Prove that $(Au)(t)=frac{d^{2}u(t)}{dt^{2}}$ is self-adjoint












1












$begingroup$


Let $D(A)={ u in L_2(0,T)| u, frac{du}{dt}$ are absolutly continuous with $frac{du}{dt} in L_2(0,T)$, $u(0)=u(T)=0}$



and, $(Au)(t)=frac{d^{2}u}{dt^{2}}$ prove that $A$ is self-adjoint.



Trial



Consider,



$langle Au(t), v(t) rangle$= $int_{0}^{T}frac{d^{2}}{dt^{2}} u(t)v(t) dt$= $int_{0}^{T}frac{d^{2}}{dt^{2}}(u(t)v(t))dt -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt- int_{0}^{T}(frac{d^2}{dt^2}v(t))u(t)$ $dt$



=$frac{dv(t)u(t)}{dt}|_{t=T}-frac{dv(t)u(t)}{dt}|_{t=0} -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt -langle u(t), Av(t) rangle$



Then I'm stuck






functional-analysis proof-verification operator-theory hilbert-spaces adjoint-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question? Is it whether your attempt is correct?
    $endgroup$
    – LordVader007
    Jan 8 at 4:06










  • $begingroup$
    yes, I forgot to mention it
    $endgroup$
    – Dreamer123
    Jan 8 at 4:12
















1












$begingroup$


Let $D(A)={ u in L_2(0,T)| u, frac{du}{dt}$ are absolutly continuous with $frac{du}{dt} in L_2(0,T)$, $u(0)=u(T)=0}$



and, $(Au)(t)=frac{d^{2}u}{dt^{2}}$ prove that $A$ is self-adjoint.



Trial



Consider,



$langle Au(t), v(t) rangle$= $int_{0}^{T}frac{d^{2}}{dt^{2}} u(t)v(t) dt$= $int_{0}^{T}frac{d^{2}}{dt^{2}}(u(t)v(t))dt -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt- int_{0}^{T}(frac{d^2}{dt^2}v(t))u(t)$ $dt$



=$frac{dv(t)u(t)}{dt}|_{t=T}-frac{dv(t)u(t)}{dt}|_{t=0} -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt -langle u(t), Av(t) rangle$



Then I'm stuck






functional-analysis proof-verification operator-theory hilbert-spaces adjoint-operators







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question? Is it whether your attempt is correct?
    $endgroup$
    – LordVader007
    Jan 8 at 4:06










  • $begingroup$
    yes, I forgot to mention it
    $endgroup$
    – Dreamer123
    Jan 8 at 4:12














1












1








1





$begingroup$


Let $D(A)={ u in L_2(0,T)| u, frac{du}{dt}$ are absolutly continuous with $frac{du}{dt} in L_2(0,T)$, $u(0)=u(T)=0}$



and, $(Au)(t)=frac{d^{2}u}{dt^{2}}$ prove that $A$ is self-adjoint.



Trial



Consider,



$langle Au(t), v(t) rangle$= $int_{0}^{T}frac{d^{2}}{dt^{2}} u(t)v(t) dt$= $int_{0}^{T}frac{d^{2}}{dt^{2}}(u(t)v(t))dt -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt- int_{0}^{T}(frac{d^2}{dt^2}v(t))u(t)$ $dt$



=$frac{dv(t)u(t)}{dt}|_{t=T}-frac{dv(t)u(t)}{dt}|_{t=0} -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt -langle u(t), Av(t) rangle$



Then I'm stuck






functional-analysis proof-verification operator-theory hilbert-spaces adjoint-operators







share|cite|improve this question











$endgroup$




Let $D(A)={ u in L_2(0,T)| u, frac{du}{dt}$ are absolutly continuous with $frac{du}{dt} in L_2(0,T)$, $u(0)=u(T)=0}$



and, $(Au)(t)=frac{d^{2}u}{dt^{2}}$ prove that $A$ is self-adjoint.



