Second variation corresponding to the functional
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I am facing difficulty to calculate the second variation to the following functional.
Define $J: W_{0}^{1,p}(Omega)tomathbb{R}$ by
$J(u)=frac{1}{p}int_{Omega}|nabla u|^p,dx$ where $p>1$.
I am able to calculate the first variation as follows:
$J'(u)phi=int_{Omega},|nabla u|^{p-2}nabla ucdotnablaphi,dx$
which I have got by using the functional $E:mathbb{R}tomathbb{R}$ defined by $E(t)=J(u+tphi)$.
But I am unable to calculate the second variation.
Any type of help is very much appreciated.
Thanks.
multivariable-calculus sobolev-spaces harmonic-functions regularity-theory-of-pdes fractional-sobolev-spaces
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up vote
2
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favorite
I am facing difficulty to calculate the second variation to the following functional.
Define $J: W_{0}^{1,p}(Omega)tomathbb{R}$ by
$J(u)=frac{1}{p}int_{Omega}|nabla u|^p,dx$ where $p>1$.
I am able to calculate the first variation as follows:
$J'(u)phi=int_{Omega},|nabla u|^{p-2}nabla ucdotnablaphi,dx$
which I have got by using the functional $E:mathbb{R}tomathbb{R}$ defined by $E(t)=J(u+tphi)$.
But I am unable to calculate the second variation.
Any type of help is very much appreciated.
Thanks.
multivariable-calculus sobolev-spaces harmonic-functions regularity-theory-of-pdes fractional-sobolev-spaces
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am facing difficulty to calculate the second variation to the following functional.
Define $J: W_{0}^{1,p}(Omega)tomathbb{R}$ by
$J(u)=frac{1}{p}int_{Omega}|nabla u|^p,dx$ where $p>1$.
I am able to calculate the first variation as follows:
$J'(u)phi=int_{Omega},|nabla u|^{p-2}nabla ucdotnablaphi,dx$
which I have got by using the functional $E:mathbb{R}tomathbb{R}$ defined by $E(t)=J(u+tphi)$.
But I am unable to calculate the second variation.
Any type of help is very much appreciated.
Thanks.
multivariable-calculus sobolev-spaces harmonic-functions regularity-theory-of-pdes fractional-sobolev-spaces
I am facing difficulty to calculate the second variation to the following functional.
Define $J: W_{0}^{1,p}(Omega)tomathbb{R}$ by
$J(u)=frac{1}{p}int_{Omega}|nabla u|^p,dx$ where $p>1$.
I am able to calculate the first variation as follows:
$J'(u)phi=int_{Omega},|nabla u|^{p-2}nabla ucdotnablaphi,dx$
which I have got by using the functional $E:mathbb{R}tomathbb{R}$ defined by $E(t)=J(u+tphi)$.
But I am unable to calculate the second variation.
Any type of help is very much appreciated.
Thanks.
multivariable-calculus sobolev-spaces harmonic-functions regularity-theory-of-pdes fractional-sobolev-spaces
multivariable-calculus sobolev-spaces harmonic-functions regularity-theory-of-pdes fractional-sobolev-spaces
edited Apr 3 at 20:25
asked Apr 3 at 18:09
Mathlover
1038
1038
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2 Answers
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Calculating $delta^2J$ by using the standard definition is a bit tedious and tricky.
