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?










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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?










    share|cite|improve this question
























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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?










      share|cite|improve this question













      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






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      asked Nov 19 at 18:22









      math.h

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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



          $$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$






          share|cite|improve this answer























          • Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
            – math.h
            Nov 20 at 11:42










          • Ah, I made an error, I will edit.
            – DaveNine
            Nov 20 at 12:55











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          up vote
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          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



          $$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$






          share|cite|improve this answer























          • Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
            – math.h
            Nov 20 at 11:42










          • Ah, I made an error, I will edit.
            – DaveNine
            Nov 20 at 12:55















          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



          $$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$






          share|cite|improve this answer























          • Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
            – math.h
            Nov 20 at 11:42










          • Ah, I made an error, I will edit.
            – DaveNine
            Nov 20 at 12:55













          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



          $$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$






          share|cite|improve this answer














          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



          $$E'(t) = 2int_{mathbb{R}^d} u_{t}(u_{tt} - Delta u) dx = 0$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 20 at 13:06

























          answered Nov 20 at 10:26









          DaveNine

          1,241914




          1,241914












          • Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
            – math.h
            Nov 20 at 11:42










          • Ah, I made an error, I will edit.
            – DaveNine
            Nov 20 at 12:55


















          • Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
            – math.h
            Nov 20 at 11:42










          • Ah, I made an error, I will edit.
            – DaveNine
            Nov 20 at 12:55
















          Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
          – math.h
          Nov 20 at 11:42




          Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= Delta u$
          – math.h
          Nov 20 at 11:42












          Ah, I made an error, I will edit.
          – DaveNine
          Nov 20 at 12:55




          Ah, I made an error, I will edit.
          – DaveNine
          Nov 20 at 12:55


















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