Trial



Consider,



$langle Au(t), v(t) rangle$= $int_{0}^{T}frac{d^{2}}{dt^{2}} u(t)v(t) dt$= $int_{0}^{T}frac{d^{2}}{dt^{2}}(u(t)v(t))dt -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt- int_{0}^{T}(frac{d^2}{dt^2}v(t))u(t)$ $dt$



=$frac{dv(t)u(t)}{dt}|_{t=T}-frac{dv(t)u(t)}{dt}|_{t=0} -2int_{0}^{T}frac{du(t)}{dt} frac{dv(t)}{dt} dt -langle u(t), Av(t) rangle$



Then I'm stuck






functional-analysis proof-verification operator-theory hilbert-spaces adjoint-operators




functional-analysis proof-verification operator-theory hilbert-spaces adjoint-operators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 4:52







Dreamer123

















asked Jan 8 at 3:54









Dreamer123Dreamer123

27629




27629












  • $begingroup$
    What is your question? Is it whether your attempt is correct?
    $endgroup$
    – LordVader007
    Jan 8 at 4:06










  • $begingroup$
    yes, I forgot to mention it
    $endgroup$
    – Dreamer123
    Jan 8 at 4:12


















  • $begingroup$
    What is your question? Is it whether your attempt is correct?
    $endgroup$
    – LordVader007
    Jan 8 at 4:06










  • $begingroup$
    yes, I forgot to mention it
    $endgroup$
    – Dreamer123
    Jan 8 at 4:12
















$begingroup$
What is your question? Is it whether your attempt is correct?
$endgroup$
– LordVader007
Jan 8 at 4:06




$begingroup$
What is your question? Is it whether your attempt is correct?
$endgroup$
– LordVader007
Jan 8 at 4:06












$begingroup$
yes, I forgot to mention it
$endgroup$
– Dreamer123
Jan 8 at 4:12




$begingroup$
yes, I forgot to mention it
$endgroup$
– Dreamer123
Jan 8 at 4:12










1 Answer
1






active

oldest

votes


















2












$begingroup$

I think you are missing a term in the second to last line. That being said, yes the operator is self-adjoint.



This is how I would have done it. Using integration by parts, we get that:



begin{align}
langle Au,v rangle &= int_0^T u''(t)v(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - int_0^T u'(t)v'(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - u(t)v'(t) bigg|_{0}^{T} +int_0^T u(t)v''(t)dt
end{align}



Apply the BC's $u(0)=u(T)=0$ and so the first two terms disappear. Note that it also works with another function $v(t)$ living inside $D(A)$. Thus,



$langle Au,v rangle = langle u,Av rangle$ indeed.



The fun part is now recognizing that any arbitrary differential operator may or may not be self adjoint. For example, try an operator like:



$Lu = iu'''(t)$, with BCs of $u(0)=u'(0) = u''(1)=0$. Is it self-adjoint?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Use langle and rangle for $langle$ and $rangle$.
    $endgroup$
    – Mattos
    Jan 8 at 4:33










  • $begingroup$
    Thanks, sorry for that. I am (still) learning LaTeX commands..
    $endgroup$
    – LordVader007
    Jan 8 at 4:35






  • 1




    $begingroup$
    No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
    $endgroup$
    – Mattos
    Jan 8 at 4:37












  • $begingroup$
    In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
    $endgroup$
    – Dreamer123
    Jan 8 at 9:24






  • 1




    $begingroup$
    In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
    $endgroup$
    – MaoWao
    Jan 8 at 13:34











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2












$begingroup$

I think you are missing a term in the second to last line. That being said, yes the operator is self-adjoint.