Notation: I assume that $Omegainmathbb{R}^n$, $nge 2$. Also, to denote the scalar product of two vectors $boldsymbol{a},boldsymbol{b}inmathbb{R}^n$ I will use both the notations $boldsymbol{a}cdotboldsymbol{b}$ and $langleboldsymbol{a},boldsymbol{b}rangle$ since the former is quickly understandable while this last one is easier to handle when its arguments are sums of vectors. Then
$$
|boldsymbol{a}|=(boldsymbol{a}cdotboldsymbol{a})^{1over 2}=langleboldsymbol{a}cdotboldsymbol{a}rangle^{1over 2}=sqrt[2]{sum_{i=1}^n a_i^2}qquadboldsymbol{a}=(a_1,dots,a_n)
$$
By using the standard definition of second variation of a functional,we have
$$
begin{split}
delta^2 J(u,phi)&=frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over p}frac{mathrm{d}}{mathrm{d}t}intlimits_Omega |nabla u+tnablaphi|^pmathrm{d}xright]_{,t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over 2}intlimits_Omega |nabla u+tnablaphi|^{p-2}frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphiranglemathrm{d}xright]_{,t=0}
end{split}label{1}tag{1}
$$
Now we have
$$
begin{split}
frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphirangle&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^n(partial_{x_i} u+tpartial_{x_i}phi)^2\
&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^nbig[(partial_{x_i} u)^2+2t(partial_{x_i} u,partial_{x_i}phi)+t^2(partial_{x_i}phi)^2big]\
&=2sum_{i=1}^nbig[(partial_{x_i} u,partial_{x_i}phi)+t(partial_{x_i}phi)^2big]=2big(nabla ucdotnablaphi+t|nablaphi|^2big)
end{split}label{2}tag{2}
$$
and by using eqref{2} in eqref{1} jointly with Leibnitz's rule we obtain
$$
begin{split}
delta^2 J(u,phi)=&frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
=&frac{mathrm{d}}{mathrm{d}t}left[intlimits_Omega |nabla u+tnablaphi|^{p-2}big(nabla ucdotnablaphi+t|nablaphi|^2big)mathrm{d}xright]_{,t=0}\
=&left[(p-2)intlimits_Omega |nabla u+tnablaphi|^{p-4}big(nabla ucdotnablaphi+t|nablaphi|^2big)^2mathrm{d}xright.\
&+left.intlimits_Omega |nabla u+tnablaphi|^{p-2}|nablaphi|^2mathrm{d}xright]_{,t=0}\
=&(p-2)intlimits_Omega |nabla u|^{p-4}(nabla ucdotnablaphi)^2mathrm{d}x+
intlimits_Omega |nabla u|^{p-2}|nablaphi|^2mathrm{d}x
end{split}label{3}tag{3}
$$
Notes
The deduction above is only formal since, without further hypotheses, if $0< p<2$ we cannot be sure that $|nabla u|neq 0$ on a set of zero Lebesgue measure non we can assume the integrability of its inverse powers $|nabla u|^{p-2}$ and $|nabla u|^{p-4}$.
Perhaps in this case, instead of the standard definition of $delta^2J$, the use of the classical definition of the second derivative of a functional (as can be seen in [1], §1.3, pp. 23-26 and also §1.4, pp. 26-29) could have eased the task: according to this definition, which dates back at least to the work of Vito Volterra,
$$
delta^2J(u,phi,psi)=frac{{partial}^2}{partial tpartial s} J(u+tphi+spsi)|_{t,s=0}
$$
This definition has the advantage that it allows the calculation of the second variation directly from the first, without having to take care of terms that will vanish at the end of calculations. Another advantage of this definition is that it works also when $u,phi$ and $psi$ are more abstract objects, that for which a concept of multiplication is not univocally defined (generalized functions and the likes). However, I preferred to follow the standard route because the OP asked so and because it is interactive.
[1] A. Ambrosetti and G. Prodi (1995), A primer of nonlinear analysis (English)
Cambridge Studies in Advanced Mathematics. 34. Cambridge: Cambridge University Press, pp. 180, ISBN: 0-521-48573-8, MR1336591, Zbl 0818.47059.
1
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
add a comment |
up vote
2
down vote
I propose a quick shortcut, based on dimensional analysis. This method does not yield the complete result, however.
The second derivative
$J''(u)phi$ must be $p-2$-homogeneous in $nabla u$ and quadratic in $nabla phi$, thus it must be that
$$
J''(u)phi =
C_1int_{Omega} |nabla u|^{p-2}|nablaphi|^2 + C_2int_{Omega} |nabla u|^{p-4}(nabla u cdot nablaphi)^2,$$
for some constants $C_1$ and $C_2$ that cannot be determined by homogeneity alone. However, for $phi=u$ it must be that
$$
begin{split}
J''(u)u =&=left.frac{d^2}{depsilon^2}right|_{epsilon=0}frac1pint_{Omega}|nabla(u+epsilon u)|^p\ &= left.frac{d^2}{depsilon^2}frac{(1+epsilon)^p}{p}right|_{epsilon=0}int_Omega|nabla u|^p=(C_1+C_2)int_Omega|nabla u|^p,
end{split}$$
so $C_1+C_2=p-1$.
