Mixing
Showing that the energy of the wave equation in $mathbb R^d$ is constant
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Exercise on Stein's/Sharkachi book chapter 6:
Let $u(x, t)$ be a smooth solution of the wave equation and let $E(t)$ denote the energy of this wave
$$E(t) = int_{mathbb R^d} bigg|frac{partial u(x,t)}{partial t}bigg|^2+sum_{j=1}^d int_{mathbb R^d}bigg | frac{partial u(x,t)}{partial x_j}bigg |^2 dx.$$
We have seen that $E(t)$ is constant using Plancherel’s formula. One can give an alternate proof
of this fact by differentiating the integral with respect to $t$ and showing that $frac{dE}{dt} = 0.$
I remember that when we were in the 1-dimensional case, to prove a similar formula (which didn't involved modules back then), we had to multiply $u_{tt}= u_{xx}$ by $u_t$, integrate with respect to $x$ and then realize integration by parts. Noticing that $1/2(u_t^2)_t = u_{tt}u_t $ and doing similarly for $u_{xx}u_t$, we would arrive that: $frac{d}{dt}(1/2int u_t^2dx+1/2int u_x^2dx ) =0 $ which implies the conservation of energy.
So the hint given for the problem in $mathbb R^d$ is the same: use integration by parts. But I dunno exactly what is the d-dimensional analog of integration by parts. How can I solve this problem?
pde
asked 2 days ago
math.h
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Exercise on Stein's/Sharkachi book chapter 6:
Let $u(x, t)$ be a smooth solution of the wave equation and let $E(t)$ denote the energy of this wave
$$E(t) = int_{mathbb R^d} bigg|frac{partial u(x,t)}{partial t}bigg|^2+sum_{j=1}^d int_{mathbb R^d}bigg | frac{partial u(x,t)}{partial x_j}bigg |^2 dx.$$
We have seen that $E(t)$ is constant using Plancherel’s formula. One can give an alternate proof
of this fact by differentiating the integral with respect to $t$ and showing that $frac{dE}{dt} = 0.$
I remember that when we were in the 1-dimensional case, to prove a similar formula (which didn't involved modules back then), we had to multiply $u_{tt}= u_{xx}$ by $u_t$, integrate with respect to $x$ and then realize integration by parts. Noticing that $1/2(u_t^2)_t = u_{tt}u_t $ and doing similarly for $u_{xx}u_t$, we would arrive that: $frac{d}{dt}(1/2int u_t^2dx+1/2int u_x^2dx ) =0 $ which implies the conservation of energy.
So the hint given for the problem in $mathbb R^d$ is the same: use integration by parts. But I dunno exactly what is the d-dimensional analog of integration by parts. How can I solve this problem?
pde
asked 2 days ago
math.h
1,065517
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Exercise on Stein's/Sharkachi book chapter 6:
Let $u(x, t)$ be a smooth solution of the wave equation and let $E(t)$ denote the energy of this wave
$$E(t) = int_{mathbb R^d} bigg|frac{partial u(x,t)}{partial t}bigg|^2+sum_{j=1}^d int_{mathbb R^d}bigg | frac{partial u(x,t)}{partial x_j}bigg |^2 dx.$$
We have seen that $E(t)$ is constant using Plancherel’s formula. One can give an alternate proof
of this fact by differentiating the integral with respect to $t$ and showing that $frac{dE}{dt} = 0.$
I remember that when we were in the 1-dimensional case, to prove a similar formula (which didn't involved modules back then), we had to multiply $u_{tt}= u_{xx}$ by $u_t$, integrate with respect to $x$ and then realize integration by parts. Noticing that $1/2(u_t^2)_t = u_{tt}u_t $ and doing similarly for $u_{xx}u_t$, we would arrive that: $frac{d}{dt}(1/2int u_t^2dx+1/2int u_x^2dx ) =0 $ which implies the conservation of energy.
So the hint given for the problem in $mathbb R^d$ is the same: use integration by parts. But I dunno exactly what is the d-dimensional analog of integration by parts. How can I solve this problem?
pde
asked 2 days ago
math.h
1,065517
Exercise on Stein's/Sharkachi book chapter 6:
Let $u(x, t)$ be a smooth solution of the wave equation and let $E(t)$ denote the energy of this wave
$$E(t) = int_{mathbb R^d} bigg|frac{partial u(x,t)}{partial t}bigg|^2+sum_{j=1}^d int_{mathbb R^d}bigg | frac{partial u(x,t)}{partial x_j}bigg |^2 dx.$$
We have seen that $E(t)$ is constant using Plancherel’s formula. One can give an alternate proof
of this fact by differentiating the integral with respect to $t$ and showing that $frac{dE}{dt} = 0.$
I remember that when we were in the 1-dimensional case, to prove a similar formula (which didn't involved modules back then), we had to multiply $u_{tt}= u_{xx}$ by $u_t$, integrate with respect to $x$ and then realize integration by parts. Noticing that $1/2(u_t^2)_t = u_{tt}u_t $ and doing similarly for $u_{xx}u_t$, we would arrive that: $frac{d}{dt}(1/2int u_t^2dx+1/2int u_x^2dx ) =0 $ which implies the conservation of energy.
