Estimate the large-time behavior of the unique viscosity solution
I am looking for some help to determine the large-time behavior of the unique solution for the equation in $mathbb R^+ times mathbb R$ $$u_t+vertnabla uvert^frac{2}{3}=0, u(0,x)=-cos x$$
More specifically, I am thinking about how to determine the behavior of u at $(2015,0)$ or $(n,0)$ for any large integer $n$, but I have no idea that where to start.
Many thanks to the help in advance.
pde hamilton-jacobi-equation viscosity-solutions
edited Dec 10 at 9:14
Harry49
5,99121031
asked Dec 9 at 9:58
Sam Wong
326110
add a comment |
I am looking for some help to determine the large-time behavior of the unique solution for the equation in $mathbb R^+ times mathbb R$ $$u_t+vertnabla uvert^frac{2}{3}=0, u(0,x)=-cos x$$
More specifically, I am thinking about how to determine the behavior of u at $(2015,0)$ or $(n,0)$ for any large integer $n$, but I have no idea that where to start.
Many thanks to the help in advance.
pde hamilton-jacobi-equation viscosity-solutions
edited Dec 10 at 9:14
Harry49
5,99121031
asked Dec 9 at 9:58
Sam Wong
326110
add a comment |
I am looking for some help to determine the large-time behavior of the unique solution for the equation in $mathbb R^+ times mathbb R$ $$u_t+vertnabla uvert^frac{2}{3}=0, u(0,x)=-cos x$$
More specifically, I am thinking about how to determine the behavior of u at $(2015,0)$ or $(n,0)$ for any large integer $n$, but I have no idea that where to start.
Many thanks to the help in advance.
pde hamilton-jacobi-equation viscosity-solutions
edited Dec 10 at 9:14
Harry49
5,99121031
asked Dec 9 at 9:58
Sam Wong
326110
I am looking for some help to determine the large-time behavior of the unique solution for the equation in $mathbb R^+ times mathbb R$ $$u_t+vertnabla uvert^frac{2}{3}=0, u(0,x)=-cos x$$
More specifically, I am thinking about how to determine the behavior of u at $(2015,0)$ or $(n,0)$ for any large integer $n$, but I have no idea that where to start.
Many thanks to the help in advance.
pde hamilton-jacobi-equation viscosity-solutions
pde hamilton-jacobi-equation viscosity-solutions
edited Dec 10 at 9:14
Harry49
5,99121031
asked Dec 9 at 9:58
Sam Wong
326110
edited Dec 10 at 9:14
Harry49
5,99121031
asked Dec 9 at 9:58
Sam Wong
326110
edited Dec 10 at 9:14
Harry49
5,99121031
edited Dec 10 at 9:14
Harry49
5,99121031
edited Dec 10 at 9:14
Harry49
5,99121031
5,99121031
asked Dec 9 at 9:58
Sam Wong
326110
asked Dec 9 at 9:58
Sam Wong
326110
asked Dec 9 at 9:58
Sam Wong
326110
326110
add a comment |
add a comment |
1 Answer
1
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oldest
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Long comment: The Hamilton-Jacobi PDE is derived from a canonical transformation involving a type-2 generating function $u(t,x)$ which makes vanish the new Hamiltonian
$K=H(u_x) + u_t$. Here, $H(p)=|p|^{2/3}$ is the original Hamiltonian.
For convex Hamiltonians $H$ which grow faster than the identity, the unique continuous viscosity solution can be obtained via the Lax-Hopf formula
$$
u(t,x) = inf_{yin Bbb R} leftlbrace t L big(tfrac{x-y}{t}big) - cos yrightrbrace
$$
where the Lagrangian $L$ is the Legendre-Fenchel transform of $H$:
$$
L(q) = sup_{pinBbb R} lbrace qp - H(p)rbrace = frac{4}{27}|q|^3 .
$$
However, the Hamiltonian is nonconvex here, which invalidates the previous approach. One may have a look at this post for complements.
answered Dec 9 at 11:13
Harry49
5,99121031
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
Long comment: The Hamilton-Jacobi PDE is derived from a canonical transformation involving a type-2 generating function $u(t,x)$ which makes vanish the new Hamiltonian
$K=H(u_x) + u_t$. Here, $H(p)=|p|^{2/3}$ is the original Hamiltonian.