I don't know how to find another equation, thus determining $C_1$ and $C_2$, so this answer is incomplete. But I do think that this method may be useful. It is always good to have these shortcuts available, when performing long computations with a high chance of mistake (or computations with high entropy, in the words of the book Street fighting mathematics).
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Calculating $delta^2J$ by using the standard definition is a bit tedious and tricky.
Notation: I assume that $Omegainmathbb{R}^n$, $nge 2$. Also, to denote the scalar product of two vectors $boldsymbol{a},boldsymbol{b}inmathbb{R}^n$ I will use both the notations $boldsymbol{a}cdotboldsymbol{b}$ and $langleboldsymbol{a},boldsymbol{b}rangle$ since the former is quickly understandable while this last one is easier to handle when its arguments are sums of vectors. Then
$$
|boldsymbol{a}|=(boldsymbol{a}cdotboldsymbol{a})^{1over 2}=langleboldsymbol{a}cdotboldsymbol{a}rangle^{1over 2}=sqrt[2]{sum_{i=1}^n a_i^2}qquadboldsymbol{a}=(a_1,dots,a_n)
$$
By using the standard definition of second variation of a functional,we have
$$
begin{split}
delta^2 J(u,phi)&=frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over p}frac{mathrm{d}}{mathrm{d}t}intlimits_Omega |nabla u+tnablaphi|^pmathrm{d}xright]_{,t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over 2}intlimits_Omega |nabla u+tnablaphi|^{p-2}frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphiranglemathrm{d}xright]_{,t=0}
end{split}label{1}tag{1}
$$
Now we have
$$
begin{split}
frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphirangle&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^n(partial_{x_i} u+tpartial_{x_i}phi)^2\
&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^nbig[(partial_{x_i} u)^2+2t(partial_{x_i} u,partial_{x_i}phi)+t^2(partial_{x_i}phi)^2big]\
&=2sum_{i=1}^nbig[(partial_{x_i} u,partial_{x_i}phi)+t(partial_{x_i}phi)^2big]=2big(nabla ucdotnablaphi+t|nablaphi|^2big)
end{split}label{2}tag{2}
$$
and by using eqref{2} in eqref{1} jointly with Leibnitz's rule we obtain
$$
begin{split}
delta^2 J(u,phi)=&frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
=&frac{mathrm{d}}{mathrm{d}t}left[intlimits_Omega |nabla u+tnablaphi|^{p-2}big(nabla ucdotnablaphi+t|nablaphi|^2big)mathrm{d}xright]_{,t=0}\
=&left[(p-2)intlimits_Omega |nabla u+tnablaphi|^{p-4}big(nabla ucdotnablaphi+t|nablaphi|^2big)^2mathrm{d}xright.\
&+left.intlimits_Omega |nabla u+tnablaphi|^{p-2}|nablaphi|^2mathrm{d}xright]_{,t=0}\
=&(p-2)intlimits_Omega |nabla u|^{p-4}(nabla ucdotnablaphi)^2mathrm{d}x+
intlimits_Omega |nabla u|^{p-2}|nablaphi|^2mathrm{d}x
end{split}label{3}tag{3}
$$
Notes
The deduction above is only formal since, without further hypotheses, if $0< p<2$ we cannot be sure that $|nabla u|neq 0$ on a set of zero Lebesgue measure non we can assume the integrability of its inverse powers $|nabla u|^{p-2}$ and $|nabla u|^{p-4}$.