So the hint given for the problem in $mathbb R^d$ is the same: use integration by parts. But I dunno exactly what is the d-dimensional analog of integration by parts. How can I solve this problem?
pde
pde
asked 2 days ago
math.h
1,065517
asked 2 days ago
math.h
1,065517
asked 2 days ago
math.h
1,065517
asked 2 days ago
math.h
1,065517
asked 2 days ago
math.h
1,065517
1,065517
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1 Answer
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This is a typical energy method approach to use this to show the wave equation have but one solution. Rewrite your integral as
$$E(t) = int_{mathbb{R}^d} u_{t}^2 + |nabla u|^2 dx$$
Then
$$E'(t) = int_{mathbb{R}^d} 2u_{t}u_{tt} + 2(nabla u cdot nabla u_t) dx$$
Here is the step where IBP is performed. Since you're curious, in higher dimension, the formula is
$$int_U u_{x_{i}}v dx = -int_U uv_{x_{i}}dx + int_{partial U}uvnu^{i} dS$$
for $i = 1,2,...n$ and $nu^i$ is the $ith$ component of the outward pointing unit normal vector of $U$, an open set bounded in $mathbb{R}^n$. Applying it here we get
$$int_{mathbb{R}^d}nabla u cdot nabla u_t ,dx = -int_{mathbb{R}^d}u_t Delta u , dx$$
since the boundary of $Bbb{R}^n$ is empty. Hence
$$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$
answered yesterday
DaveNine
1,209914
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is a typical energy method approach to use this to show the wave equation have but one solution. Rewrite your integral as
$$E(t) = int_{mathbb{R}^d} u_{t}^2 + |nabla u|^2 dx$$
Then
$$E'(t) = int_{mathbb{R}^d} 2u_{t}u_{tt} + 2(nabla u cdot nabla u_t) dx$$
Here is the step where IBP is performed. Since you're curious, in higher dimension, the formula is
$$int_U u_{x_{i}}v dx = -int_U uv_{x_{i}}dx + int_{partial U}uvnu^{i} dS$$
for $i = 1,2,...n$ and $nu^i$ is the $ith$ component of the outward pointing unit normal vector of $U$, an open set bounded in $mathbb{R}^n$. Applying it here we get
$$int_{mathbb{R}^d}nabla u cdot nabla u_t ,dx = -int_{mathbb{R}^d}u_t Delta u , dx$$
since the boundary of $Bbb{R}^n$ is empty. Hence
$$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$
answered yesterday
DaveNine
1,209914
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
add a comment |
up vote
1
down vote
This is a typical energy method approach to use this to show the wave equation have but one solution. Rewrite your integral as
$$E(t) = int_{mathbb{R}^d} u_{t}^2 + |nabla u|^2 dx$$
Then
$$E'(t) = int_{mathbb{R}^d} 2u_{t}u_{tt} + 2(nabla u cdot nabla u_t) dx$$
Here is the step where IBP is performed. Since you're curious, in higher dimension, the formula is
$$int_U u_{x_{i}}v dx = -int_U uv_{x_{i}}dx + int_{partial U}uvnu^{i} dS$$
for $i = 1,2,...n$ and $nu^i$ is the $ith$ component of the outward pointing unit normal vector of $U$, an open set bounded in $mathbb{R}^n$. Applying it here we get
$$int_{mathbb{R}^d}nabla u cdot nabla u_t ,dx = -int_{mathbb{R}^d}u_t Delta u , dx$$
since the boundary of $Bbb{R}^n$ is empty. Hence
$$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$
answered yesterday
DaveNine
1,209914
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
This is a typical energy method approach to use this to show the wave equation have but one solution. Rewrite your integral as
$$E(t) = int_{mathbb{R}^d} u_{t}^2 + |nabla u|^2 dx$$
Then
$$E'(t) = int_{mathbb{R}^d} 2u_{t}u_{tt} + 2(nabla u cdot nabla u_t) dx$$
Here is the step where IBP is performed. Since you're curious, in higher dimension, the formula is
$$int_U u_{x_{i}}v dx = -int_U uv_{x_{i}}dx + int_{partial U}uvnu^{i} dS$$
for $i = 1,2,...n$ and $nu^i$ is the $ith$ component of the outward pointing unit normal vector of $U$, an open set bounded in $mathbb{R}^n$. Applying it here we get
$$int_{mathbb{R}^d}nabla u cdot nabla u_t ,dx = -int_{mathbb{R}^d}u_t Delta u , dx$$
since the boundary of $Bbb{R}^n$ is empty. Hence
$$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$
answered yesterday
DaveNine
1,209914
This is a typical energy method approach to use this to show the wave equation have but one solution. Rewrite your integral as
$$E(t) = int_{mathbb{R}^d} u_{t}^2 + |nabla u|^2 dx$$
Then
$$E'(t) = int_{mathbb{R}^d} 2u_{t}u_{tt} + 2(nabla u cdot nabla u_t) dx$$
Here is the step where IBP is performed. Since you're curious, in higher dimension, the formula is
$$int_U u_{x_{i}}v dx = -int_U uv_{x_{i}}dx + int_{partial U}uvnu^{i} dS$$
for $i = 1,2,...n$ and $nu^i$ is the $ith$ component of the outward pointing unit normal vector of $U$, an open set bounded in $mathbb{R}^n$. Applying it here we get
$$int_{mathbb{R}^d}nabla u cdot nabla u_t ,dx = -int_{mathbb{R}^d}u_t Delta u , dx$$
since the boundary of $Bbb{R}^n$ is empty. Hence
$$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$
answered yesterday
DaveNine
1,209914
edited yesterday
answered yesterday
DaveNine
1,209914
answered yesterday
DaveNine
1,209914
answered yesterday
DaveNine
1,209914
1,209914
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
add a comment |
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
– math.h
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
Ah, I made an error, I will edit.
– DaveNine
yesterday
add a comment |
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