Greens first identity












1












$begingroup$


Good afternoon all.



Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity



$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$



enter image description here



enter image description here



My answer is :



enter image description here






finite-element-method greens-function greens-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 15:37






  • 1




    $begingroup$
    It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:15








  • 1




    $begingroup$
    This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:20


















1












$begingroup$


Good afternoon all.



Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity



$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$



enter image description here



enter image description here



My answer is :



enter image description here






finite-element-method greens-function greens-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 15:37






  • 1




    $begingroup$
    It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:15








  • 1




    $begingroup$
    This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:20
















1












1








1





$begingroup$


Good afternoon all.



Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity



$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$



enter image description here



enter image description here



My answer is :



enter image description here






finite-element-method greens-function greens-theorem







share|cite|improve this question











$endgroup$




Good afternoon all.



Using the greens first identity I am finding weak formulation. Can someone check I did this correct with below mentioned greens first identity



$$int_U (nabla cdot nabla mathbf{u}) cdot , vec{omega} , dU = int_{partial U} frac{partial {mathbf{u}}}{partial n} cdot vec{omega}, dS - int_U nabla mathbf{u} : nabla vec{omega} , dU $$



enter image description here



enter image description here



My answer is :



enter image description here






finite-element-method greens-function greens-theorem




finite-element-method greens-function greens-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 14 '16 at 16:37







Nithish

















asked Feb 14 '16 at 14:28









NithishNithish

5110




5110












  • $begingroup$
    I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 15:37






  • 1




    $begingroup$
    It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:15








  • 1




    $begingroup$
    This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:20




















  • $begingroup$
    I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 15:37






  • 1




    $begingroup$
    It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:15








  • 1




    $begingroup$
    This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:20


















$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37




$begingroup$
I think you need to use the scalar Green's first identity: en.wikipedia.org/wiki/Green%27s_identities
$endgroup$
– Svetoslav
Feb 14 '16 at 15:37




1




1




$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15






$begingroup$
It is correct, but if you want to get a symmetric variational problem, then you should split the term $alphafrac{partial^2u}{partial xpartial y}=frac{alpha}{2}frac{partial^2u}{partial xpartial y}+frac{alpha}{2}frac{partial^2u}{partial ypartial x}=frac{alpha}{2}frac{partial }{partial x}left(frac{partial u}{partial y}right)+frac{alpha}{2}frac{partial }{partial y}left(frac{partial u}{partial x}right)$
$endgroup$
– Svetoslav
Feb 14 '16 at 17:15






1




1




$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20






$begingroup$
This means to write your initial PDE in divergence form (because $alpha,beta inmathbb R$ and so you can insert them in the derivatives)
$endgroup$
– Svetoslav
Feb 14 '16 at 17:20












1 Answer
1






active

oldest

votes


















1












$begingroup$

You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:38










  • $begingroup$
    Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 16:57










  • $begingroup$
    I updated my solution
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:59






  • 1




    $begingroup$
    Here is another weak PDE math.stackexchange.com/questions/1653737/…
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:36











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%2f1654470%2fgreens-first-identity%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 have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:38










  • $begingroup$
    Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 16:57










  • $begingroup$
    I updated my solution
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:59






  • 1




    $begingroup$
    Here is another weak PDE math.stackexchange.com/questions/1653737/…
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:36
















1












$begingroup$

You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:38










  • $begingroup$
    Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 16:57










  • $begingroup$
    I updated my solution
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:59






  • 1




    $begingroup$
    Here is another weak PDE math.stackexchange.com/questions/1653737/…
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:36














1












1








1





$begingroup$

You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .






share|cite|improve this answer









$endgroup$



You have the equation:
$$-9u_{xx}-5u_{yy}=f$$
Now multiplying by a test function $vin H_0^1(Omega)$ both sides and integrating over $Omega$ you get.
$$-9intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}-5intlimits_{Omega}{frac{partial}{partial y}left(frac{partial u}{partial y}right)vdxdy}=intlimits_{Omega}{fvdxdy}$$
$$9intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}+5intlimits_{Omega}{frac{partial u}{partial y}frac{partial v}{partial y}dxdy}=intlimits_{Omega}{fvdxdy}quadquad (*)$$
which is the weak formulation that you should get.
We used $$-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}right)vdxdy}=-intlimits_{Omega}{frac{partial}{partial x}left(frac{partial u}{partial x}vright)dxdy}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=-underbrace{intlimits_{partialOmega}{frac{partial u}{partial x}vn_1ds}}_{=0}+intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}=intlimits_{Omega}{frac{partial u}{partial x}frac{partial v}{partial x}dxdy}$$
Here, the tool that we used is the divergence theorem (with which is actually derived the Green's first identity). Note that the surface integral is $0$ because $v$ is zero on $partial Omega$ (to be more speciffic, it is zero in the trace sense). Or if you want to avoid the use of the divergence formula in the weak sense, then you can start by multiplying by a smooth test function $vin C_0^infty(Omega)$, derive the equation $(*)$ above, and then use density argument to show that $(*)$ is actually satisfied also for all $vin H_0^1(Omega)$. The integral with the $y$-derivative is done in similar way, just having in mind that in the surface integral there will be the second component $n_2$ of the normal vector $n$ to $partialOmega$ .







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 14 '16 at 16:05









SvetoslavSvetoslav

4,5672828




4,5672828












  • $begingroup$
    I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:38










  • $begingroup$
    Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 16:57










  • $begingroup$
    I updated my solution
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:59






  • 1




    $begingroup$
    Here is another weak PDE math.stackexchange.com/questions/1653737/…
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:36


















  • $begingroup$
    I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:38










  • $begingroup$
    Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 16:57










  • $begingroup$
    I updated my solution
    $endgroup$
    – Nithish
    Feb 14 '16 at 16:59






  • 1




    $begingroup$
    Here is another weak PDE math.stackexchange.com/questions/1653737/…
    $endgroup$
    – Svetoslav
    Feb 14 '16 at 17:36
















$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38




$begingroup$
I understood the concept mentoined above, which is pretty much similar to integration by parts. I have confusion about the xy terms together if you can check with my updated question please
$endgroup$
– Nithish
Feb 14 '16 at 16:38












$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57




$begingroup$
Yes, it is actually just integration by parts. I will check your solution and write here possibly in several hours.
$endgroup$
– Svetoslav
Feb 14 '16 at 16:57












$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59




$begingroup$
I updated my solution
$endgroup$
– Nithish
Feb 14 '16 at 16:59




1




1




$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36




$begingroup$
Here is another weak PDE math.stackexchange.com/questions/1653737/…
$endgroup$
– Svetoslav
Feb 14 '16 at 17:36


















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%2f1654470%2fgreens-first-identity%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

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

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