Perhaps in this case, instead of the standard definition of $delta^2J$, the use of the classical definition of the second derivative of a functional (as can be seen in [1], §1.3, pp. 23-26 and also §1.4, pp. 26-29) could have eased the task: according to this definition, which dates back at least to the work of Vito Volterra,
$$
delta^2J(u,phi,psi)=frac{{partial}^2}{partial tpartial s} J(u+tphi+spsi)|_{t,s=0}
$$
This definition has the advantage that it allows the calculation of the second variation directly from the first, without having to take care of terms that will vanish at the end of calculations. Another advantage of this definition is that it works also when $u,phi$ and $psi$ are more abstract objects, that for which a concept of multiplication is not univocally defined (generalized functions and the likes). However, I preferred to follow the standard route because the OP asked so and because it is interactive.
[1] A. Ambrosetti and G. Prodi (1995), A primer of nonlinear analysis (English)
Cambridge Studies in Advanced Mathematics. 34. Cambridge: Cambridge University Press, pp. 180, ISBN: 0-521-48573-8, MR1336591, Zbl 0818.47059.
1
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
add a comment |
up vote
2
down vote
Calculating $delta^2J$ by using the standard definition is a bit tedious and tricky.
Notation: I assume that $Omegainmathbb{R}^n$, $nge 2$. Also, to denote the scalar product of two vectors $boldsymbol{a},boldsymbol{b}inmathbb{R}^n$ I will use both the notations $boldsymbol{a}cdotboldsymbol{b}$ and $langleboldsymbol{a},boldsymbol{b}rangle$ since the former is quickly understandable while this last one is easier to handle when its arguments are sums of vectors. Then
$$
|boldsymbol{a}|=(boldsymbol{a}cdotboldsymbol{a})^{1over 2}=langleboldsymbol{a}cdotboldsymbol{a}rangle^{1over 2}=sqrt[2]{sum_{i=1}^n a_i^2}qquadboldsymbol{a}=(a_1,dots,a_n)
$$
By using the standard definition of second variation of a functional,we have
$$
begin{split}
delta^2 J(u,phi)&=frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over p}frac{mathrm{d}}{mathrm{d}t}intlimits_Omega |nabla u+tnablaphi|^pmathrm{d}xright]_{,t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over 2}intlimits_Omega |nabla u+tnablaphi|^{p-2}frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphiranglemathrm{d}xright]_{,t=0}
end{split}label{1}tag{1}
$$
Now we have
$$
begin{split}
frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphirangle&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^n(partial_{x_i} u+tpartial_{x_i}phi)^2\
&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^nbig[(partial_{x_i} u)^2+2t(partial_{x_i} u,partial_{x_i}phi)+t^2(partial_{x_i}phi)^2big]\
&=2sum_{i=1}^nbig[(partial_{x_i} u,partial_{x_i}phi)+t(partial_{x_i}phi)^2big]=2big(nabla ucdotnablaphi+t|nablaphi|^2big)
end{split}label{2}tag{2}
$$
and by using eqref{2} in eqref{1} jointly with Leibnitz's rule we obtain
$$
begin{split}
delta^2 J(u,phi)=&frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
=&frac{mathrm{d}}{mathrm{d}t}left[intlimits_Omega |nabla u+tnablaphi|^{p-2}big(nabla ucdotnablaphi+t|nablaphi|^2big)mathrm{d}xright]_{,t=0}\
=&left[(p-2)intlimits_Omega |nabla u+tnablaphi|^{p-4}big(nabla ucdotnablaphi+t|nablaphi|^2big)^2mathrm{d}xright.\
&+left.intlimits_Omega |nabla u+tnablaphi|^{p-2}|nablaphi|^2mathrm{d}xright]_{,t=0}\
=&(p-2)intlimits_Omega |nabla u|^{p-4}(nabla ucdotnablaphi)^2mathrm{d}x+
intlimits_Omega |nabla u|^{p-2}|nablaphi|^2mathrm{d}x
end{split}label{3}tag{3}
$$
Notes
The deduction above is only formal since, without further hypotheses, if $0< p<2$ we cannot be sure that $|nabla u|neq 0$ on a set of zero Lebesgue measure non we can assume the integrability of its inverse powers $|nabla u|^{p-2}$ and $|nabla u|^{p-4}$.
