Mixing
Use Energy Method to prove the uniqueness of the initial boundary value problem
$begingroup$
I am really stuck in this proof. Any hint or suggestion from you would be appreciated.
Let $U subset Bbb{R^{n}}$ be open, bounded, with smooth boundary $partial U$, and $ T gt 0$. Use energy method to prove that the initial boundary-value problem
$u_{t}-Delta u =f$ in $U_{T}$
$frac{partial u}{partial nu}=0$ on $partial U times [0,T]$
$u=g$ on $U times {t=0}$
,where $f=f(x,t),g=g(x) $ are given smooth functions, has at most one solution $u in C^{2}_{1}$ $(overline U_{T})$
pde
edited Jan 25 at 7:55
dmtri
1,6792521
asked Dec 6 '13 at 15:25
YangYang
206417
$endgroup$
add a comment |
$begingroup$
I am really stuck in this proof. Any hint or suggestion from you would be appreciated.
Let $U subset Bbb{R^{n}}$ be open, bounded, with smooth boundary $partial U$, and $ T gt 0$. Use energy method to prove that the initial boundary-value problem
$u_{t}-Delta u =f$ in $U_{T}$
$frac{partial u}{partial nu}=0$ on $partial U times [0,T]$
$u=g$ on $U times {t=0}$
,where $f=f(x,t),g=g(x) $ are given smooth functions, has at most one solution $u in C^{2}_{1}$ $(overline U_{T})$
pde
edited Jan 25 at 7:55
dmtri
1,6792521
asked Dec 6 '13 at 15:25
YangYang
206417
$endgroup$
add a comment |
$begingroup$
I am really stuck in this proof. Any hint or suggestion from you would be appreciated.
Let $U subset Bbb{R^{n}}$ be open, bounded, with smooth boundary $partial U$, and $ T gt 0$. Use energy method to prove that the initial boundary-value problem
$u_{t}-Delta u =f$ in $U_{T}$
$frac{partial u}{partial nu}=0$ on $partial U times [0,T]$
$u=g$ on $U times {t=0}$
,where $f=f(x,t),g=g(x) $ are given smooth functions, has at most one solution $u in C^{2}_{1}$ $(overline U_{T})$
pde
edited Jan 25 at 7:55
dmtri
1,6792521
asked Dec 6 '13 at 15:25
YangYang
206417
$endgroup$
I am really stuck in this proof. Any hint or suggestion from you would be appreciated.
Let $U subset Bbb{R^{n}}$ be open, bounded, with smooth boundary $partial U$, and $ T gt 0$. Use energy method to prove that the initial boundary-value problem
$u_{t}-Delta u =f$ in $U_{T}$
$frac{partial u}{partial nu}=0$ on $partial U times [0,T]$
$u=g$ on $U times {t=0}$
,where $f=f(x,t),g=g(x) $ are given smooth functions, has at most one solution $u in C^{2}_{1}$ $(overline U_{T})$
pde
pde
edited Jan 25 at 7:55
dmtri
1,6792521
asked Dec 6 '13 at 15:25
YangYang
206417
edited Jan 25 at 7:55
dmtri
1,6792521
asked Dec 6 '13 at 15:25
YangYang
206417
edited Jan 25 at 7:55
dmtri
1,6792521
edited Jan 25 at 7:55
dmtri
1,6792521
edited Jan 25 at 7:55
dmtri
1,6792521
1,6792521
asked Dec 6 '13 at 15:25
YangYang
206417
asked Dec 6 '13 at 15:25
YangYang
206417
asked Dec 6 '13 at 15:25
YangYang
206417
206417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-Delta u$ against $u$, for example, and use something like $ u Delta u = nablacdot (unabla u)-nabla ucdotnabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.
