Mixing
Gradient of time dependent function
$begingroup$
Let $uin L^2(bar{Omega})$ be a differentiable function which is also time-dependent, where $Omegasubset mathbb{R}^d$. Is there any connection between $(nabla u_t,nabla u)$ and $(nabla u,nabla u)$. My final problem is to show the positivity of $(nabla u_t,nabla u)$.
I worked with a simple example of $d=1$ and taking $u(x,t)=x^2t$. Here, I get it as $(nabla u_t,nabla u)=t(nabla u,nabla u)$ with $Omega=(0,1)$. Can we say it in general this holds? I mean can I say $(nabla u_t,nabla u)geq0$
calculus multivariable-calculus
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
$endgroup$
add a comment |
$begingroup$
Let $uin L^2(bar{Omega})$ be a differentiable function which is also time-dependent, where $Omegasubset mathbb{R}^d$. Is there any connection between $(nabla u_t,nabla u)$ and $(nabla u,nabla u)$. My final problem is to show the positivity of $(nabla u_t,nabla u)$.
I worked with a simple example of $d=1$ and taking $u(x,t)=x^2t$. Here, I get it as $(nabla u_t,nabla u)=t(nabla u,nabla u)$ with $Omega=(0,1)$. Can we say it in general this holds? I mean can I say $(nabla u_t,nabla u)geq0$
calculus multivariable-calculus
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
$endgroup$
add a comment |
$begingroup$
Let $uin L^2(bar{Omega})$ be a differentiable function which is also time-dependent, where $Omegasubset mathbb{R}^d$. Is there any connection between $(nabla u_t,nabla u)$ and $(nabla u,nabla u)$. My final problem is to show the positivity of $(nabla u_t,nabla u)$.
I worked with a simple example of $d=1$ and taking $u(x,t)=x^2t$. Here, I get it as $(nabla u_t,nabla u)=t(nabla u,nabla u)$ with $Omega=(0,1)$. Can we say it in general this holds? I mean can I say $(nabla u_t,nabla u)geq0$
calculus multivariable-calculus
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
$endgroup$
Let $uin L^2(bar{Omega})$ be a differentiable function which is also time-dependent, where $Omegasubset mathbb{R}^d$. Is there any connection between $(nabla u_t,nabla u)$ and $(nabla u,nabla u)$. My final problem is to show the positivity of $(nabla u_t,nabla u)$.
I worked with a simple example of $d=1$ and taking $u(x,t)=x^2t$. Here, I get it as $(nabla u_t,nabla u)=t(nabla u,nabla u)$ with $Omega=(0,1)$. Can we say it in general this holds? I mean can I say $(nabla u_t,nabla u)geq0$
calculus multivariable-calculus
calculus multivariable-calculus
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
asked Jan 30 at 11:19
Abhinav JhaAbhinav Jha
2741211
2741211
add a comment |
add a comment |
1 Answer
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No. Take $u(x,t)=e^{-t}x^2$ instead of $u(x,t)=x^2t$ in your example. Then
$$
(nabla u_t,nabla u)=-(nabla u,nabla u).
$$
answered Jan 30 at 12:21
jobejobe
1,109615
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
No. Take $u(x,t)=e^{-t}x^2$ instead of $u(x,t)=x^2t$ in your example. Then
$$
(nabla u_t,nabla u)=-(nabla u,nabla u).
$$
answered Jan 30 at 12:21
jobejobe
1,109615
$endgroup$
add a comment |
$begingroup$
No. Take $u(x,t)=e^{-t}x^2$ instead of $u(x,t)=x^2t$ in your example. Then
$$
(nabla u_t,nabla u)=-(nabla u,nabla u).
$$
answered Jan 30 at 12:21
jobejobe
1,109615
$endgroup$
add a comment |
$begingroup$
No. Take $u(x,t)=e^{-t}x^2$ instead of $u(x,t)=x^2t$ in your example. Then
$$
(nabla u_t,nabla u)=-(nabla u,nabla u).
$$
answered Jan 30 at 12:21
jobejobe
1,109615
$endgroup$
No. Take $u(x,t)=e^{-t}x^2$ instead of $u(x,t)=x^2t$ in your example. Then
$$
(nabla u_t,nabla u)=-(nabla u,nabla u).
$$
answered Jan 30 at 12:21
jobejobe
1,109615
answered Jan 30 at 12:21
jobejobe
1,109615
answered Jan 30 at 12:21
jobejobe
1,109615
answered Jan 30 at 12:21
jobejobe
1,109615
1,109615
add a comment |
add a comment |
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