Sturm-Liouville eigenvalues have lower bound
$begingroup$
I am looking to find a proof that eigenvalues are bounded below for the the general Sturm Liouville equation on an open region $ Omega subset mathfrak R^d $ (perhaps with compact closure),
$Ly -lambda rho(mathbf x) y = -nabla cdot big( p(mathbf x) nabla y big) + q(mathbf x) y - lambda rho(mathbf x) y = 0, quad mathbf x in Omega $
where $y $ is twice differentiable, $ p, rho > 0 $ and boundary conditions on $ partialOmega $,
$alpha(mathbf s) y + beta(mathbf s) frac{partial y}{partial n} = 0 $
($alpha, beta$ not both zero).
Can anyone help me find a clear treatment handling of this question?
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors calculus-of-variations sturm-liouville
asked Dec 17 '18 at 10:00
WA DonWA Don
261
$endgroup$
add a comment |
$begingroup$
I am looking to find a proof that eigenvalues are bounded below for the the general Sturm Liouville equation on an open region $ Omega subset mathfrak R^d $ (perhaps with compact closure),
$Ly -lambda rho(mathbf x) y = -nabla cdot big( p(mathbf x) nabla y big) + q(mathbf x) y - lambda rho(mathbf x) y = 0, quad mathbf x in Omega $
where $y $ is twice differentiable, $ p, rho > 0 $ and boundary conditions on $ partialOmega $,
$alpha(mathbf s) y + beta(mathbf s) frac{partial y}{partial n} = 0 $
($alpha, beta$ not both zero).
Can anyone help me find a clear treatment handling of this question?
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors calculus-of-variations sturm-liouville
asked Dec 17 '18 at 10:00
WA DonWA Don
261
$endgroup$
$begingroup$
For any non-trivial solution $(λ,y)$ you get $$ newcommand{vol}{{rm vol}} newcommand{area}{{rm area}} λ=frac{ int_Ωp(x)|∇y(x)|^2,dvol(x) +int_Ωq(x)|y(x)|^2,dvol(x) - int_{∂Ω}y(x)p(x)langle ∇y(x), n(x)rangle,darea(x) }{ int_Ωρ(x)|y(x)|^2,dvol(x) } $$ so essentially you "only" have to treat the last term in the numerator.
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– LutzL
Dec 17 '18 at 11:45
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Thanks so much, and the last term simplified to zero if either of $ alpha $ or $ beta $ is zero and if neither is, it becomes $ int_{partial Omega} p frac{beta}{alpha} left| frac{partial y}{partial n}right|^2 ds = int_{partialOmega} p frac{alpha}{beta} left| y right|^2 ds $ which is where I get stuck. I am wondering if it is not true without restrictions on $alpha, beta$ - for instance both must have the same sign when non-zero.
$endgroup$
– WA Don
Dec 18 '18 at 12:27
add a comment |
$begingroup$
I am looking to find a proof that eigenvalues are bounded below for the the general Sturm Liouville equation on an open region $ Omega subset mathfrak R^d $ (perhaps with compact closure),
$Ly -lambda rho(mathbf x) y = -nabla cdot big( p(mathbf x) nabla y big) + q(mathbf x) y - lambda rho(mathbf x) y = 0, quad mathbf x in Omega $
where $y $ is twice differentiable, $ p, rho > 0 $ and boundary conditions on $ partialOmega $,
$alpha(mathbf s) y + beta(mathbf s) frac{partial y}{partial n} = 0 $
($alpha, beta$ not both zero).
Can anyone help me find a clear treatment handling of this question?
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors calculus-of-variations sturm-liouville
asked Dec 17 '18 at 10:00
WA DonWA Don
261
$endgroup$
I am looking to find a proof that eigenvalues are bounded below for the the general Sturm Liouville equation on an open region $ Omega subset mathfrak R^d $ (perhaps with compact closure),
$Ly -lambda rho(mathbf x) y = -nabla cdot big( p(mathbf x) nabla y big) + q(mathbf x) y - lambda rho(mathbf x) y = 0, quad mathbf x in Omega $
where $y $ is twice differentiable, $ p, rho > 0 $ and boundary conditions on $ partialOmega $,
$alpha(mathbf s) y + beta(mathbf s) frac{partial y}{partial n} = 0 $
($alpha, beta$ not both zero).
Can anyone help me find a clear treatment handling of this question?