So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies
$$
u_{t}-Delta u = 0,;;; frac{partial u}{partialnu}=0 mbox{ on } partial U times [0,T], ;;; u = 0 mbox{ on } U times { t= 0}.
$$
Therefore,
$$
0=int_{U}uu_{t}-uDelta u,dx = int_{U}uu_{t}-nablacdot(unabla u)+|nabla u|^{2},dx
$$
$$ = int_{U}uu_{t}+|nabla u|^{2},dx-int_{partial U}ufrac{partial u}{partialnu},dS(x),
$$
where $dS$ is the surface area element on $partial U$. (This is Gauss' Divergence Theorem.) Because $frac{partial u}{partial nu}=0$ on $partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes
$$
0 = int_{U}frac{1}{2}frac{partial}{partial t}u^{2}+|nabla u|^{2}dx
$$
Now integrate the above on $[0,t']$ for any $0 le t' le T$, and use the fact that $u(x,t)|_{t=0}=0$:
$$
0 = int_{U}frac{1}{2}|u(x,t')|^{2},dx +int_{0}^{t'}int_{U}|nabla u(x,t)|^{2}dxdt.
$$
Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x in U$ and $0 le t' le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
add a comment |
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1 Answer
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1 Answer
1
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oldest
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votes
$begingroup$
HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-Delta u$ against $u$, for example, and use something like $ u Delta u = nablacdot (unabla u)-nabla ucdotnabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.
So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies
$$
u_{t}-Delta u = 0,;;; frac{partial u}{partialnu}=0 mbox{ on } partial U times [0,T], ;;; u = 0 mbox{ on } U times { t= 0}.
$$
Therefore,
$$
0=int_{U}uu_{t}-uDelta u,dx = int_{U}uu_{t}-nablacdot(unabla u)+|nabla u|^{2},dx
$$
$$ = int_{U}uu_{t}+|nabla u|^{2},dx-int_{partial U}ufrac{partial u}{partialnu},dS(x),
$$
where $dS$ is the surface area element on $partial U$. (This is Gauss' Divergence Theorem.) Because $frac{partial u}{partial nu}=0$ on $partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes
$$
0 = int_{U}frac{1}{2}frac{partial}{partial t}u^{2}+|nabla u|^{2}dx
$$
Now integrate the above on $[0,t']$ for any $0 le t' le T$, and use the fact that $u(x,t)|_{t=0}=0$:
$$
0 = int_{U}frac{1}{2}|u(x,t')|^{2},dx +int_{0}^{t'}int_{U}|nabla u(x,t)|^{2}dxdt.
$$
Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x in U$ and $0 le t' le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
add a comment |
$begingroup$
HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-Delta u$ against $u$, for example, and use something like $ u Delta u = nablacdot (unabla u)-nabla ucdotnabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.
So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies
$$
u_{t}-Delta u = 0,;;; frac{partial u}{partialnu}=0 mbox{ on } partial U times [0,T], ;;; u = 0 mbox{ on } U times { t= 0}.
$$
Therefore,
$$
0=int_{U}uu_{t}-uDelta u,dx = int_{U}uu_{t}-nablacdot(unabla u)+|nabla u|^{2},dx
$$
$$ = int_{U}uu_{t}+|nabla u|^{2},dx-int_{partial U}ufrac{partial u}{partialnu},dS(x),
$$
where $dS$ is the surface area element on $partial U$. (This is Gauss' Divergence Theorem.) Because $frac{partial u}{partial nu}=0$ on $partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes
$$
0 = int_{U}frac{1}{2}frac{partial}{partial t}u^{2}+|nabla u|^{2}dx
$$
Now integrate the above on $[0,t']$ for any $0 le t' le T$, and use the fact that $u(x,t)|_{t=0}=0$:
$$
0 = int_{U}frac{1}{2}|u(x,t')|^{2},dx +int_{0}^{t'}int_{U}|nabla u(x,t)|^{2}dxdt.
$$
Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x in U$ and $0 le t' le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
add a comment |
$begingroup$
HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-Delta u$ against $u$, for example, and use something like $ u Delta u = nablacdot (unabla u)-nabla ucdotnabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.
