Exercise about the Sturm-Liouville problems
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Under what condition on the constants $c$ and $c'$ are the boundary conditions $$f(b)=cf(a)$$ and $$f'(b)=c'f'(a)$$ self-adjoint for the operator the operator $$mathcal{L}f=frac{d}{dx}left(p_0(x)frac{df}{dx}right)+p_2(x)f$$ on $[a,b]$? ($p_0(x),p_2(x)$ are reals)
I don't know how to impose the boundary conditions to the problem, actually I don't know what to do to find that constants.
Using the Lagrange identity I found $$leftlangle f|mathcal{L}|frightrangle=langle f|mathcal{L}^{dagger}|frangle+left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}$$
and this condition requires that
$$left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}=0$$
so
$$p_0(b)left(frac{df}{dx}(b)f^{*}(b)-f(b)frac{df^{*}}{dx}(b)right)=p_0(a)left(frac{df}{dx}(a)f^{*}(a)-f(a)frac{df^{*}}{dx}(a)right),.$$
The answer is $${cc'}^{*}=frac{p_0(a)}{p_0(b)}$$
where did I go wrong?
differential-equations hilbert-spaces mathematical-physics differential-operators sturm-liouville
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
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add a comment |
$begingroup$
Under what condition on the constants $c$ and $c'$ are the boundary conditions $$f(b)=cf(a)$$ and $$f'(b)=c'f'(a)$$ self-adjoint for the operator the operator $$mathcal{L}f=frac{d}{dx}left(p_0(x)frac{df}{dx}right)+p_2(x)f$$ on $[a,b]$? ($p_0(x),p_2(x)$ are reals)
I don't know how to impose the boundary conditions to the problem, actually I don't know what to do to find that constants.
Using the Lagrange identity I found $$leftlangle f|mathcal{L}|frightrangle=langle f|mathcal{L}^{dagger}|frangle+left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}$$
and this condition requires that
$$left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}=0$$
so
$$p_0(b)left(frac{df}{dx}(b)f^{*}(b)-f(b)frac{df^{*}}{dx}(b)right)=p_0(a)left(frac{df}{dx}(a)f^{*}(a)-f(a)frac{df^{*}}{dx}(a)right),.$$
The answer is $${cc'}^{*}=frac{p_0(a)}{p_0(b)}$$
where did I go wrong?
differential-equations hilbert-spaces mathematical-physics differential-operators sturm-liouville
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
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$begingroup$
Is that the answer provided, or the one you came up with?
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– AlexanderJ93
Dec 6 '18 at 0:33
$begingroup$
This is the provided answer
$endgroup$
– Marcos Pereira
Dec 6 '18 at 1:41
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Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case.
$endgroup$
– DisintegratingByParts
Dec 6 '18 at 17:42
add a comment |
$begingroup$
Under what condition on the constants $c$ and $c'$ are the boundary conditions $$f(b)=cf(a)$$ and $$f'(b)=c'f'(a)$$ self-adjoint for the operator the operator $$mathcal{L}f=frac{d}{dx}left(p_0(x)frac{df}{dx}right)+p_2(x)f$$ on $[a,b]$? ($p_0(x),p_2(x)$ are reals)
I don't know how to impose the boundary conditions to the problem, actually I don't know what to do to find that constants.
Using the Lagrange identity I found $$leftlangle f|mathcal{L}|frightrangle=langle f|mathcal{L}^{dagger}|frangle+left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}$$
and this condition requires that
$$left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}=0$$
so
$$p_0(b)left(frac{df}{dx}(b)f^{*}(b)-f(b)frac{df^{*}}{dx}(b)right)=p_0(a)left(frac{df}{dx}(a)f^{*}(a)-f(a)frac{df^{*}}{dx}(a)right),.$$
The answer is $${cc'}^{*}=frac{p_0(a)}{p_0(b)}$$
where did I go wrong?
differential-equations hilbert-spaces mathematical-physics differential-operators sturm-liouville
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
$endgroup$
Under what condition on the constants $c$ and $c'$ are the boundary conditions $$f(b)=cf(a)$$ and $$f'(b)=c'f'(a)$$ self-adjoint for the operator the operator $$mathcal{L}f=frac{d}{dx}left(p_0(x)frac{df}{dx}right)+p_2(x)f$$ on $[a,b]$? ($p_0(x),p_2(x)$ are reals)
I don't know how to impose the boundary conditions to the problem, actually I don't know what to do to find that constants.
Using the Lagrange identity I found $$leftlangle f|mathcal{L}|frightrangle=langle f|mathcal{L}^{dagger}|frangle+left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}$$
and this condition requires that
$$left[p_0(x)left(frac{df}{dx}f^{*}(x)-f(x)frac{df^{*}}{dx}right)right]_{a}^{b}=0$$
so
$$p_0(b)left(frac{df}{dx}(b)f^{*}(b)-f(b)frac{df^{*}}{dx}(b)right)=p_0(a)left(frac{df}{dx}(a)f^{*}(a)-f(a)frac{df^{*}}{dx}(a)right),.$$
The answer is $${cc'}^{*}=frac{p_0(a)}{p_0(b)}$$
where did I go wrong?
differential-equations hilbert-spaces mathematical-physics differential-operators sturm-liouville
differential-equations hilbert-spaces mathematical-physics differential-operators sturm-liouville
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
asked Dec 5 '18 at 22:05
Marcos PereiraMarcos Pereira
162
162
$begingroup$
Is that the answer provided, or the one you came up with?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:33
$begingroup$
This is the provided answer
$endgroup$
– Marcos Pereira
Dec 6 '18 at 1:41
$begingroup$
Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case.
$endgroup$
– DisintegratingByParts
Dec 6 '18 at 17:42
add a comment |
$begingroup$
Is that the answer provided, or the one you came up with?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:33
$begingroup$
This is the provided answer
$endgroup$
– Marcos Pereira
Dec 6 '18 at 1:41
$begingroup$
Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case.
$endgroup$
– DisintegratingByParts
Dec 6 '18 at 17:42
$begingroup$
Is that the answer provided, or the one you came up with?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:33
$begingroup$
Is that the answer provided, or the one you came up with?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:33
$begingroup$
This is the provided answer
$endgroup$
– Marcos Pereira
Dec 6 '18 at 1:41
$begingroup$
This is the provided answer
$endgroup$
– Marcos Pereira
Dec 6 '18 at 1:41
$begingroup$
Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case.
$endgroup$
– DisintegratingByParts
Dec 6 '18 at 17:42
$begingroup$
Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case.
$endgroup$
– DisintegratingByParts
Dec 6 '18 at 17:42
add a comment |
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$begingroup$
Is that the answer provided, or the one you came up with?
$endgroup$
– AlexanderJ93
Dec 6 '18 at 0:33
$begingroup$
This is the provided answer
$endgroup$
– Marcos Pereira
Dec 6 '18 at 1:41
$begingroup$
Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case.
$endgroup$
– DisintegratingByParts
Dec 6 '18 at 17:42