Mixing
Determining a variational formulation of $u^{(4)} = f$ with $3$-rd order BC.
$begingroup$
Below is a problem from a recent exam and I have some questions about it.
Given the boundary value problem:$$dfrac{partial^4 u}{partial x^4} = f,quad fin L^2(0,1)$$
$$u(0) = u''(0) = u'(1) = u'''(1) = 0,$$
derive a variational formulation: $$a(u,v) = l(v)$$
by defining the appropriate Hilbert space $V$ and symmetric, continuous, coercive bilinear form $a(cdot, cdot),$ along with a continuous linear form $L(cdot)$ on $V.$
So I let $$V = H_0^4(0,1) = {vin L^2(0,1)vert, v',v'',v''',v^{(4)}in L^2(0,1)text{ and } v(0) = v''(0) = v'(1) = v'''(1) = 0}$$ and equip it with the norm $||cdot||_V$ which I will get to later. Given the boundary conditions and starting with $u^{(4)} = f,$ it's fairly straightforward to derive:
$$(f,v)=(u^{(4)},v) = -(u''',v) = (u'',v'') = -(u',v'''),,, forall vin V.$$
This clearly suggests that we take:
$$l(v) = (f,v)quadtext{ and }quad a(u,v) = (u'',v'') = int_{0}^1u''(x)v''(x)dx.$$
With this choice, the continuity of $a(cdot, cdot)$ and $l(cdot)$ are just simple Cauchy-Schwartz. However, the coerciveness condition states:
$$exists alpha >0 :a(v,v)geqalpha||v||_V^2,,,forall vin V.quad (1)$$
This is where my confusions are. For instance, if I pick the standard norm on the Sobolev space, that is: $$||v||_V = left(int_0^1v^2+(v')^2+(v'')^2+(v''')^2+(v^{(4)})^2dxright)^{frac 12},$$
then I am not sure how to establish $(1).$On the other hand, I can improvise and take the norm:
$$||v||_2 =left(int_0^1v^2+(v')^2+(v'')^2dxright)^{frac 12},$$
then I can easily establish $(1)$ with the choice $alpha = dfrac 13.$ This follows from proving:
$$||v||_{L^2(0,1)}leq ||v'||_{L^2(0,1)}leq ||v''||_{L^2(0,1)}.$$
So to sum up, my questions are:
$1)$ Is there a problem with picking a norm of my choice to make the inequality easier like I did?
$2)$ If I have to go with the standard norm $||cdot||_V,$ then is there an easy way to establish the coerciviness?
$3)$ Are my choices of $a(u,v)$ and $V$ correct?
Thanks in advance.
ordinary-differential-equations computational-mathematics finite-element-method variational-analysis
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
$endgroup$
add a comment |
$begingroup$
Below is a problem from a recent exam and I have some questions about it.
Given the boundary value problem:$$dfrac{partial^4 u}{partial x^4} = f,quad fin L^2(0,1)$$
$$u(0) = u''(0) = u'(1) = u'''(1) = 0,$$
derive a variational formulation: $$a(u,v) = l(v)$$
by defining the appropriate Hilbert space $V$ and symmetric, continuous, coercive bilinear form $a(cdot, cdot),$ along with a continuous linear form $L(cdot)$ on $V.$
So I let $$V = H_0^4(0,1) = {vin L^2(0,1)vert, v',v'',v''',v^{(4)}in L^2(0,1)text{ and } v(0) = v''(0) = v'(1) = v'''(1) = 0}$$ and equip it with the norm $||cdot||_V$ which I will get to later. Given the boundary conditions and starting with $u^{(4)} = f,$ it's fairly straightforward to derive:
$$(f,v)=(u^{(4)},v) = -(u''',v) = (u'',v'') = -(u',v'''),,, forall vin V.$$
This clearly suggests that we take:
$$l(v) = (f,v)quadtext{ and }quad a(u,v) = (u'',v'') = int_{0}^1u''(x)v''(x)dx.$$
With this choice, the continuity of $a(cdot, cdot)$ and $l(cdot)$ are just simple Cauchy-Schwartz. However, the coerciveness condition states:
$$exists alpha >0 :a(v,v)geqalpha||v||_V^2,,,forall vin V.quad (1)$$
This is where my confusions are. For instance, if I pick the standard norm on the Sobolev space, that is: $$||v||_V = left(int_0^1v^2+(v')^2+(v'')^2+(v''')^2+(v^{(4)})^2dxright)^{frac 12},$$
then I am not sure how to establish $(1).$On the other hand, I can improvise and take the norm:
$$||v||_2 =left(int_0^1v^2+(v')^2+(v'')^2dxright)^{frac 12},$$
then I can easily establish $(1)$ with the choice $alpha = dfrac 13.$ This follows from proving:
$$||v||_{L^2(0,1)}leq ||v'||_{L^2(0,1)}leq ||v''||_{L^2(0,1)}.$$
So to sum up, my questions are:
$1)$ Is there a problem with picking a norm of my choice to make the inequality easier like I did?