This is how I would have done it. Using integration by parts, we get that:



begin{align}
langle Au,v rangle &= int_0^T u''(t)v(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - int_0^T u'(t)v'(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - u(t)v'(t) bigg|_{0}^{T} +int_0^T u(t)v''(t)dt
end{align}



Apply the BC's $u(0)=u(T)=0$ and so the first two terms disappear. Note that it also works with another function $v(t)$ living inside $D(A)$. Thus,



$langle Au,v rangle = langle u,Av rangle$ indeed.



The fun part is now recognizing that any arbitrary differential operator may or may not be self adjoint. For example, try an operator like:



$Lu = iu'''(t)$, with BCs of $u(0)=u'(0) = u''(1)=0$. Is it self-adjoint?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Use langle and rangle for $langle$ and $rangle$.
    $endgroup$
    – Mattos
    Jan 8 at 4:33










  • $begingroup$
    Thanks, sorry for that. I am (still) learning LaTeX commands..
    $endgroup$
    – LordVader007
    Jan 8 at 4:35






  • 1




    $begingroup$
    No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
    $endgroup$
    – Mattos
    Jan 8 at 4:37












  • $begingroup$
    In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
    $endgroup$
    – Dreamer123
    Jan 8 at 9:24






  • 1




    $begingroup$
    In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
    $endgroup$
    – MaoWao
    Jan 8 at 13:34
















2












$begingroup$

I think you are missing a term in the second to last line. That being said, yes the operator is self-adjoint.



This is how I would have done it. Using integration by parts, we get that:



begin{align}
langle Au,v rangle &= int_0^T u''(t)v(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - int_0^T u'(t)v'(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - u(t)v'(t) bigg|_{0}^{T} +int_0^T u(t)v''(t)dt
end{align}



Apply the BC's $u(0)=u(T)=0$ and so the first two terms disappear. Note that it also works with another function $v(t)$ living inside $D(A)$. Thus,



$langle Au,v rangle = langle u,Av rangle$ indeed.



The fun part is now recognizing that any arbitrary differential operator may or may not be self adjoint. For example, try an operator like:



$Lu = iu'''(t)$, with BCs of $u(0)=u'(0) = u''(1)=0$. Is it self-adjoint?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Use langle and rangle for $langle$ and $rangle$.
    $endgroup$
    – Mattos
    Jan 8 at 4:33










  • $begingroup$
    Thanks, sorry for that. I am (still) learning LaTeX commands..
    $endgroup$
    – LordVader007
    Jan 8 at 4:35






  • 1




    $begingroup$
    No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
    $endgroup$
    – Mattos
    Jan 8 at 4:37












  • $begingroup$
    In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
    $endgroup$
    – Dreamer123
    Jan 8 at 9:24






  • 1




    $begingroup$
    In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
    $endgroup$
    – MaoWao
    Jan 8 at 13:34














2












2








2





$begingroup$

I think you are missing a term in the second to last line. That being said, yes the operator is self-adjoint.



This is how I would have done it. Using integration by parts, we get that:



begin{align}
langle Au,v rangle &= int_0^T u''(t)v(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - int_0^T u'(t)v'(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - u(t)v'(t) bigg|_{0}^{T} +int_0^T u(t)v''(t)dt
end{align}



Apply the BC's $u(0)=u(T)=0$ and so the first two terms disappear. Note that it also works with another function $v(t)$ living inside $D(A)$. Thus,



$langle Au,v rangle = langle u,Av rangle$ indeed.



The fun part is now recognizing that any arbitrary differential operator may or may not be self adjoint. For example, try an operator like:



$Lu = iu'''(t)$, with BCs of $u(0)=u'(0) = u''(1)=0$. Is it self-adjoint?






share|cite|improve this answer











$endgroup$



I think you are missing a term in the second to last line. That being said, yes the operator is self-adjoint.



