weak solution of PDE and apply Lax-Milgram












-2














Can someone help me for this problem?




Write the weak formulation of:
$$left{begin{align}
-frac{partial^2u}{partial x^2}-5frac{partial^2u}{partial y^2}=fquad&text{in}quadOmegasubsetmathbb R^2\
u=0 quad&text{in}quadpartialOmega
end{align}right.$$
Then apply Lax-Milgram to show existence of weak solution.




Using Green identity I found the following weak formulation:



Find $uin H_0^1(Omega)$ solution of
$$int_Omega u_x v_x+5int_Omega u_y v_y =int_Omega fv,quadforall vin H_0^1(Omega).$$



Is this correct?
How can I now apply Lax-Milgram?



Can I transform this equation using gradient or Laplace operator?
Thanks






pde weak-derivatives







share|cite|improve this question
























  • Please, check if my edits in your post are correct.
    – Pedro
    Nov 26 '18 at 0:03










  • Also, see this can help.
    – Pedro
    Nov 26 '18 at 0:09
















-2














Can someone help me for this problem?




Write the weak formulation of:
$$left{begin{align}
-frac{partial^2u}{partial x^2}-5frac{partial^2u}{partial y^2}=fquad&text{in}quadOmegasubsetmathbb R^2\
u=0 quad&text{in}quadpartialOmega
end{align}right.$$
Then apply Lax-Milgram to show existence of weak solution.




Using Green identity I found the following weak formulation:



Find $uin H_0^1(Omega)$ solution of
$$int_Omega u_x v_x+5int_Omega u_y v_y =int_Omega fv,quadforall vin H_0^1(Omega).$$



Is this correct?
How can I now apply Lax-Milgram?



Can I transform this equation using gradient or Laplace operator?
Thanks






pde weak-derivatives







share|cite|improve this question
























  • Please, check if my edits in your post are correct.
    – Pedro
    Nov 26 '18 at 0:03










  • Also, see this can help.
    – Pedro
    Nov 26 '18 at 0:09














-2












-2








-2


2





Can someone help me for this problem?




Write the weak formulation of:
$$left{begin{align}
-frac{partial^2u}{partial x^2}-5frac{partial^2u}{partial y^2}=fquad&text{in}quadOmegasubsetmathbb R^2\
u=0 quad&text{in}quadpartialOmega
end{align}right.$$
Then apply Lax-Milgram to show existence of weak solution.




Using Green identity I found the following weak formulation:



Find $uin H_0^1(Omega)$ solution of
$$int_Omega u_x v_x+5int_Omega u_y v_y =int_Omega fv,quadforall vin H_0^1(Omega).$$



Is this correct?
How can I now apply Lax-Milgram?



Can I transform this equation using gradient or Laplace operator?
Thanks






pde weak-derivatives







share|cite|improve this question















Can someone help me for this problem?




Write the weak formulation of:
$$left{begin{align}
-frac{partial^2u}{partial x^2}-5frac{partial^2u}{partial y^2}=fquad&text{in}quadOmegasubsetmathbb R^2\
u=0 quad&text{in}quadpartialOmega
end{align}right.$$
Then apply Lax-Milgram to show existence of weak solution.




Using Green identity I found the following weak formulation:



Find $uin H_0^1(Omega)$ solution of
$$int_Omega u_x v_x+5int_Omega u_y v_y =int_Omega fv,quadforall vin H_0^1(Omega).$$



Is this correct?
How can I now apply Lax-Milgram?



Can I transform this equation using gradient or Laplace operator?
Thanks






pde weak-derivatives




pde weak-derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 0:00









Pedro

10.2k23068




10.2k23068










asked Nov 25 '18 at 23:34









Mary

4




4












  • Please, check if my edits in your post are correct.
    – Pedro
    Nov 26 '18 at 0:03










  • Also, see this can help.
    – Pedro
    Nov 26 '18 at 0:09


















  • Please, check if my edits in your post are correct.
    – Pedro
    Nov 26 '18 at 0:03










  • Also, see this can help.
    – Pedro
    Nov 26 '18 at 0:09
















Please, check if my edits in your post are correct.
– Pedro
Nov 26 '18 at 0:03




Please, check if my edits in your post are correct.
– Pedro
Nov 26 '18 at 0:03












Also, see this can help.
– Pedro
Nov 26 '18 at 0:09




Also, see this can help.
– Pedro
Nov 26 '18 at 0:09










1 Answer
1






active

oldest

votes


















1















How can I now apply Lax-Milgram?




You have to identify what are the bilinear form and the linear functional which correspond to your weak formulation.



In this case, the bilinear form is $B:H_0^1times H_0^1tomathbb R$ defined by
$$B[u,v]=int_Omega u_x v_x+5int_Omega u_y v_y.$$



And the linear functional is $Lambda:H_0^1tomathbb R$ defined by
$$Lambda[v]=int_Omega fv.$$



Now, in order to prove existence of weak solution, you have to show that $B$ is continuous and coercive.






share|cite|improve this answer





















  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:20












  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:26










  • Your first equality is not correct because there are other term.
    – Pedro
    Nov 26 '18 at 0:27










  • Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
    – Mary
    Nov 26 '18 at 0:34










  • @Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
    – Pedro
    Nov 26 '18 at 0:44













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1















How can I now apply Lax-Milgram?