$2)$ If I have to go with the standard norm $||cdot||_V,$ then is there an easy way to establish the coerciviness?
$3)$ Are my choices of $a(u,v)$ and $V$ correct?
Thanks in advance.
ordinary-differential-equations computational-mathematics finite-element-method variational-analysis
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
$endgroup$
add a comment |
$begingroup$
Below is a problem from a recent exam and I have some questions about it.
Given the boundary value problem:$$dfrac{partial^4 u}{partial x^4} = f,quad fin L^2(0,1)$$
$$u(0) = u''(0) = u'(1) = u'''(1) = 0,$$
derive a variational formulation: $$a(u,v) = l(v)$$
by defining the appropriate Hilbert space $V$ and symmetric, continuous, coercive bilinear form $a(cdot, cdot),$ along with a continuous linear form $L(cdot)$ on $V.$
So I let $$V = H_0^4(0,1) = {vin L^2(0,1)vert, v',v'',v''',v^{(4)}in L^2(0,1)text{ and } v(0) = v''(0) = v'(1) = v'''(1) = 0}$$ and equip it with the norm $||cdot||_V$ which I will get to later. Given the boundary conditions and starting with $u^{(4)} = f,$ it's fairly straightforward to derive:
$$(f,v)=(u^{(4)},v) = -(u''',v) = (u'',v'') = -(u',v'''),,, forall vin V.$$
This clearly suggests that we take:
$$l(v) = (f,v)quadtext{ and }quad a(u,v) = (u'',v'') = int_{0}^1u''(x)v''(x)dx.$$
With this choice, the continuity of $a(cdot, cdot)$ and $l(cdot)$ are just simple Cauchy-Schwartz. However, the coerciveness condition states:
$$exists alpha >0 :a(v,v)geqalpha||v||_V^2,,,forall vin V.quad (1)$$
This is where my confusions are. For instance, if I pick the standard norm on the Sobolev space, that is: $$||v||_V = left(int_0^1v^2+(v')^2+(v'')^2+(v''')^2+(v^{(4)})^2dxright)^{frac 12},$$
then I am not sure how to establish $(1).$On the other hand, I can improvise and take the norm:
$$||v||_2 =left(int_0^1v^2+(v')^2+(v'')^2dxright)^{frac 12},$$
then I can easily establish $(1)$ with the choice $alpha = dfrac 13.$ This follows from proving:
$$||v||_{L^2(0,1)}leq ||v'||_{L^2(0,1)}leq ||v''||_{L^2(0,1)}.$$
So to sum up, my questions are:
$1)$ Is there a problem with picking a norm of my choice to make the inequality easier like I did?
$2)$ If I have to go with the standard norm $||cdot||_V,$ then is there an easy way to establish the coerciviness?
$3)$ Are my choices of $a(u,v)$ and $V$ correct?
Thanks in advance.
ordinary-differential-equations computational-mathematics finite-element-method variational-analysis
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
$endgroup$
Below is a problem from a recent exam and I have some questions about it.
Given the boundary value problem:$$dfrac{partial^4 u}{partial x^4} = f,quad fin L^2(0,1)$$
$$u(0) = u''(0) = u'(1) = u'''(1) = 0,$$
derive a variational formulation: $$a(u,v) = l(v)$$
by defining the appropriate Hilbert space $V$ and symmetric, continuous, coercive bilinear form $a(cdot, cdot),$ along with a continuous linear form $L(cdot)$ on $V.$
So I let $$V = H_0^4(0,1) = {vin L^2(0,1)vert, v',v'',v''',v^{(4)}in L^2(0,1)text{ and } v(0) = v''(0) = v'(1) = v'''(1) = 0}$$ and equip it with the norm $||cdot||_V$ which I will get to later. Given the boundary conditions and starting with $u^{(4)} = f,$ it's fairly straightforward to derive:
$$(f,v)=(u^{(4)},v) = -(u''',v) = (u'',v'') = -(u',v'''),,, forall vin V.$$
This clearly suggests that we take:
$$l(v) = (f,v)quadtext{ and }quad a(u,v) = (u'',v'') = int_{0}^1u''(x)v''(x)dx.$$
With this choice, the continuity of $a(cdot, cdot)$ and $l(cdot)$ are just simple Cauchy-Schwartz. However, the coerciveness condition states:
$$exists alpha >0 :a(v,v)geqalpha||v||_V^2,,,forall vin V.quad (1)$$
This is where my confusions are. For instance, if I pick the standard norm on the Sobolev space, that is: $$||v||_V = left(int_0^1v^2+(v')^2+(v'')^2+(v''')^2+(v^{(4)})^2dxright)^{frac 12},$$
then I am not sure how to establish $(1).$On the other hand, I can improvise and take the norm:
$$||v||_2 =left(int_0^1v^2+(v')^2+(v'')^2dxright)^{frac 12},$$
then I can easily establish $(1)$ with the choice $alpha = dfrac 13.$ This follows from proving:
$$||v||_{L^2(0,1)}leq ||v'||_{L^2(0,1)}leq ||v''||_{L^2(0,1)}.$$
So to sum up, my questions are:
$1)$ Is there a problem with picking a norm of my choice to make the inequality easier like I did?