This is how I would have done it. Using integration by parts, we get that:



begin{align}
langle Au,v rangle &= int_0^T u''(t)v(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - int_0^T u'(t)v'(t)dt \
&= u'(t)v(t) bigg|_{0}^{T} - u(t)v'(t) bigg|_{0}^{T} +int_0^T u(t)v''(t)dt
end{align}



Apply the BC's $u(0)=u(T)=0$ and so the first two terms disappear. Note that it also works with another function $v(t)$ living inside $D(A)$. Thus,



$langle Au,v rangle = langle u,Av rangle$ indeed.



The fun part is now recognizing that any arbitrary differential operator may or may not be self adjoint. For example, try an operator like:



$Lu = iu'''(t)$, with BCs of $u(0)=u'(0) = u''(1)=0$. Is it self-adjoint?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 4:38









Mattos

2,73721321




2,73721321










answered Jan 8 at 4:26









LordVader007LordVader007

402313




402313








  • 1




    $begingroup$
    Use langle and rangle for $langle$ and $rangle$.
    $endgroup$
    – Mattos
    Jan 8 at 4:33










  • $begingroup$
    Thanks, sorry for that. I am (still) learning LaTeX commands..
    $endgroup$
    – LordVader007
    Jan 8 at 4:35






  • 1




    $begingroup$
    No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
    $endgroup$
    – Mattos
    Jan 8 at 4:37












  • $begingroup$
    In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
    $endgroup$
    – Dreamer123
    Jan 8 at 9:24






  • 1




    $begingroup$
    In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
    $endgroup$
    – MaoWao
    Jan 8 at 13:34














  • 1




    $begingroup$
    Use langle and rangle for $langle$ and $rangle$.
    $endgroup$
    – Mattos
    Jan 8 at 4:33










  • $begingroup$
    Thanks, sorry for that. I am (still) learning LaTeX commands..
    $endgroup$
    – LordVader007
    Jan 8 at 4:35






  • 1




    $begingroup$
    No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
    $endgroup$
    – Mattos
    Jan 8 at 4:37












  • $begingroup$
    In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
    $endgroup$
    – Dreamer123
    Jan 8 at 9:24






  • 1




    $begingroup$
    In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
    $endgroup$
    – MaoWao
    Jan 8 at 13:34








1




1




$begingroup$
Use langle and rangle for $langle$ and $rangle$.
$endgroup$
– Mattos
Jan 8 at 4:33




$begingroup$
Use langle and rangle for $langle$ and $rangle$.
$endgroup$
– Mattos
Jan 8 at 4:33












$begingroup$
Thanks, sorry for that. I am (still) learning LaTeX commands..
$endgroup$
– LordVader007
Jan 8 at 4:35




$begingroup$
Thanks, sorry for that. I am (still) learning LaTeX commands..
$endgroup$
– LordVader007
Jan 8 at 4:35




1




1




$begingroup$
No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
$endgroup$
– Mattos
Jan 8 at 4:37






$begingroup$
No worries, it looks much better now. Also, you can use begin {align}, end {align} and &= to format equal signs underneath each other (like I just edited for you). It makes it easier to read.
$endgroup$
– Mattos
Jan 8 at 4:37














$begingroup$
In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
$endgroup$
– Dreamer123
Jan 8 at 9:24




$begingroup$
In the 5th line, we have $u'(T)v(T) - u'(0)v(0) - u(T)v'(T) +u(0)v'(0)$, $u(0)=u(T)=0$ leaves us with two terms $u'(T)v(T)-u'(0)v(0)$ so for A to be self-adjoint we need $v(t)$ to be living in $D(A)$ so that $v(T)=v(0)=0$?
$endgroup$
– Dreamer123
Jan 8 at 9:24




1




1




$begingroup$
In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
$endgroup$
– MaoWao
Jan 8 at 13:34




$begingroup$
In addition two Dreamer123's comment, this argument only shows the symmetry of $A$. Proving $D(A)=D(A^ast)$ is (as usual) the more delicate part.
$endgroup$
– MaoWao
Jan 8 at 13:34


















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Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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Has there ever been an instance of an active nuclear power plant within or near a war zone?

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36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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