Perhaps in this case, instead of the standard definition of $delta^2J$, the use of the classical definition of the second derivative of a functional (as can be seen in [1], §1.3, pp. 23-26 and also §1.4, pp. 26-29) could have eased the task: according to this definition, which dates back at least to the work of Vito Volterra,
$$
delta^2J(u,phi,psi)=frac{{partial}^2}{partial tpartial s} J(u+tphi+spsi)|_{t,s=0}
$$
This definition has the advantage that it allows the calculation of the second variation directly from the first, without having to take care of terms that will vanish at the end of calculations. Another advantage of this definition is that it works also when $u,phi$ and $psi$ are more abstract objects, that for which a concept of multiplication is not univocally defined (generalized functions and the likes). However, I preferred to follow the standard route because the OP asked so and because it is interactive.
[1] A. Ambrosetti and G. Prodi (1995), A primer of nonlinear analysis (English)
Cambridge Studies in Advanced Mathematics. 34. Cambridge: Cambridge University Press, pp. 180, ISBN: 0-521-48573-8, MR1336591, Zbl 0818.47059.
1
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
add a comment |
up vote
2
down vote
up vote
2
down vote
Calculating $delta^2J$ by using the standard definition is a bit tedious and tricky.
Notation: I assume that $Omegainmathbb{R}^n$, $nge 2$. Also, to denote the scalar product of two vectors $boldsymbol{a},boldsymbol{b}inmathbb{R}^n$ I will use both the notations $boldsymbol{a}cdotboldsymbol{b}$ and $langleboldsymbol{a},boldsymbol{b}rangle$ since the former is quickly understandable while this last one is easier to handle when its arguments are sums of vectors. Then
$$
|boldsymbol{a}|=(boldsymbol{a}cdotboldsymbol{a})^{1over 2}=langleboldsymbol{a}cdotboldsymbol{a}rangle^{1over 2}=sqrt[2]{sum_{i=1}^n a_i^2}qquadboldsymbol{a}=(a_1,dots,a_n)
$$
By using the standard definition of second variation of a functional,we have
$$
begin{split}
delta^2 J(u,phi)&=frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over p}frac{mathrm{d}}{mathrm{d}t}intlimits_Omega |nabla u+tnablaphi|^pmathrm{d}xright]_{,t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over 2}intlimits_Omega |nabla u+tnablaphi|^{p-2}frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphiranglemathrm{d}xright]_{,t=0}
end{split}label{1}tag{1}
$$
Now we have
$$
begin{split}
frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphirangle&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^n(partial_{x_i} u+tpartial_{x_i}phi)^2\
&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^nbig[(partial_{x_i} u)^2+2t(partial_{x_i} u,partial_{x_i}phi)+t^2(partial_{x_i}phi)^2big]\
&=2sum_{i=1}^nbig[(partial_{x_i} u,partial_{x_i}phi)+t(partial_{x_i}phi)^2big]=2big(nabla ucdotnablaphi+t|nablaphi|^2big)
end{split}label{2}tag{2}
$$
and by using eqref{2} in eqref{1} jointly with Leibnitz's rule we obtain
$$
begin{split}
delta^2 J(u,phi)=&frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
=&frac{mathrm{d}}{mathrm{d}t}left[intlimits_Omega |nabla u+tnablaphi|^{p-2}big(nabla ucdotnablaphi+t|nablaphi|^2big)mathrm{d}xright]_{,t=0}\
=&left[(p-2)intlimits_Omega |nabla u+tnablaphi|^{p-4}big(nabla ucdotnablaphi+t|nablaphi|^2big)^2mathrm{d}xright.\
&+left.intlimits_Omega |nabla u+tnablaphi|^{p-2}|nablaphi|^2mathrm{d}xright]_{,t=0}\
=&(p-2)intlimits_Omega |nabla u|^{p-4}(nabla ucdotnablaphi)^2mathrm{d}x+
intlimits_Omega |nabla u|^{p-2}|nablaphi|^2mathrm{d}x
end{split}label{3}tag{3}
$$
Notes
The deduction above is only formal since, without further hypotheses, if $0< p<2$ we cannot be sure that $|nabla u|neq 0$ on a set of zero Lebesgue measure non we can assume the integrability of its inverse powers $|nabla u|^{p-2}$ and $|nabla u|^{p-4}$.
