What's wrong with this distributional laplacian?
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Let $f$ be a locally integrable function on an open set $G$ of $mathbb{R}^{n}$ and $ ngeq2 $. Suppose $theta$ is in $ C^{infty}(G) $ such that its laplacian $ Delta theta=1 $ everywhere in $mathbb{R}^{n}$, and let $phiin C_{c}^{infty}(G)$. Clearly $phi thetain C_{c}^{infty}(G).$ We know that the distributional laplacian of $f$ is defined and is given by
$$ langle Delta f,(phi theta)rangle=int_{G}f(x)Delta[phi(x) theta(x)]dx. $$
Now, if I write that this integral is equal to
$$ int_{G}f(x)phi(x)dx+int_{G}f(x)[Deltaphi(x)] theta(x)dx +sumlimits^{n}_{j=1}int_{G}f(x)frac{partial phi(x)}{partial x_{j}}frac{partialtheta(x) }{partial x_{j}} dx,$$ I obtain absurd results by using some specific functions for $phi$. There must be somthing wrong with the first integral (by the way I used here the fact that $Deltathetaequiv1$). What's wrong with that?
real-analysis complex-analysis distribution-theory
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
$endgroup$
|
show 3 more comments
$begingroup$
Let $f$ be a locally integrable function on an open set $G$ of $mathbb{R}^{n}$ and $ ngeq2 $. Suppose $theta$ is in $ C^{infty}(G) $ such that its laplacian $ Delta theta=1 $ everywhere in $mathbb{R}^{n}$, and let $phiin C_{c}^{infty}(G)$. Clearly $phi thetain C_{c}^{infty}(G).$ We know that the distributional laplacian of $f$ is defined and is given by
$$ langle Delta f,(phi theta)rangle=int_{G}f(x)Delta[phi(x) theta(x)]dx. $$
Now, if I write that this integral is equal to
$$ int_{G}f(x)phi(x)dx+int_{G}f(x)[Deltaphi(x)] theta(x)dx +sumlimits^{n}_{j=1}int_{G}f(x)frac{partial phi(x)}{partial x_{j}}frac{partialtheta(x) }{partial x_{j}} dx,$$ I obtain absurd results by using some specific functions for $phi$. There must be somthing wrong with the first integral (by the way I used here the fact that $Deltathetaequiv1$). What's wrong with that?
real-analysis complex-analysis distribution-theory
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
$endgroup$
$begingroup$
Is it $Delta f = 1$ or $Delta f = delta$ (Dirac distribution) ?
$endgroup$
– Jean Marie
Dec 15 '18 at 4:15
$begingroup$
$f$ is a locally integrable function, but not smooth.
$endgroup$
– M. Rahmat
Dec 15 '18 at 4:35
$begingroup$
Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $delta$ ? This is not at all the same.
$endgroup$
– Jean Marie
Dec 15 '18 at 5:16
$begingroup$
Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution?
$endgroup$
– M. Rahmat
Dec 15 '18 at 6:51
$begingroup$
I mean the RHS of your first equation $Delta theta = 1$.
$endgroup$
– Jean Marie
Dec 15 '18 at 9:24
|
show 3 more comments
$begingroup$
Let $f$ be a locally integrable function on an open set $G$ of $mathbb{R}^{n}$ and $ ngeq2 $. Suppose $theta$ is in $ C^{infty}(G) $ such that its laplacian $ Delta theta=1 $ everywhere in $mathbb{R}^{n}$, and let $phiin C_{c}^{infty}(G)$. Clearly $phi thetain C_{c}^{infty}(G).$ We know that the distributional laplacian of $f$ is defined and is given by
$$ langle Delta f,(phi theta)rangle=int_{G}f(x)Delta[phi(x) theta(x)]dx. $$
Now, if I write that this integral is equal to
$$ int_{G}f(x)phi(x)dx+int_{G}f(x)[Deltaphi(x)] theta(x)dx +sumlimits^{n}_{j=1}int_{G}f(x)frac{partial phi(x)}{partial x_{j}}frac{partialtheta(x) }{partial x_{j}} dx,$$ I obtain absurd results by using some specific functions for $phi$. There must be somthing wrong with the first integral (by the way I used here the fact that $Deltathetaequiv1$). What's wrong with that?
