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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020412%2fintegral-of-pde%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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
















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

11.2k21438




11.2k21438












  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020412%2fintegral-of-pde%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more