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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1 Answer
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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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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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Display a custom attribute below product name in the front-end Magento 1.9.3.8

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0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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