Perhaps in this case, instead of the standard definition of $delta^2J$, the use of the classical definition of the second derivative of a functional (as can be seen in [1], §1.3, pp. 23-26 and also §1.4, pp. 26-29) could have eased the task: according to this definition, which dates back at least to the work of Vito Volterra,
$$
delta^2J(u,phi,psi)=frac{{partial}^2}{partial tpartial s} J(u+tphi+spsi)|_{t,s=0}
$$
This definition has the advantage that it allows the calculation of the second variation directly from the first, without having to take care of terms that will vanish at the end of calculations. Another advantage of this definition is that it works also when $u,phi$ and $psi$ are more abstract objects, that for which a concept of multiplication is not univocally defined (generalized functions and the likes). However, I preferred to follow the standard route because the OP asked so and because it is interactive.
[1] A. Ambrosetti and G. Prodi (1995), A primer of nonlinear analysis (English)
Cambridge Studies in Advanced Mathematics. 34. Cambridge: Cambridge University Press, pp. 180, ISBN: 0-521-48573-8, MR1336591, Zbl 0818.47059.
Calculating $delta^2J$ by using the standard definition is a bit tedious and tricky.
Notation: I assume that $Omegainmathbb{R}^n$, $nge 2$. Also, to denote the scalar product of two vectors $boldsymbol{a},boldsymbol{b}inmathbb{R}^n$ I will use both the notations $boldsymbol{a}cdotboldsymbol{b}$ and $langleboldsymbol{a},boldsymbol{b}rangle$ since the former is quickly understandable while this last one is easier to handle when its arguments are sums of vectors. Then
$$
|boldsymbol{a}|=(boldsymbol{a}cdotboldsymbol{a})^{1over 2}=langleboldsymbol{a}cdotboldsymbol{a}rangle^{1over 2}=sqrt[2]{sum_{i=1}^n a_i^2}qquadboldsymbol{a}=(a_1,dots,a_n)
$$
By using the standard definition of second variation of a functional,we have
$$
begin{split}
delta^2 J(u,phi)&=frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over p}frac{mathrm{d}}{mathrm{d}t}intlimits_Omega |nabla u+tnablaphi|^pmathrm{d}xright]_{,t=0}\
&=frac{mathrm{d}}{mathrm{d}t}left[{1over 2}intlimits_Omega |nabla u+tnablaphi|^{p-2}frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphiranglemathrm{d}xright]_{,t=0}
end{split}label{1}tag{1}
$$
Now we have
$$
begin{split}
frac{mathrm{d}}{mathrm{d}t}langlenabla u+tnablaphi,nabla u+tnablaphirangle&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^n(partial_{x_i} u+tpartial_{x_i}phi)^2\
&=frac{mathrm{d}}{mathrm{d}t}sum_{i=1}^nbig[(partial_{x_i} u)^2+2t(partial_{x_i} u,partial_{x_i}phi)+t^2(partial_{x_i}phi)^2big]\
&=2sum_{i=1}^nbig[(partial_{x_i} u,partial_{x_i}phi)+t(partial_{x_i}phi)^2big]=2big(nabla ucdotnablaphi+t|nablaphi|^2big)
end{split}label{2}tag{2}
$$
and by using eqref{2} in eqref{1} jointly with Leibnitz's rule we obtain
$$
begin{split}
delta^2 J(u,phi)=&frac{mathrm{d}^2}{mathrm{d}t^2} J(u+tphi)|_{t=0}\
=&frac{mathrm{d}}{mathrm{d}t}left[intlimits_Omega |nabla u+tnablaphi|^{p-2}big(nabla ucdotnablaphi+t|nablaphi|^2big)mathrm{d}xright]_{,t=0}\
=&left[(p-2)intlimits_Omega |nabla u+tnablaphi|^{p-4}big(nabla ucdotnablaphi+t|nablaphi|^2big)^2mathrm{d}xright.\
&+left.intlimits_Omega |nabla u+tnablaphi|^{p-2}|nablaphi|^2mathrm{d}xright]_{,t=0}\
=&(p-2)intlimits_Omega |nabla u|^{p-4}(nabla ucdotnablaphi)^2mathrm{d}x+
intlimits_Omega |nabla u|^{p-2}|nablaphi|^2mathrm{d}x
end{split}label{3}tag{3}
$$
Notes
The deduction above is only formal since, without further hypotheses, if $0< p<2$ we cannot be sure that $|nabla u|neq 0$ on a set of zero Lebesgue measure non we can assume the integrability of its inverse powers $|nabla u|^{p-2}$ and $|nabla u|^{p-4}$.
