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
pde
|
show 2 more comments
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
pde
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
|
show 2 more comments
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
pde
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
pde
pde
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
1
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up vote
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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.
Thank you, actually this is from a seismic book published by Springer
– JIM BOY
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Thank you, actually this is from a seismic book published by Springer
– JIM BOY
yesterday
add a comment |
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.
Thank you, actually this is from a seismic book published by Springer
– JIM BOY
yesterday
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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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