boldsymbol{u}|boldsymbol{u}rangle=int -Delta u, u + int u_t^2
$$
is constant along the wave flow, which is a fundamental fact known as conservation of energy. (The first summand equals $int |nabla u|^2$, which is how it usually appears).
This is all very basic, but already shows that, sometimes, this kind of manipulation can be useful.
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
$endgroup$
This is connected to the Hamiltonian formalism, and is also done in PDEs sometimes. I am far from being an expert, so let me just present the case of the wave equation $u_{tt} = Delta u$. Introduce the two-component vector (position-momentum):
$$
boldsymbol{u}:=begin{bmatrix} u \ u_tend{bmatrix}.$$
Then the wave equation is rewritten as
$$
partial_t boldsymbol{u} = J H boldsymbol{u},qquad J=begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}, qquad H=begin{bmatrix} -Delta & 0 \ 0 & 1 end{bmatrix}.$$
Is this useful? Well, to begin with, it lets the abstract theory of semigroups come into play. (That theory is the generalization to the infinite-dimensional case of the equation $dot{x}= A x$, where $A$ is replaced with a linear operator on some Banach space). It also reveals a conservation law; if $boldsymbol{u}$ solves the wave equation then
$$
begin{split}
partial_t langle Hboldsymbol{u}(t), boldsymbol{u}(t)rangle &= langle HJHboldsymbol{u}| boldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle \ &= -langle Hboldsymbol{u}|JHboldsymbol{u}rangle + langle Hboldsymbol{u}|JHboldsymbol{u}rangle =0,
end{split}
$$
where
$$
langle (f_1, g_1) |(f_2, g_2)rangle:= int f_1f_2 + int g_1g_2,$$
and where we used the fact that $H$ is a symmetric operator, while $J$ is an anti-symmetric one. We have thus shown that
$$
langle H boldsymbol{u}|boldsymbol{u}rangle=int -Delta u, u + int u_t^2
$$
is constant along the wave flow, which is a fundamental fact known as conservation of energy. (The first summand equals $int |nabla u|^2$, which is how it usually appears).
This is all very basic, but already shows that, sometimes, this kind of manipulation can be useful.
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
answered Jan 23 at 15:30
Giuseppe NegroGiuseppe Negro
17.4k332126
17.4k332126
add a comment |
add a comment |
$begingroup$
It's not clear to me how this would work for PDE's. For ODE's:
Yes, any higher-order ODE is equivalent to a system of first-order ODE's. For example $$y''=f(y,y',t)$$is equivalent to the system $$y_1'=y_2, y_2'=f(y_1,y_2,t).$$I don't know that this is important for solving ODE's, but it's certainly useful in proving things about them. Because that system of two ODE"s is just $$Y'=F(Y,t)$$for a suitable $F$, where $Y=(y_1,y_2)$. Proofs of things for ODE's often work just as well for "vector-valued" ODE's, which then have corollaries for higher-order (scalar-valued) ODE's. Hmm, I suppose the same holds for numerical methods for approximate solutions.
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
add a comment |
$begingroup$
It's not clear to me how this would work for PDE's. For ODE's:
Yes, any higher-order ODE is equivalent to a system of first-order ODE's. For example $$y''=f(y,y',t)$$is equivalent to the system $$y_1'=y_2, y_2'=f(y_1,y_2,t).$$I don't know that this is important for solving ODE's, but it's certainly useful in proving things about them. Because that system of two ODE"s is just $$Y'=F(Y,t)$$for a suitable $F$, where $Y=(y_1,y_2)$. Proofs of things for ODE's often work just as well for "vector-valued" ODE's, which then have corollaries for higher-order (scalar-valued) ODE's. Hmm, I suppose the same holds for numerical methods for approximate solutions.
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
add a comment |
$begingroup$
It's not clear to me how this would work for PDE's. For ODE's:
Yes, any higher-order ODE is equivalent to a system of first-order ODE's. For example $$y''=f(y,y',t)$$is equivalent to the system $$y_1'=y_2, y_2'=f(y_1,y_2,t).$$I don't know that this is important for solving ODE's, but it's certainly useful in proving things about them. Because that system of two ODE"s is just $$Y'=F(Y,t)$$for a suitable $F$, where $Y=(y_1,y_2)$. Proofs of things for ODE's often work just as well for "vector-valued" ODE's, which then have corollaries for higher-order (scalar-valued) ODE's. Hmm, I suppose the same holds for numerical methods for approximate solutions.
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
It's not clear to me how this would work for PDE's. For ODE's:
Yes, any higher-order ODE is equivalent to a system of first-order ODE's. For example $$y''=f(y,y',t)$$is equivalent to the system $$y_1'=y_2, y_2'=f(y_1,y_2,t).$$I don't know that this is important for solving ODE's, but it's certainly useful in proving things about them. Because that system of two ODE"s is just $$Y'=F(Y,t)$$for a suitable $F$, where $Y=(y_1,y_2)$. Proofs of things for ODE's often work just as well for "vector-valued" ODE's, which then have corollaries for higher-order (scalar-valued) ODE's. Hmm, I suppose the same holds for numerical methods for approximate solutions.
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 23 at 15:03
David C. UllrichDavid C. Ullrich
61.2k43994
61.2k43994
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
add a comment |
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
They do this in evolution PDEs also, when they are interpreted as infinite-dimensional ODEs. (It can get messy, though...)
$endgroup$
– Giuseppe Negro
Jan 23 at 15:46
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
$begingroup$
This downvote (without explanation) is unfair. The question has been substantially edited after this answer.
$endgroup$
– Giuseppe Negro
Jan 23 at 18:42
add a comment |
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