Perhaps in this case, instead of the standard definition of $delta^2J$, the use of the classical definition of the second derivative of a functional (as can be seen in [1], §1.3, pp. 23-26 and also §1.4, pp. 26-29) could have eased the task: according to this definition, which dates back at least to the work of Vito Volterra,
$$
delta^2J(u,phi,psi)=frac{{partial}^2}{partial tpartial s} J(u+tphi+spsi)|_{t,s=0}
$$
This definition has the advantage that it allows the calculation of the second variation directly from the first, without having to take care of terms that will vanish at the end of calculations. Another advantage of this definition is that it works also when $u,phi$ and $psi$ are more abstract objects, that for which a concept of multiplication is not univocally defined (generalized functions and the likes). However, I preferred to follow the standard route because the OP asked so and because it is interactive.
[1] A. Ambrosetti and G. Prodi (1995), A primer of nonlinear analysis (English)
Cambridge Studies in Advanced Mathematics. 34. Cambridge: Cambridge University Press, pp. 180, ISBN: 0-521-48573-8, MR1336591, Zbl 0818.47059.
edited Nov 14 at 5:51
answered Nov 13 at 22:15
Daniele Tampieri
1,5371619
1,5371619
1
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
add a comment |
1
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
1
1
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
Dear Daniele, can you please have a look at my answer too? There is a disagreement between our answers, maybe you will spot where's the mistake. Thanks!
– Giuseppe Negro
Nov 13 at 22:33
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
@GiuseppeNegro: there is one error and one typo: the error is mine, since I erroneously wrote $${1over p-2}text{ instead of } p-2,$$ while the typo is yours, since in your deduction you considered $$pJ(u)text{ instead of }J(u).$$ In such cases I console myself recalling that even Poincaré did many errors in simple calculations. And do not forget to correct your answer so I'll upvote it.
– Daniele Tampieri
Nov 14 at 6:36
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
It is done. And thank you for mentioning Poincaré. I also do many computational errors. Does it mean that we are as good as Poincaré? :-)
– Giuseppe Negro
Nov 14 at 8:56
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
Let's hope so, while working hard enjoying our work! :D
– Daniele Tampieri
Nov 14 at 9:05
add a comment |
up vote
2
down vote
I propose a quick shortcut, based on dimensional analysis. This method does not yield the complete result, however.
The second derivative
$J''(u)phi$ must be $p-2$-homogeneous in $nabla u$ and quadratic in $nabla phi$, thus it must be that
$$
J''(u)phi =
C_1int_{Omega} |nabla u|^{p-2}|nablaphi|^2 + C_2int_{Omega} |nabla u|^{p-4}(nabla u cdot nablaphi)^2,$$
for some constants $C_1$ and $C_2$ that cannot be determined by homogeneity alone. However, for $phi=u$ it must be that
$$
begin{split}
J''(u)u =&=left.frac{d^2}{depsilon^2}right|_{epsilon=0}frac1pint_{Omega}|nabla(u+epsilon u)|^p\ &= left.frac{d^2}{depsilon^2}frac{(1+epsilon)^p}{p}right|_{epsilon=0}int_Omega|nabla u|^p=(C_1+C_2)int_Omega|nabla u|^p,
end{split}$$
so $C_1+C_2=p-1$.
I don't know how to find another equation, thus determining $C_1$ and $C_2$, so this answer is incomplete. But I do think that this method may be useful. It is always good to have these shortcuts available, when performing long computations with a high chance of mistake (or computations with high entropy, in the words of the book Street fighting mathematics).
add a comment |
up vote
2
down vote
I propose a quick shortcut, based on dimensional analysis. This method does not yield the complete result, however.
