Integral of PDE












0












$begingroup$


how can we compute the integral of PDE if both the PDE and integral is with respect to $x$?



For example:



$int w(x) , partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



I know we have to do integration by parts, but how do we do integration of:



$int partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



The book says that the result is:



$int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$



I cannot get that result.



Thanks



enter image description here






pde







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $int partial_x f(x) dx = f(x) + c$?
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:16










  • $begingroup$
    Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:28








  • 1




    $begingroup$
    thats a little sloppy... but i will say $partial x$ and $dx$ has no mathematical difference, its just notation to remind us that there are two++ variables
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:29










  • $begingroup$
    Okay, but how we can get $int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$ ? How $w(x)$ ends up inside $partial_x$
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:40












  • $begingroup$
    You should refresh yourself on integration by parts, the derivative on $mu partial_x u $ goes to the other term in the integrand which is $w$
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:56
















0












$begingroup$


how can we compute the integral of PDE if both the PDE and integral is with respect to $x$?



For example:



$int w(x) , partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



I know we have to do integration by parts, but how do we do integration of:



$int partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



The book says that the result is:



$int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$



I cannot get that result.



Thanks



enter image description here






pde







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $int partial_x f(x) dx = f(x) + c$?
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:16










  • $begingroup$
    Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:28








  • 1




    $begingroup$
    thats a little sloppy... but i will say $partial x$ and $dx$ has no mathematical difference, its just notation to remind us that there are two++ variables
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:29










  • $begingroup$
    Okay, but how we can get $int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$ ? How $w(x)$ ends up inside $partial_x$
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:40












  • $begingroup$
    You should refresh yourself on integration by parts, the derivative on $mu partial_x u $ goes to the other term in the integrand which is $w$
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:56














0












0








0





$begingroup$


how can we compute the integral of PDE if both the PDE and integral is with respect to $x$?



For example:



$int w(x) , partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



I know we have to do integration by parts, but how do we do integration of:



$int partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



The book says that the result is:



$int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$



I cannot get that result.



Thanks



enter image description here






pde







share|cite|improve this question









$endgroup$




how can we compute the integral of PDE if both the PDE and integral is with respect to $x$?



For example:



$int w(x) , partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



I know we have to do integration by parts, but how do we do integration of:



$int partial_{x} , [mu(x),partial_{x},u(x,t)]dx$



The book says that the result is:



$int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$



I cannot get that result.



Thanks



enter image description here






pde




pde






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 17:59









JIM BOYJIM BOY

356




356








  • 1




    $begingroup$
    $int partial_x f(x) dx = f(x) + c$?
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:16










  • $begingroup$
    Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:28








  • 1




    $begingroup$
    thats a little sloppy... but i will say $partial x$ and $dx$ has no mathematical difference, its just notation to remind us that there are two++ variables
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:29










  • $begingroup$
    Okay, but how we can get $int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$ ? How $w(x)$ ends up inside $partial_x$
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:40












  • $begingroup$
    You should refresh yourself on integration by parts, the derivative on $mu partial_x u $ goes to the other term in the integrand which is $w$
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:56














  • 1




    $begingroup$
    $int partial_x f(x) dx = f(x) + c$?
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:16










  • $begingroup$
    Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:28








  • 1




    $begingroup$
    thats a little sloppy... but i will say $partial x$ and $dx$ has no mathematical difference, its just notation to remind us that there are two++ variables
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:29










  • $begingroup$
    Okay, but how we can get $int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$ ? How $w(x)$ ends up inside $partial_x$
    $endgroup$
    – JIM BOY
    Nov 30 '18 at 18:40












  • $begingroup$
    You should refresh yourself on integration by parts, the derivative on $mu partial_x u $ goes to the other term in the integrand which is $w$
    $endgroup$
    – Calvin Khor
    Nov 30 '18 at 18:56








1




1




$begingroup$
$int partial_x f(x) dx = f(x) + c$?
$endgroup$
– Calvin Khor
Nov 30 '18 at 18:16




$begingroup$
$int partial_x f(x) dx = f(x) + c$?
$endgroup$
– Calvin Khor
Nov 30 '18 at 18:16












$begingroup$
Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
$endgroup$
– JIM BOY
Nov 30 '18 at 18:28






$begingroup$
Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
$endgroup$
– JIM BOY
Nov 30 '18 at 18:28






1




1




$begingroup$
thats a little sloppy... but i will say $partial x$ and $dx$ has no mathematical difference, its just notation to remind us that there are two++ variables
$endgroup$
– Calvin Khor
Nov 30 '18 at 18:29




$begingroup$
thats a little sloppy... but i will say $partial x$ and $dx$ has no mathematical difference, its just notation to remind us that there are two++ variables
$endgroup$
– Calvin Khor
Nov 30 '18 at 18:29












$begingroup$
Okay, but how we can get $int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$ ? How $w(x)$ ends up inside $partial_x$
$endgroup$
– JIM BOY
Nov 30 '18 at 18:40






$begingroup$
Okay, but how we can get $int mu(x) , partial_{x} ,w(x),partial_{x},u(x,t)dx$ ? How $w(x)$ ends up inside $partial_x$
$endgroup$
– JIM BOY
Nov 30 '18 at 18:40