For convex Hamiltonians $H$ which grow faster than the identity, the unique continuous viscosity solution can be obtained via the Lax-Hopf formula
$$
u(t,x) = inf_{yin Bbb R} leftlbrace t L big(tfrac{x-y}{t}big) - cos yrightrbrace
$$
where the Lagrangian $L$ is the Legendre-Fenchel transform of $H$:
$$
L(q) = sup_{pinBbb R} lbrace qp - H(p)rbrace = frac{4}{27}|q|^3 .
$$
However, the Hamiltonian is nonconvex here, which invalidates the previous approach. One may have a look at this post for complements.
answered Dec 9 at 11:13
Harry49
5,99121031
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
|
show 3 more comments
Long comment: The Hamilton-Jacobi PDE is derived from a canonical transformation involving a type-2 generating function $u(t,x)$ which makes vanish the new Hamiltonian
$K=H(u_x) + u_t$. Here, $H(p)=|p|^{2/3}$ is the original Hamiltonian.
For convex Hamiltonians $H$ which grow faster than the identity, the unique continuous viscosity solution can be obtained via the Lax-Hopf formula
$$
u(t,x) = inf_{yin Bbb R} leftlbrace t L big(tfrac{x-y}{t}big) - cos yrightrbrace
$$
where the Lagrangian $L$ is the Legendre-Fenchel transform of $H$:
$$
L(q) = sup_{pinBbb R} lbrace qp - H(p)rbrace = frac{4}{27}|q|^3 .
$$
However, the Hamiltonian is nonconvex here, which invalidates the previous approach. One may have a look at this post for complements.
answered Dec 9 at 11:13
Harry49
5,99121031
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
|
show 3 more comments
Long comment: The Hamilton-Jacobi PDE is derived from a canonical transformation involving a type-2 generating function $u(t,x)$ which makes vanish the new Hamiltonian
$K=H(u_x) + u_t$. Here, $H(p)=|p|^{2/3}$ is the original Hamiltonian.
For convex Hamiltonians $H$ which grow faster than the identity, the unique continuous viscosity solution can be obtained via the Lax-Hopf formula
$$
u(t,x) = inf_{yin Bbb R} leftlbrace t L big(tfrac{x-y}{t}big) - cos yrightrbrace
$$
where the Lagrangian $L$ is the Legendre-Fenchel transform of $H$:
$$
L(q) = sup_{pinBbb R} lbrace qp - H(p)rbrace = frac{4}{27}|q|^3 .
$$
However, the Hamiltonian is nonconvex here, which invalidates the previous approach. One may have a look at this post for complements.
answered Dec 9 at 11:13
Harry49
5,99121031
Long comment: The Hamilton-Jacobi PDE is derived from a canonical transformation involving a type-2 generating function $u(t,x)$ which makes vanish the new Hamiltonian
$K=H(u_x) + u_t$. Here, $H(p)=|p|^{2/3}$ is the original Hamiltonian.
For convex Hamiltonians $H$ which grow faster than the identity, the unique continuous viscosity solution can be obtained via the Lax-Hopf formula
$$
u(t,x) = inf_{yin Bbb R} leftlbrace t L big(tfrac{x-y}{t}big) - cos yrightrbrace
$$
where the Lagrangian $L$ is the Legendre-Fenchel transform of $H$:
$$
L(q) = sup_{pinBbb R} lbrace qp - H(p)rbrace = frac{4}{27}|q|^3 .
$$
However, the Hamiltonian is nonconvex here, which invalidates the previous approach. One may have a look at this post for complements.
answered Dec 9 at 11:13
Harry49
5,99121031
edited Dec 20 at 9:20
answered Dec 9 at 11:13
Harry49
5,99121031
answered Dec 9 at 11:13
Harry49
5,99121031
answered Dec 9 at 11:13
Harry49
5,99121031
5,99121031
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
|
show 3 more comments
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:)
– Sam Wong
Dec 10 at 2:27
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Thank you so much. It's clear to me now :)
– Sam Wong
Dec 10 at 9:21
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $nge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong?
– Sam Wong
Dec 10 at 22:39
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thanks for your hint, I will reconsider this problem :)
– Sam Wong
Dec 11 at 10:08
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
Thank you so much:)
– Sam Wong
Dec 11 at 14:03
|
show 3 more comments
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