a question of density in showing that the variational question of Laplace equation [on hold]











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Here is a statement in showing that the variational question of Laplace equation, I'm confused with the following step:



if $int_Omega (nabla u+f)phi dxdy=0$ for every $phiin C_c^{infty}Omega$，



then according to the statement that $C_c^{infty}Omega$ is dense in $L^2(Omega)$，we have:
$nabla u=-f$



why? how to prove this?



Any help would be appreciated, thanks in advance!






functional-analysis pde sobolev-spaces







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put on hold as off-topic by Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $nabla u = -f $ surely?
    – Richard Martin
    yesterday










  • I think you want a proof more formal than this. But imagine $phi$ being a small bump (like a Gaussian or whatever) and make it thinner and thinner like a delta function. Then you will get your result.
    – Richard Martin
    yesterday










  • yes, $nabla u=-f$. but i still confused of this, i don't really understand your explanation, where we use the density property of $C_c^{infty}(Omega)$ in $L^2(Omega)$
    – chloe hj
    yesterday

















up vote
-3
down vote

favorite












Here is a statement in showing that the variational question of Laplace equation, I'm confused with the following step:



if $int_Omega (nabla u+f)phi dxdy=0$ for every $phiin C_c^{infty}Omega$，



then according to the statement that $C_c^{infty}Omega$ is dense in $L^2(Omega)$，we have:
$nabla u=-f$



why? how to prove this?



Any help would be appreciated, thanks in advance!






functional-analysis pde sobolev-spaces







share|cite|improve this question









New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $nabla u = -f $ surely?
    – Richard Martin
    yesterday










  • I think you want a proof more formal than this. But imagine $phi$ being a small bump (like a Gaussian or whatever) and make it thinner and thinner like a delta function. Then you will get your result.
    – Richard Martin
    yesterday










  • yes, $nabla u=-f$. but i still confused of this, i don't really understand your explanation, where we use the density property of $C_c^{infty}(Omega)$ in $L^2(Omega)$
    – chloe hj
    yesterday















up vote
-3
down vote

favorite









up vote
-3
down vote

favorite











Here is a statement in showing that the variational question of Laplace equation, I'm confused with the following step:



if $int_Omega (nabla u+f)phi dxdy=0$ for every $phiin C_c^{infty}Omega$，



then according to the statement that $C_c^{infty}Omega$ is dense in $L^2(Omega)$，we have:
$nabla u=-f$



why? how to prove this?



Any help would be appreciated, thanks in advance!






functional-analysis pde sobolev-spaces







share|cite|improve this question









New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Here is a statement in showing that the variational question of Laplace equation, I'm confused with the following step:



if $int_Omega (nabla u+f)phi dxdy=0$ for every $phiin C_c^{infty}Omega$，



then according to the statement that $C_c^{infty}Omega$ is dense in $L^2(Omega)$，we have:
$nabla u=-f$



why? how to prove this?



Any help would be appreciated, thanks in advance!






functional-analysis pde sobolev-spaces




functional-analysis pde sobolev-spaces






share|cite|improve this question









New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 17 hours ago





















New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









chloe hj

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535




New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, José Carlos Santos, amWhy, GNUSupporter 8964民主女神 地下教會, RRL

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $nabla u = -f $ surely?
    – Richard Martin
    yesterday










  • I think you want a proof more formal than this. But imagine $phi$ being a small bump (like a Gaussian or whatever) and make it thinner and thinner like a delta function. Then you will get your result.
    – Richard Martin
    yesterday










  • yes, $nabla u=-f$. but i still confused of this, i don't really understand your explanation, where we use the density property of $C_c^{infty}(Omega)$ in $L^2(Omega)$
    – chloe hj
    yesterday




















  • $nabla u = -f $ surely?
    – Richard Martin
    yesterday










  • I think you want a proof more formal than this. But imagine $phi$ being a small bump (like a Gaussian or whatever) and make it thinner and thinner like a delta function. Then you will get your result.
    – Richard Martin
    yesterday










  • yes, $nabla u=-f$. but i still confused of this, i don't really understand your explanation, where we use the density property of $C_c^{infty}(Omega)$ in $L^2(Omega)$
    – chloe hj
    yesterday


















$nabla u = -f $ surely?
– Richard Martin
yesterday




$nabla u = -f $ surely?
– Richard Martin
yesterday












I think you want a proof more formal than this. But imagine $phi$ being a small bump (like a Gaussian or whatever) and make it thinner and thinner like a delta function. Then you will get your result.
– Richard Martin
yesterday




I think you want a proof more formal than this. But imagine $phi$ being a small bump (like a Gaussian or whatever) and make it thinner and thinner like a delta function. Then you will get your result.
– Richard Martin
yesterday












yes, $nabla u=-f$. but i still confused of this, i don't really understand your explanation, where we use the density property of $C_c^{infty}(Omega)$ in $L^2(Omega)$
– chloe hj
yesterday






yes, $nabla u=-f$. but i still confused of this, i don't really understand your explanation, where we use the density property of $C_c^{infty}(Omega)$ in $L^2(Omega)$
– chloe hj
yesterday

















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