$begingroup$
You should refresh yourself on integration by parts, the derivative on $mu partial_x u $ goes to the other term in the integrand which is $w$
$endgroup$
– Calvin Khor
Nov 30 '18 at 18:56




$begingroup$
You should refresh yourself on integration by parts, the derivative on $mu partial_x u $ goes to the other term in the integrand which is $w$
$endgroup$
– Calvin Khor
Nov 30 '18 at 18:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are integrating only in $x$. As such you should treat $y$ constant. Let
$y$ be fixed, and set $U(x) := u(x,y)$. Then we have $partial_x u = U'$, and the integral is
$$ int_0^L w (mu u')' dx$$



Integration by parts:
$$ int_0^L w (mu U')' dx = w mu U'Big|_0^L - int_0^L w' (mu U') dx$$
the boundary term disappears using the Neumann boundary condition.



PS note that $w:Gtomathbb R$ is a one-variable function so some people would say you cannot write $partial_x w$, only $w'$. The person who wrote your notes is not this kind of person, so you should get used to treating $partial_x$ as a synonym for $d/dx$ if you want to continue reading.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, actually this is from a seismic book published by Springer
    $endgroup$
    – JIM BOY
    Dec 1 '18 at 14:19











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1 Answer
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1 Answer
1






active

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active

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1












$begingroup$

You are integrating only in $x$. As such you should treat $y$ constant. Let
$y$ be fixed, and set $U(x) := u(x,y)$. Then we have $partial_x u = U'$, and the integral is
$$ int_0^L w (mu u')' dx$$



Integration by parts:
$$ int_0^L w (mu U')' dx = w mu U'Big|_0^L - int_0^L w' (mu U') dx$$
the boundary term disappears using the Neumann boundary condition.



PS note that $w:Gtomathbb R$ is a one-variable function so some people would say you cannot write $partial_x w$, only $w'$. The person who wrote your notes is not this kind of person, so you should get used to treating $partial_x$ as a synonym for $d/dx$ if you want to continue reading.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, actually this is from a seismic book published by Springer
    $endgroup$
    – JIM BOY
    Dec 1 '18 at 14:19
















1












$begingroup$

You are integrating only in $x$. As such you should treat $y$ constant. Let
$y$ be fixed, and set $U(x) := u(x,y)$. Then we have $partial_x u = U'$, and the integral is
$$ int_0^L w (mu u')' dx$$



Integration by parts:
$$ int_0^L w (mu U')' dx = w mu U'Big|_0^L - int_0^L w' (mu U') dx$$
the boundary term disappears using the Neumann boundary condition.



PS note that $w:Gtomathbb R$ is a one-variable function so some people would say you cannot write $partial_x w$, only $w'$. The person who wrote your notes is not this kind of person, so you should get used to treating $partial_x$ as a synonym for $d/dx$ if you want to continue reading.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, actually this is from a seismic book published by Springer
    $endgroup$
    – JIM BOY
    Dec 1 '18 at 14:19














1












1








1





$begingroup$

You are integrating only in $x$. As such you should treat $y$ constant. Let
$y$ be fixed, and set $U(x) := u(x,y)$. Then we have $partial_x u = U'$, and the integral is
$$ int_0^L w (mu u')' dx$$



Integration by parts:
$$ int_0^L w (mu U')' dx = w mu U'Big|_0^L - int_0^L w' (mu U') dx$$
the boundary term disappears using the Neumann boundary condition.



PS note that $w:Gtomathbb R$ is a one-variable function so some people would say you cannot write $partial_x w$, only $w'$. The person who wrote your notes is not this kind of person, so you should get used to treating $partial_x$ as a synonym for $d/dx$ if you want to continue reading.






share|cite|improve this answer











$endgroup$



You are integrating only in $x$. As such you should treat $y$ constant. Let
$y$ be fixed, and set $U(x) := u(x,y)$. Then we have $partial_x u = U'$, and the integral is
$$ int_0^L w (mu u')' dx$$



Integration by parts:
$$ int_0^L w (mu U')' dx = w mu U'Big|_0^L - int_0^L w' (mu U') dx$$
the boundary term disappears using the Neumann boundary condition.



PS note that $w:Gtomathbb R$ is a one-variable function so some people would say you cannot write $partial_x w$, only $w'$. The person who wrote your notes is not this kind of person, so you should get used to treating $partial_x$ as a synonym for $d/dx$ if you want to continue reading.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 19:35

























answered Nov 30 '18 at 19:28









Calvin KhorCalvin Khor

11.3k21438




11.3k21438












  • $begingroup$
    Thank you, actually this is from a seismic book published by Springer
    $endgroup$
    – JIM BOY
    Dec 1 '18 at 14:19


















  • $begingroup$
    Thank you, actually this is from a seismic book published by Springer
    $endgroup$
    – JIM BOY
    Dec 1 '18 at 14:19
















$begingroup$
Thank you, actually this is from a seismic book published by Springer
$endgroup$
– JIM BOY
Dec 1 '18 at 14:19




$begingroup$
Thank you, actually this is from a seismic book published by Springer
$endgroup$
– JIM BOY
Dec 1 '18 at 14:19


















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