The second derivative
$J''(u)phi$ must be $p-2$-homogeneous in $nabla u$ and quadratic in $nabla phi$, thus it must be that
$$
J''(u)phi =
C_1int_{Omega} |nabla u|^{p-2}|nablaphi|^2 + C_2int_{Omega} |nabla u|^{p-4}(nabla u cdot nablaphi)^2,$$
for some constants $C_1$ and $C_2$ that cannot be determined by homogeneity alone. However, for $phi=u$ it must be that
$$
begin{split}
J''(u)u =&=left.frac{d^2}{depsilon^2}right|_{epsilon=0}frac1pint_{Omega}|nabla(u+epsilon u)|^p\ &= left.frac{d^2}{depsilon^2}frac{(1+epsilon)^p}{p}right|_{epsilon=0}int_Omega|nabla u|^p=(C_1+C_2)int_Omega|nabla u|^p,
end{split}$$
so $C_1+C_2=p-1$.
I don't know how to find another equation, thus determining $C_1$ and $C_2$, so this answer is incomplete. But I do think that this method may be useful. It is always good to have these shortcuts available, when performing long computations with a high chance of mistake (or computations with high entropy, in the words of the book Street fighting mathematics).
add a comment |
up vote
2
down vote
up vote
2
down vote
I propose a quick shortcut, based on dimensional analysis. This method does not yield the complete result, however.
The second derivative
$J''(u)phi$ must be $p-2$-homogeneous in $nabla u$ and quadratic in $nabla phi$, thus it must be that
$$
J''(u)phi =
C_1int_{Omega} |nabla u|^{p-2}|nablaphi|^2 + C_2int_{Omega} |nabla u|^{p-4}(nabla u cdot nablaphi)^2,$$
for some constants $C_1$ and $C_2$ that cannot be determined by homogeneity alone. However, for $phi=u$ it must be that
$$
begin{split}
J''(u)u =&=left.frac{d^2}{depsilon^2}right|_{epsilon=0}frac1pint_{Omega}|nabla(u+epsilon u)|^p\ &= left.frac{d^2}{depsilon^2}frac{(1+epsilon)^p}{p}right|_{epsilon=0}int_Omega|nabla u|^p=(C_1+C_2)int_Omega|nabla u|^p,
end{split}$$
so $C_1+C_2=p-1$.
I don't know how to find another equation, thus determining $C_1$ and $C_2$, so this answer is incomplete. But I do think that this method may be useful. It is always good to have these shortcuts available, when performing long computations with a high chance of mistake (or computations with high entropy, in the words of the book Street fighting mathematics).
I propose a quick shortcut, based on dimensional analysis. This method does not yield the complete result, however.
The second derivative
$J''(u)phi$ must be $p-2$-homogeneous in $nabla u$ and quadratic in $nabla phi$, thus it must be that
$$
J''(u)phi =
C_1int_{Omega} |nabla u|^{p-2}|nablaphi|^2 + C_2int_{Omega} |nabla u|^{p-4}(nabla u cdot nablaphi)^2,$$
for some constants $C_1$ and $C_2$ that cannot be determined by homogeneity alone. However, for $phi=u$ it must be that
$$
begin{split}
J''(u)u =&=left.frac{d^2}{depsilon^2}right|_{epsilon=0}frac1pint_{Omega}|nabla(u+epsilon u)|^p\ &= left.frac{d^2}{depsilon^2}frac{(1+epsilon)^p}{p}right|_{epsilon=0}int_Omega|nabla u|^p=(C_1+C_2)int_Omega|nabla u|^p,
end{split}$$
so $C_1+C_2=p-1$.
I don't know how to find another equation, thus determining $C_1$ and $C_2$, so this answer is incomplete. But I do think that this method may be useful. It is always good to have these shortcuts available, when performing long computations with a high chance of mistake (or computations with high entropy, in the words of the book Street fighting mathematics).
edited Nov 14 at 15:09
answered Nov 13 at 22:32
Giuseppe Negro
17k329121
17k329121
add a comment |
add a comment |
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