FEM: Testing basis functions in a subspace V_h
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In the finite element method, at a certain point we arrive at the following Galerkian problem where it is desired to find the solution $u_h space in V_h$ that solves the following equation:
$$ a(u_h,v_h)=L(v_h) space space space forall v_h in V_h $$
where $a$ and $L$ are, respectively, a bilinear and linear operators. I cannot understand why is normally stated that it is enough to test against a set of basis functions $Phi_i in V_h$ (which are linearly combined to form $u_h$)and not against all functions $vin V_h$
Thank you very much in advance and I hope you may help me understanding this issue.
Kind regards
finite-element-method galerkin-methods
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
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add a comment |
$begingroup$
In the finite element method, at a certain point we arrive at the following Galerkian problem where it is desired to find the solution $u_h space in V_h$ that solves the following equation:
$$ a(u_h,v_h)=L(v_h) space space space forall v_h in V_h $$
where $a$ and $L$ are, respectively, a bilinear and linear operators. I cannot understand why is normally stated that it is enough to test against a set of basis functions $Phi_i in V_h$ (which are linearly combined to form $u_h$)and not against all functions $vin V_h$
Thank you very much in advance and I hope you may help me understanding this issue.
Kind regards
finite-element-method galerkin-methods
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
$endgroup$
add a comment |
$begingroup$
In the finite element method, at a certain point we arrive at the following Galerkian problem where it is desired to find the solution $u_h space in V_h$ that solves the following equation:
$$ a(u_h,v_h)=L(v_h) space space space forall v_h in V_h $$
where $a$ and $L$ are, respectively, a bilinear and linear operators. I cannot understand why is normally stated that it is enough to test against a set of basis functions $Phi_i in V_h$ (which are linearly combined to form $u_h$)and not against all functions $vin V_h$
Thank you very much in advance and I hope you may help me understanding this issue.
Kind regards
finite-element-method galerkin-methods
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
$endgroup$
In the finite element method, at a certain point we arrive at the following Galerkian problem where it is desired to find the solution $u_h space in V_h$ that solves the following equation:
$$ a(u_h,v_h)=L(v_h) space space space forall v_h in V_h $$
where $a$ and $L$ are, respectively, a bilinear and linear operators. I cannot understand why is normally stated that it is enough to test against a set of basis functions $Phi_i in V_h$ (which are linearly combined to form $u_h$)and not against all functions $vin V_h$
Thank you very much in advance and I hope you may help me understanding this issue.
Kind regards
finite-element-method galerkin-methods
finite-element-method galerkin-methods
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
asked Jan 17 at 15:48
Antonio SilvestreAntonio Silvestre
31
31
add a comment |
add a comment |
1 Answer
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The idea of FEM is to find a finite set of equations after getting the Galkerin form. Once we have reached $$a(u_h,v_h)=L(v_h),$$ we want to reduce it to a system of equations of the form $Au=L$. Now an easy way to do this is if we choose $v_h={phi_i}$ because we have finite dimension and hence finite ${phi_i}$, and the biggest thing is that we know these elements.
Now, the question is why only ${phi_i}$ and not any other $n$ elements of $V_h$. Well if we choose $phi_i$ then we can easily compute our matrix $A={a_{ij}}$ as then we have $a_{ij}=a(phi_i,phi_j)$ and we already know these elements.
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
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1 Answer
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1 Answer
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$begingroup$
The idea of FEM is to find a finite set of equations after getting the Galkerin form. Once we have reached $$a(u_h,v_h)=L(v_h),$$ we want to reduce it to a system of equations of the form $Au=L$. Now an easy way to do this is if we choose $v_h={phi_i}$ because we have finite dimension and hence finite ${phi_i}$, and the biggest thing is that we know these elements.
Now, the question is why only ${phi_i}$ and not any other $n$ elements of $V_h$. Well if we choose $phi_i$ then we can easily compute our matrix $A={a_{ij}}$ as then we have $a_{ij}=a(phi_i,phi_j)$ and we already know these elements.
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
$endgroup$
add a comment |
$begingroup$
The idea of FEM is to find a finite set of equations after getting the Galkerin form. Once we have reached $$a(u_h,v_h)=L(v_h),$$ we want to reduce it to a system of equations of the form $Au=L$. Now an easy way to do this is if we choose $v_h={phi_i}$ because we have finite dimension and hence finite ${phi_i}$, and the biggest thing is that we know these elements.
Now, the question is why only ${phi_i}$ and not any other $n$ elements of $V_h$. Well if we choose $phi_i$ then we can easily compute our matrix $A={a_{ij}}$ as then we have $a_{ij}=a(phi_i,phi_j)$ and we already know these elements.
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
$endgroup$
add a comment |
$begingroup$
The idea of FEM is to find a finite set of equations after getting the Galkerin form. Once we have reached $$a(u_h,v_h)=L(v_h),$$ we want to reduce it to a system of equations of the form $Au=L$. Now an easy way to do this is if we choose $v_h={phi_i}$ because we have finite dimension and hence finite ${phi_i}$, and the biggest thing is that we know these elements.
Now, the question is why only ${phi_i}$ and not any other $n$ elements of $V_h$. Well if we choose $phi_i$ then we can easily compute our matrix $A={a_{ij}}$ as then we have $a_{ij}=a(phi_i,phi_j)$ and we already know these elements.
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
$endgroup$
The idea of FEM is to find a finite set of equations after getting the Galkerin form. Once we have reached $$a(u_h,v_h)=L(v_h),$$ we want to reduce it to a system of equations of the form $Au=L$. Now an easy way to do this is if we choose $v_h={phi_i}$ because we have finite dimension and hence finite ${phi_i}$, and the biggest thing is that we know these elements.
Now, the question is why only ${phi_i}$ and not any other $n$ elements of $V_h$. Well if we choose $phi_i$ then we can easily compute our matrix $A={a_{ij}}$ as then we have $a_{ij}=a(phi_i,phi_j)$ and we already know these elements.
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
answered Jan 22 at 11:13
Abhinav JhaAbhinav Jha
2741211
2741211
add a comment |
add a comment |
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