You have to identify what are the bilinear form and the linear functional which correspond to your weak formulation.



In this case, the bilinear form is $B:H_0^1times H_0^1tomathbb R$ defined by
$$B[u,v]=int_Omega u_x v_x+5int_Omega u_y v_y.$$



And the linear functional is $Lambda:H_0^1tomathbb R$ defined by
$$Lambda[v]=int_Omega fv.$$



Now, in order to prove existence of weak solution, you have to show that $B$ is continuous and coercive.






share|cite|improve this answer





















  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:20












  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:26










  • Your first equality is not correct because there are other term.
    – Pedro
    Nov 26 '18 at 0:27










  • Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
    – Mary
    Nov 26 '18 at 0:34










  • @Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
    – Pedro
    Nov 26 '18 at 0:44


















1















How can I now apply Lax-Milgram?




You have to identify what are the bilinear form and the linear functional which correspond to your weak formulation.



In this case, the bilinear form is $B:H_0^1times H_0^1tomathbb R$ defined by
$$B[u,v]=int_Omega u_x v_x+5int_Omega u_y v_y.$$



And the linear functional is $Lambda:H_0^1tomathbb R$ defined by
$$Lambda[v]=int_Omega fv.$$



Now, in order to prove existence of weak solution, you have to show that $B$ is continuous and coercive.






share|cite|improve this answer





















  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:20












  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:26










  • Your first equality is not correct because there are other term.
    – Pedro
    Nov 26 '18 at 0:27










  • Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
    – Mary
    Nov 26 '18 at 0:34










  • @Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
    – Pedro
    Nov 26 '18 at 0:44
















1












1








1







How can I now apply Lax-Milgram?




You have to identify what are the bilinear form and the linear functional which correspond to your weak formulation.



In this case, the bilinear form is $B:H_0^1times H_0^1tomathbb R$ defined by
$$B[u,v]=int_Omega u_x v_x+5int_Omega u_y v_y.$$



And the linear functional is $Lambda:H_0^1tomathbb R$ defined by
$$Lambda[v]=int_Omega fv.$$



Now, in order to prove existence of weak solution, you have to show that $B$ is continuous and coercive.






share|cite|improve this answer













How can I now apply Lax-Milgram?




You have to identify what are the bilinear form and the linear functional which correspond to your weak formulation.



In this case, the bilinear form is $B:H_0^1times H_0^1tomathbb R$ defined by
$$B[u,v]=int_Omega u_x v_x+5int_Omega u_y v_y.$$



And the linear functional is $Lambda:H_0^1tomathbb R$ defined by
$$Lambda[v]=int_Omega fv.$$



Now, in order to prove existence of weak solution, you have to show that $B$ is continuous and coercive.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 23:53









Pedro

10.2k23068




10.2k23068












  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:20












  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:26










  • Your first equality is not correct because there are other term.
    – Pedro
    Nov 26 '18 at 0:27










  • Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
    – Mary
    Nov 26 '18 at 0:34










  • @Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
    – Pedro
    Nov 26 '18 at 0:44




















  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:20












  • i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
    – Mary
    Nov 26 '18 at 0:26










  • Your first equality is not correct because there are other term.
    – Pedro
    Nov 26 '18 at 0:27










  • Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
    – Mary
    Nov 26 '18 at 0:34










  • @Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
    – Pedro
    Nov 26 '18 at 0:44


















i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
– Mary
Nov 26 '18 at 0:20






i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||u_x ||_(L^2 ) |v_x ||_(L^2 ) $ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
– Mary
Nov 26 '18 at 0:20














i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
– Mary
Nov 26 '18 at 0:26




i know this B[u,v]=$∫u_x v_x=(u_x ,v_x )≤||〖u_x ||〗_(L^2 ) |〖v_x ||〗_(L^2 )$ can I use that $||u_x ||_(L^2) |v_x ||_(L^2 )≤||∇u||_(L^2) ||∇v||_(L^2) $
– Mary
Nov 26 '18 at 0:26












Your first equality is not correct because there are other term.
– Pedro
Nov 26 '18 at 0:27




Your first equality is not correct because there are other term.
– Pedro
Nov 26 '18 at 0:27












Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
– Mary
Nov 26 '18 at 0:34




Can I say that the norm of a component of the vector is smaller than the norm of the whole vector? (in this case the gradient) ||vx||<=||gradv||
– Mary
Nov 26 '18 at 0:34












@Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
– Pedro
Nov 26 '18 at 0:44






@Mary According to the usual notation, $$|nabla v|_{L^2}^2:=int_Omega |nabla v|^2=int_Omega (v_x^2+v_y^2)=|v_x|_{L^2}^2+|v_y|^2_{L^2}geq |v_x|^2_{L^2}.$$ Therefore, $|v_x|_{L^2}leq |nabla v|_{L^2}$. Analogously, $|v_y|_{L^2}leq |nabla v|_{L^2}$.
– Pedro
Nov 26 '18 at 0:44




















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