real-analysis complex-analysis distribution-theory
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
$endgroup$
Let $f$ be a locally integrable function on an open set $G$ of $mathbb{R}^{n}$ and $ ngeq2 $. Suppose $theta$ is in $ C^{infty}(G) $ such that its laplacian $ Delta theta=1 $ everywhere in $mathbb{R}^{n}$, and let $phiin C_{c}^{infty}(G)$. Clearly $phi thetain C_{c}^{infty}(G).$ We know that the distributional laplacian of $f$ is defined and is given by
$$ langle Delta f,(phi theta)rangle=int_{G}f(x)Delta[phi(x) theta(x)]dx. $$
Now, if I write that this integral is equal to
$$ int_{G}f(x)phi(x)dx+int_{G}f(x)[Deltaphi(x)] theta(x)dx +sumlimits^{n}_{j=1}int_{G}f(x)frac{partial phi(x)}{partial x_{j}}frac{partialtheta(x) }{partial x_{j}} dx,$$ I obtain absurd results by using some specific functions for $phi$. There must be somthing wrong with the first integral (by the way I used here the fact that $Deltathetaequiv1$). What's wrong with that?
real-analysis complex-analysis distribution-theory
real-analysis complex-analysis distribution-theory
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
asked Dec 15 '18 at 2:53
M. RahmatM. Rahmat
332212
332212
$begingroup$
Is it $Delta f = 1$ or $Delta f = delta$ (Dirac distribution) ?
$endgroup$
– Jean Marie
Dec 15 '18 at 4:15
$begingroup$
$f$ is a locally integrable function, but not smooth.
$endgroup$
– M. Rahmat
Dec 15 '18 at 4:35
$begingroup$
Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $delta$ ? This is not at all the same.
$endgroup$
– Jean Marie
Dec 15 '18 at 5:16
$begingroup$
Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution?
$endgroup$
– M. Rahmat
Dec 15 '18 at 6:51
$begingroup$
I mean the RHS of your first equation $Delta theta = 1$.
$endgroup$
– Jean Marie
Dec 15 '18 at 9:24
|
show 3 more comments
$begingroup$
Is it $Delta f = 1$ or $Delta f = delta$ (Dirac distribution) ?
$endgroup$
– Jean Marie
Dec 15 '18 at 4:15
$begingroup$
$f$ is a locally integrable function, but not smooth.
$endgroup$
– M. Rahmat
Dec 15 '18 at 4:35
$begingroup$
Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $delta$ ? This is not at all the same.
$endgroup$
– Jean Marie
Dec 15 '18 at 5:16
$begingroup$
Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution?
$endgroup$
– M. Rahmat
Dec 15 '18 at 6:51
$begingroup$
I mean the RHS of your first equation $Delta theta = 1$.
$endgroup$
– Jean Marie
Dec 15 '18 at 9:24
$begingroup$
Is it $Delta f = 1$ or $Delta f = delta$ (Dirac distribution) ?
$endgroup$
– Jean Marie
Dec 15 '18 at 4:15
$begingroup$
Is it $Delta f = 1$ or $Delta f = delta$ (Dirac distribution) ?
$endgroup$
– Jean Marie
Dec 15 '18 at 4:15
$begingroup$
$f$ is a locally integrable function, but not smooth.
$endgroup$
– M. Rahmat
Dec 15 '18 at 4:35
$begingroup$
$f$ is a locally integrable function, but not smooth.
$endgroup$
– M. Rahmat
Dec 15 '18 at 4:35
$begingroup$
Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $delta$ ? This is not at all the same.
$endgroup$
– Jean Marie
Dec 15 '18 at 5:16
$begingroup$
Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $delta$ ? This is not at all the same.
$endgroup$
– Jean Marie
Dec 15 '18 at 5:16
$begingroup$
Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution?
$endgroup$
– M. Rahmat
Dec 15 '18 at 6:51
$begingroup$
Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution?
$endgroup$
– M. Rahmat
Dec 15 '18 at 6:51
$begingroup$
I mean the RHS of your first equation $Delta theta = 1$.
$endgroup$
– Jean Marie
Dec 15 '18 at 9:24
$begingroup$
I mean the RHS of your first equation $Delta theta = 1$.
$endgroup$
– Jean Marie
Dec 15 '18 at 9:24
|
show 3 more comments
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$begingroup$
Is it $Delta f = 1$ or $Delta f = delta$ (Dirac distribution) ?
$endgroup$
– Jean Marie
Dec 15 '18 at 4:15
$begingroup$
$f$ is a locally integrable function, but not smooth.
$endgroup$
– M. Rahmat
Dec 15 '18 at 4:35
$begingroup$
Be kind enough to answer my question : is the RHS constant function $1$ or a Dirac peak $delta$ ? This is not at all the same.
$endgroup$
– Jean Marie
Dec 15 '18 at 5:16
$begingroup$
Sorry! I don't understand your question! What does it mean that the integral on the right equals 1? Or equals the Dirac distribution?
$endgroup$
– M. Rahmat
Dec 15 '18 at 6:51
$begingroup$
I mean the RHS of your first equation $Delta theta = 1$.
$endgroup$
– Jean Marie
Dec 15 '18 at 9:24