$2)$ If I have to go with the standard norm $||cdot||_V,$ then is there an easy way to establish the coerciviness?
$3)$ Are my choices of $a(u,v)$ and $V$ correct?
Thanks in advance.
ordinary-differential-equations computational-mathematics finite-element-method variational-analysis
ordinary-differential-equations computational-mathematics finite-element-method variational-analysis
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
asked Jan 12 at 22:05
dezdichadodezdichado
6,3711929
6,3711929
add a comment |
add a comment |
1 Answer
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$begingroup$
So, your choice of $V$ is wrong. The space $V=H_0^2(0,1)$. The idea of FEM is to reduce the differentiability condition of the strong solution. Here, as the problem is $4^{th}$ order after using integration by parts twice, we get a $2^{nd}$ order weak formulation.
The original problem is reduced to
$$a(u,v)=(u'',v'')$$
where $u$ needs to be only second order differentiable and not fourth. Hence, $V=H_0^2(0,1)$.
edited Jan 14 at 17:11
dezdichado
6,3711929
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
$endgroup$
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
So, your choice of $V$ is wrong. The space $V=H_0^2(0,1)$. The idea of FEM is to reduce the differentiability condition of the strong solution. Here, as the problem is $4^{th}$ order after using integration by parts twice, we get a $2^{nd}$ order weak formulation.
The original problem is reduced to
$$a(u,v)=(u'',v'')$$
where $u$ needs to be only second order differentiable and not fourth. Hence, $V=H_0^2(0,1)$.
edited Jan 14 at 17:11
dezdichado
6,3711929
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
$endgroup$
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
add a comment |
$begingroup$
So, your choice of $V$ is wrong. The space $V=H_0^2(0,1)$. The idea of FEM is to reduce the differentiability condition of the strong solution. Here, as the problem is $4^{th}$ order after using integration by parts twice, we get a $2^{nd}$ order weak formulation.
The original problem is reduced to
$$a(u,v)=(u'',v'')$$
where $u$ needs to be only second order differentiable and not fourth. Hence, $V=H_0^2(0,1)$.
edited Jan 14 at 17:11
dezdichado
6,3711929
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
$endgroup$
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
add a comment |
$begingroup$
So, your choice of $V$ is wrong. The space $V=H_0^2(0,1)$. The idea of FEM is to reduce the differentiability condition of the strong solution. Here, as the problem is $4^{th}$ order after using integration by parts twice, we get a $2^{nd}$ order weak formulation.
The original problem is reduced to
$$a(u,v)=(u'',v'')$$
where $u$ needs to be only second order differentiable and not fourth. Hence, $V=H_0^2(0,1)$.
edited Jan 14 at 17:11
dezdichado
6,3711929
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
$endgroup$
So, your choice of $V$ is wrong. The space $V=H_0^2(0,1)$. The idea of FEM is to reduce the differentiability condition of the strong solution. Here, as the problem is $4^{th}$ order after using integration by parts twice, we get a $2^{nd}$ order weak formulation.
The original problem is reduced to
$$a(u,v)=(u'',v'')$$
where $u$ needs to be only second order differentiable and not fourth. Hence, $V=H_0^2(0,1)$.
edited Jan 14 at 17:11
dezdichado
6,3711929
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
edited Jan 14 at 17:11
dezdichado
6,3711929
edited Jan 14 at 17:11
dezdichado
6,3711929
edited Jan 14 at 17:11
dezdichado
6,3711929
6,3711929
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
answered Jan 14 at 14:49
Abhinav JhaAbhinav Jha
2721211
2721211
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
add a comment |
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
$begingroup$
yeah I realized that shortly after I posted the question.
$endgroup$
– dezdichado
Jan 14 at 17:10
add a comment |
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