Integral of PDE











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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










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  • 1




    $int partial_x f(x) dx = f(x) + c$?
    – Calvin Khor
    2 days ago










  • Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    – JIM BOY
    2 days ago








  • 1




    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
    – Calvin Khor
    2 days ago










  • 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 $partial_x$
    – JIM BOY
    2 days ago












  • 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 $w$
    – Calvin Khor
    2 days ago















up vote
0
down vote

favorite












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










share|cite|improve this question


















  • 1




    $int partial_x f(x) dx = f(x) + c$?
    – Calvin Khor
    2 days ago










  • Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    – JIM BOY
    2 days ago








  • 1




    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
    – Calvin Khor
    2 days ago










  • 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 $partial_x$
    – JIM BOY
    2 days ago












  • 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 $w$
    – Calvin Khor
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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










share|cite|improve this question













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






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asked 2 days ago









JIM BOY

356




356








  • 1




    $int partial_x f(x) dx = f(x) + c$?
    – Calvin Khor
    2 days ago










  • Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    – JIM BOY
    2 days ago








  • 1




    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
    – Calvin Khor
    2 days ago










  • 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 $partial_x$
    – JIM BOY
    2 days ago












  • 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 $w$
    – Calvin Khor
    2 days ago














  • 1




    $int partial_x f(x) dx = f(x) + c$?
    – Calvin Khor
    2 days ago










  • Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
    – JIM BOY
    2 days ago








  • 1




    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
    – Calvin Khor
    2 days ago










  • 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 $partial_x$
    – JIM BOY
    2 days ago












  • 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 $w$
    – Calvin Khor
    2 days ago








1




1




$int partial_x f(x) dx = f(x) + c$?
– Calvin Khor
2 days ago




$int partial_x f(x) dx = f(x) + c$?
– Calvin Khor
2 days ago












Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
– JIM BOY
2 days ago






Why is it like that? Is it because $frac{partial f(x)}{partial x}dx$ and $partial x$ will be eliminated by $dx$ ($frac{partial x}{dx}=1$)?
– JIM BOY
2 days ago






1




1




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
– Calvin Khor
2 days ago




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
– Calvin Khor
2 days ago












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 $partial_x$
– JIM BOY
2 days ago






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 $partial_x$
– JIM BOY
2 days ago














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 $w$
– Calvin Khor
2 days ago




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 $w$
– Calvin Khor
2 days ago










1 Answer
1






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1
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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























  • Thank you, actually this is from a seismic book published by Springer
    – JIM BOY
    yesterday











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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























  • Thank you, actually this is from a seismic book published by Springer
    – JIM BOY
    yesterday















up vote
1
down vote













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























  • Thank you, actually this is from a seismic book published by Springer
    – JIM BOY
    yesterday













up vote
1
down vote










up vote
1
down vote









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














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 2 days ago

























answered 2 days ago









Calvin Khor

10.8k21437




10.8k21437












  • Thank you, actually this is from a seismic book published by Springer
    – JIM BOY
    yesterday


















  • Thank you, actually this is from a seismic book published by Springer
    – JIM BOY
    yesterday
















Thank you, actually this is from a seismic book published by Springer
– JIM BOY
yesterday




Thank you, actually this is from a seismic book published by Springer
– JIM BOY
yesterday


















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