So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies
$$
u_{t}-Delta u = 0,;;; frac{partial u}{partialnu}=0 mbox{ on } partial U times [0,T], ;;; u = 0 mbox{ on } U times { t= 0}.
$$
Therefore,
$$
0=int_{U}uu_{t}-uDelta u,dx = int_{U}uu_{t}-nablacdot(unabla u)+|nabla u|^{2},dx
$$
$$ = int_{U}uu_{t}+|nabla u|^{2},dx-int_{partial U}ufrac{partial u}{partialnu},dS(x),
$$
where $dS$ is the surface area element on $partial U$. (This is Gauss' Divergence Theorem.) Because $frac{partial u}{partial nu}=0$ on $partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes
$$
0 = int_{U}frac{1}{2}frac{partial}{partial t}u^{2}+|nabla u|^{2}dx
$$
Now integrate the above on $[0,t']$ for any $0 le t' le T$, and use the fact that $u(x,t)|_{t=0}=0$:
$$
0 = int_{U}frac{1}{2}|u(x,t')|^{2},dx +int_{0}^{t'}int_{U}|nabla u(x,t)|^{2}dxdt.
$$
Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x in U$ and $0 le t' le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-Delta u$ against $u$, for example, and use something like $ u Delta u = nablacdot (unabla u)-nabla ucdotnabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.
So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies
$$
u_{t}-Delta u = 0,;;; frac{partial u}{partialnu}=0 mbox{ on } partial U times [0,T], ;;; u = 0 mbox{ on } U times { t= 0}.
$$
Therefore,
$$
0=int_{U}uu_{t}-uDelta u,dx = int_{U}uu_{t}-nablacdot(unabla u)+|nabla u|^{2},dx
$$
$$ = int_{U}uu_{t}+|nabla u|^{2},dx-int_{partial U}ufrac{partial u}{partialnu},dS(x),
$$
where $dS$ is the surface area element on $partial U$. (This is Gauss' Divergence Theorem.) Because $frac{partial u}{partial nu}=0$ on $partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes
$$
0 = int_{U}frac{1}{2}frac{partial}{partial t}u^{2}+|nabla u|^{2}dx
$$
Now integrate the above on $[0,t']$ for any $0 le t' le T$, and use the fact that $u(x,t)|_{t=0}=0$:
$$
0 = int_{U}frac{1}{2}|u(x,t')|^{2},dx +int_{0}^{t'}int_{U}|nabla u(x,t)|^{2}dxdt.
$$
Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x in U$ and $0 le t' le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
edited Dec 8 '13 at 8:20
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Dec 6 '13 at 16:02
DisintegratingByPartsDisintegratingByParts
59.8k42681
59.8k42681
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
add a comment |
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Thanks. However, I have a question. I know how to show $u=u_1-u_2 =0$ in $U_{T}$ and $U times {t=0}$. Can you give me some ideas on how to show u is zero on another boundary. Thank you@T.A.E.
$endgroup$
– Yang
Dec 6 '13 at 16:18
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
Once you know that the integral of a continuous function is 0 over the open region, then you know that the function must be 0. For example, if you can show that the square of the length of the gradient is 0 over the open region, then the gradient must be 0 because it is assumed continuous. So you conclude that the function must be a constant because all of its first order derivatives are 0. You can use line integrals to conclude this. Etc.
$endgroup$
– DisintegratingByParts
Dec 6 '13 at 19:51
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
I noticed that I wrote the integral of a continuous function when I meant integral of the square of a continuous function. Big difference, and the latter applies to your case.
$endgroup$
– DisintegratingByParts
Dec 7 '13 at 22:24
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
Can I ask you how I can apply the energy method to do this proof? Thanks@T.A.E.
$endgroup$
– Yang
Dec 8 '13 at 4:14
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
$begingroup$
I added more detail for you.
$endgroup$
– DisintegratingByParts
Dec 8 '13 at 7:01
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
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Post as a guest
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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