Mixing
Relation between energy functionals of stationary and dynamic version of a PDE.
$begingroup$
Actually, I am dealing with a very specific equation, but It is rather complex and goes along with many notations and long calculations so I would like to ask a question in a conceptual spirit.
Given a PDE in some bounded Domain $Omega subseteq mathbb{R}^n$
$$
u_{tt} + A(u) = f(u) quad mbox{ in } Omega times (0, infty)\
B u = 0 quad mbox{ on } partial Omegatimes(0, infty)\
u(0) = u_0, u_t(0) = u_1 quad mbox{ in } Omega .
$$
where $A$ is a nonlinear differential operator in a function space, lets just say $H^2$ for example. And let the operator $B$ describe some boundary condition.
Additionally look at the stationary problem
$$
A(u) = f(u) quad mbox{ in } Omega\
B u = 0 quad mbox{ on } partial Omega.
$$
In order to derive an energy functional for the first problem(the time dependent one) just multiply the first equation with $u_t$ and integrate in $Omega$ and in time and use the standard calculations to end up with an energy of the form
$$
E_{{tiny dynamic}}(t) = h_0(u_t(t)) + h_1(u(t))
$$
where $h_0(u_t(t))$ and $h_1(u(t))$ just denote terms that depend on $u_t$ or $u$ respectively.
$underline{Question:}$ In order to get a suitable energy(compatible to the topology of space for weak solutions, ...) for the stationary problem is it okay just to delete the $h_0(u(t))$ term? I.e. to set
$$
E_{{tiny stationary}} = h_1(u).
$$
Should it be, at least topologically, the same result as when one multiplies the stationary problem with $u$ instead of $u_t$ and ends through usual computations up with
$$
tilde{E}_{{tiny stationary}} = h(u)
$$
where $h(u)$ is a term depending on $u$?
$underline{Motivation}$: People around me use the method of deleting the terms which are zero in the stationary case, starting off with the energy derived for the dynamical problem via multiplication by $u_t$. But I, when starting of with the stationary problem, and multiplying with $u$, end up with a different outcome then them.
$underline{Summary}$: According to my calculations the two methods
- multiply the dynamical problem with $u_t$ and then delete terms depending on $u_t$.
- multiply the stationary problem with $u$
can lead in the case of a given nonlinear(!) equation to different results.
Is that really true? Or should both approaches in general lead to the same result(at least in the sense of equivalent induced topologies by the two "energies")? Is there anyone who works with weak formulations and energy methods and maybe knows something to answer this question on the conceptual level?
Thanks a lot in advance for any input!
pde weak-topology
asked Jan 16 at 13:34
lessirlessir
816
$endgroup$
add a comment |
$begingroup$
Actually, I am dealing with a very specific equation, but It is rather complex and goes along with many notations and long calculations so I would like to ask a question in a conceptual spirit.
Given a PDE in some bounded Domain $Omega subseteq mathbb{R}^n$
$$
u_{tt} + A(u) = f(u) quad mbox{ in } Omega times (0, infty)\
B u = 0 quad mbox{ on } partial Omegatimes(0, infty)\
u(0) = u_0, u_t(0) = u_1 quad mbox{ in } Omega .
$$
where $A$ is a nonlinear differential operator in a function space, lets just say $H^2$ for example. And let the operator $B$ describe some boundary condition.
Additionally look at the stationary problem
$$
A(u) = f(u) quad mbox{ in } Omega\
B u = 0 quad mbox{ on } partial Omega.
$$
In order to derive an energy functional for the first problem(the time dependent one) just multiply the first equation with $u_t$ and integrate in $Omega$ and in time and use the standard calculations to end up with an energy of the form
$$
E_{{tiny dynamic}}(t) = h_0(u_t(t)) + h_1(u(t))
$$
where $h_0(u_t(t))$ and $h_1(u(t))$ just denote terms that depend on $u_t$ or $u$ respectively.
$underline{Question:}$ In order to get a suitable energy(compatible to the topology of space for weak solutions, ...) for the stationary problem is it okay just to delete the $h_0(u(t))$ term? I.e. to set
$$
E_{{tiny stationary}} = h_1(u).
$$
Should it be, at least topologically, the same result as when one multiplies the stationary problem with $u$ instead of $u_t$ and ends through usual computations up with
$$
tilde{E}_{{tiny stationary}} = h(u)
$$
where $h(u)$ is a term depending on $u$?
$underline{Motivation}$: People around me use the method of deleting the terms which are zero in the stationary case, starting off with the energy derived for the dynamical problem via multiplication by $u_t$. But I, when starting of with the stationary problem, and multiplying with $u$, end up with a different outcome then them.
