weak solution of PDE and apply Lax-Milgram
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
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
asked Nov 25 '18 at 23:34
Mary
4
add a comment |
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
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
asked Nov 25 '18 at 23:34
Mary
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
add a comment |
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
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
asked Nov 25 '18 at 23:34
Mary
4
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
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
asked Nov 25 '18 at 23:34
Mary
4
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
asked Nov 25 '18 at 23:34
Mary
4
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
edited Nov 26 '18 at 0:00
Pedro
10.2k23068
10.2k23068
asked Nov 25 '18 at 23:34
Mary
4
asked Nov 25 '18 at 23:34
Mary
4
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
add a comment |
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
add a comment |
1 Answer
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votes
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.
answered Nov 25 '18 at 23:53
Pedro
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
|
show 1 more comment
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1 Answer
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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.
answered Nov 25 '18 at 23:53
Pedro
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
|
show 1 more comment
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.
answered Nov 25 '18 at 23:53
Pedro
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
|
show 1 more comment
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.
answered Nov 25 '18 at 23:53
Pedro
10.2k23068
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.
answered Nov 25 '18 at 23:53
Pedro
10.2k23068
answered Nov 25 '18 at 23:53
Pedro
10.2k23068
answered Nov 25 '18 at 23:53
Pedro
10.2k23068
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
|
show 1 more comment
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
|
show 1 more comment
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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