$underline{Summary}$: According to my calculations the two methods
- multiply the dynamical problem with $u_t$ and then delete terms depending on $u_t$.
- multiply the stationary problem with $u$
can lead in the case of a given nonlinear(!) equation to different results.
Is that really true? Or should both approaches in general lead to the same result(at least in the sense of equivalent induced topologies by the two "energies")? Is there anyone who works with weak formulations and energy methods and maybe knows something to answer this question on the conceptual level?
Thanks a lot in advance for any input!
pde weak-topology
asked Jan 16 at 13:34
lessirlessir
816
$endgroup$
$begingroup$
Is the the answer maybe: In the stationary case the energy should be a functional that, when taking the Frechet derivative in $u$ results in the weak formulation, while an admissible energy functional in the dynamical case coincides directly with the weak formulation with testfunction taken to be $u_t$? So in the end it is legit to obtain the stationary energy from the dynamical problem by deleting $u_t$ terms but on the other hand it is not enough to just multiply the stationary equation by $u$ in order to get the stationary energy?
$endgroup$
– lessir
Jan 17 at 3:04
add a comment |
$begingroup$
Actually, I am dealing with a very specific equation, but It is rather complex and goes along with many notations and long calculations so I would like to ask a question in a conceptual spirit.
Given a PDE in some bounded Domain $Omega subseteq mathbb{R}^n$
$$
u_{tt} + A(u) = f(u) quad mbox{ in } Omega times (0, infty)\
B u = 0 quad mbox{ on } partial Omegatimes(0, infty)\
u(0) = u_0, u_t(0) = u_1 quad mbox{ in } Omega .
$$
where $A$ is a nonlinear differential operator in a function space, lets just say $H^2$ for example. And let the operator $B$ describe some boundary condition.
Additionally look at the stationary problem
$$
A(u) = f(u) quad mbox{ in } Omega\
B u = 0 quad mbox{ on } partial Omega.
$$
In order to derive an energy functional for the first problem(the time dependent one) just multiply the first equation with $u_t$ and integrate in $Omega$ and in time and use the standard calculations to end up with an energy of the form
$$
E_{{tiny dynamic}}(t) = h_0(u_t(t)) + h_1(u(t))
$$
where $h_0(u_t(t))$ and $h_1(u(t))$ just denote terms that depend on $u_t$ or $u$ respectively.
$underline{Question:}$ In order to get a suitable energy(compatible to the topology of space for weak solutions, ...) for the stationary problem is it okay just to delete the $h_0(u(t))$ term? I.e. to set
$$
E_{{tiny stationary}} = h_1(u).
$$
Should it be, at least topologically, the same result as when one multiplies the stationary problem with $u$ instead of $u_t$ and ends through usual computations up with
$$
tilde{E}_{{tiny stationary}} = h(u)
$$
where $h(u)$ is a term depending on $u$?
$underline{Motivation}$: People around me use the method of deleting the terms which are zero in the stationary case, starting off with the energy derived for the dynamical problem via multiplication by $u_t$. But I, when starting of with the stationary problem, and multiplying with $u$, end up with a different outcome then them.
$underline{Summary}$: According to my calculations the two methods
- multiply the dynamical problem with $u_t$ and then delete terms depending on $u_t$.
- multiply the stationary problem with $u$
can lead in the case of a given nonlinear(!) equation to different results.
Is that really true? Or should both approaches in general lead to the same result(at least in the sense of equivalent induced topologies by the two "energies")? Is there anyone who works with weak formulations and energy methods and maybe knows something to answer this question on the conceptual level?
Thanks a lot in advance for any input!
pde weak-topology
asked Jan 16 at 13:34
lessirlessir
816
$endgroup$
Actually, I am dealing with a very specific equation, but It is rather complex and goes along with many notations and long calculations so I would like to ask a question in a conceptual spirit.
Given a PDE in some bounded Domain $Omega subseteq mathbb{R}^n$
$$
u_{tt} + A(u) = f(u) quad mbox{ in } Omega times (0, infty)\
B u = 0 quad mbox{ on } partial Omegatimes(0, infty)\
u(0) = u_0, u_t(0) = u_1 quad mbox{ in } Omega .
$$
where $A$ is a nonlinear differential operator in a function space, lets just say $H^2$ for example. And let the operator $B$ describe some boundary condition.
Additionally look at the stationary problem
$$
A(u) = f(u) quad mbox{ in } Omega\
B u = 0 quad mbox{ on } partial Omega.
$$
In order to derive an energy functional for the first problem(the time dependent one) just multiply the first equation with $u_t$ and integrate in $Omega$ and in time and use the standard calculations to end up with an energy of the form
$$
E_{{tiny dynamic}}(t) = h_0(u_t(t)) + h_1(u(t))
$$
where $h_0(u_t(t))$ and $h_1(u(t))$ just denote terms that depend on $u_t$ or $u$ respectively.
$underline{Question:}$ In order to get a suitable energy(compatible to the topology of space for weak solutions, ...) for the stationary problem is it okay just to delete the $h_0(u(t))$ term? I.e. to set
$$
E_{{tiny stationary}} = h_1(u).
$$
Should it be, at least topologically, the same result as when one multiplies the stationary problem with $u$ instead of $u_t$ and ends through usual computations up with
$$
tilde{E}_{{tiny stationary}} = h(u)
$$
where $h(u)$ is a term depending on $u$?
$underline{Motivation}$: People around me use the method of deleting the terms which are zero in the stationary case, starting off with the energy derived for the dynamical problem via multiplication by $u_t$. But I, when starting of with the stationary problem, and multiplying with $u$, end up with a different outcome then them.
$underline{Summary}$: According to my calculations the two methods
- multiply the dynamical problem with $u_t$ and then delete terms depending on $u_t$.
- multiply the stationary problem with $u$
can lead in the case of a given nonlinear(!) equation to different results.
Is that really true? Or should both approaches in general lead to the same result(at least in the sense of equivalent induced topologies by the two "energies")? Is there anyone who works with weak formulations and energy methods and maybe knows something to answer this question on the conceptual level?
Thanks a lot in advance for any input!
pde weak-topology
pde weak-topology
asked Jan 16 at 13:34
lessirlessir
816
asked Jan 16 at 13:34
lessirlessir
816
asked Jan 16 at 13:34
lessirlessir
816
asked Jan 16 at 13:34
lessirlessir
816
asked Jan 16 at 13:34
lessirlessir
816
816
$begingroup$
Is the the answer maybe: In the stationary case the energy should be a functional that, when taking the Frechet derivative in $u$ results in the weak formulation, while an admissible energy functional in the dynamical case coincides directly with the weak formulation with testfunction taken to be $u_t$? So in the end it is legit to obtain the stationary energy from the dynamical problem by deleting $u_t$ terms but on the other hand it is not enough to just multiply the stationary equation by $u$ in order to get the stationary energy?
$endgroup$
– lessir
Jan 17 at 3:04
add a comment |
$begingroup$
Is the the answer maybe: In the stationary case the energy should be a functional that, when taking the Frechet derivative in $u$ results in the weak formulation, while an admissible energy functional in the dynamical case coincides directly with the weak formulation with testfunction taken to be $u_t$? So in the end it is legit to obtain the stationary energy from the dynamical problem by deleting $u_t$ terms but on the other hand it is not enough to just multiply the stationary equation by $u$ in order to get the stationary energy?
$endgroup$
– lessir
Jan 17 at 3:04
$begingroup$
Is the the answer maybe: In the stationary case the energy should be a functional that, when taking the Frechet derivative in $u$ results in the weak formulation, while an admissible energy functional in the dynamical case coincides directly with the weak formulation with testfunction taken to be $u_t$? So in the end it is legit to obtain the stationary energy from the dynamical problem by deleting $u_t$ terms but on the other hand it is not enough to just multiply the stationary equation by $u$ in order to get the stationary energy?
$endgroup$
– lessir
Jan 17 at 3:04
$begingroup$
Is the the answer maybe: In the stationary case the energy should be a functional that, when taking the Frechet derivative in $u$ results in the weak formulation, while an admissible energy functional in the dynamical case coincides directly with the weak formulation with testfunction taken to be $u_t$? So in the end it is legit to obtain the stationary energy from the dynamical problem by deleting $u_t$ terms but on the other hand it is not enough to just multiply the stationary equation by $u$ in order to get the stationary energy?
$endgroup$
– lessir
Jan 17 at 3:04
add a comment |
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$begingroup$
Is the the answer maybe: In the stationary case the energy should be a functional that, when taking the Frechet derivative in $u$ results in the weak formulation, while an admissible energy functional in the dynamical case coincides directly with the weak formulation with testfunction taken to be $u_t$? So in the end it is legit to obtain the stationary energy from the dynamical problem by deleting $u_t$ terms but on the other hand it is not enough to just multiply the stationary equation by $u$ in order to get the stationary energy?
$endgroup$
– lessir
Jan 17 at 3:04