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors calculus-of-variations sturm-liouville
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors calculus-of-variations sturm-liouville
asked Dec 17 '18 at 10:00
WA DonWA Don
261
asked Dec 17 '18 at 10:00
WA DonWA Don
261
asked Dec 17 '18 at 10:00
WA DonWA Don
261
asked Dec 17 '18 at 10:00
WA DonWA Don
261
asked Dec 17 '18 at 10:00
WA DonWA Don
261
261
$begingroup$
For any non-trivial solution $(λ,y)$ you get $$ newcommand{vol}{{rm vol}} newcommand{area}{{rm area}} λ=frac{ int_Ωp(x)|∇y(x)|^2,dvol(x) +int_Ωq(x)|y(x)|^2,dvol(x) - int_{∂Ω}y(x)p(x)langle ∇y(x), n(x)rangle,darea(x) }{ int_Ωρ(x)|y(x)|^2,dvol(x) } $$ so essentially you "only" have to treat the last term in the numerator.
$endgroup$
– LutzL
Dec 17 '18 at 11:45
$begingroup$
Thanks so much, and the last term simplified to zero if either of $ alpha $ or $ beta $ is zero and if neither is, it becomes $ int_{partial Omega} p frac{beta}{alpha} left| frac{partial y}{partial n}right|^2 ds = int_{partialOmega} p frac{alpha}{beta} left| y right|^2 ds $ which is where I get stuck. I am wondering if it is not true without restrictions on $alpha, beta$ - for instance both must have the same sign when non-zero.
$endgroup$
– WA Don
Dec 18 '18 at 12:27
add a comment |
$begingroup$
For any non-trivial solution $(λ,y)$ you get $$ newcommand{vol}{{rm vol}} newcommand{area}{{rm area}} λ=frac{ int_Ωp(x)|∇y(x)|^2,dvol(x) +int_Ωq(x)|y(x)|^2,dvol(x) - int_{∂Ω}y(x)p(x)langle ∇y(x), n(x)rangle,darea(x) }{ int_Ωρ(x)|y(x)|^2,dvol(x) } $$ so essentially you "only" have to treat the last term in the numerator.
$endgroup$
– LutzL
Dec 17 '18 at 11:45
$begingroup$
Thanks so much, and the last term simplified to zero if either of $ alpha $ or $ beta $ is zero and if neither is, it becomes $ int_{partial Omega} p frac{beta}{alpha} left| frac{partial y}{partial n}right|^2 ds = int_{partialOmega} p frac{alpha}{beta} left| y right|^2 ds $ which is where I get stuck. I am wondering if it is not true without restrictions on $alpha, beta$ - for instance both must have the same sign when non-zero.
$endgroup$
– WA Don
Dec 18 '18 at 12:27
$begingroup$
For any non-trivial solution $(λ,y)$ you get $$ newcommand{vol}{{rm vol}} newcommand{area}{{rm area}} λ=frac{ int_Ωp(x)|∇y(x)|^2,dvol(x) +int_Ωq(x)|y(x)|^2,dvol(x) - int_{∂Ω}y(x)p(x)langle ∇y(x), n(x)rangle,darea(x) }{ int_Ωρ(x)|y(x)|^2,dvol(x) } $$ so essentially you "only" have to treat the last term in the numerator.
$endgroup$
– LutzL
Dec 17 '18 at 11:45
$begingroup$
For any non-trivial solution $(λ,y)$ you get $$ newcommand{vol}{{rm vol}} newcommand{area}{{rm area}} λ=frac{ int_Ωp(x)|∇y(x)|^2,dvol(x) +int_Ωq(x)|y(x)|^2,dvol(x) - int_{∂Ω}y(x)p(x)langle ∇y(x), n(x)rangle,darea(x) }{ int_Ωρ(x)|y(x)|^2,dvol(x) } $$ so essentially you "only" have to treat the last term in the numerator.
$endgroup$
– LutzL
Dec 17 '18 at 11:45
$begingroup$
Thanks so much, and the last term simplified to zero if either of $ alpha $ or $ beta $ is zero and if neither is, it becomes $ int_{partial Omega} p frac{beta}{alpha} left| frac{partial y}{partial n}right|^2 ds = int_{partialOmega} p frac{alpha}{beta} left| y right|^2 ds $ which is where I get stuck. I am wondering if it is not true without restrictions on $alpha, beta$ - for instance both must have the same sign when non-zero.
$endgroup$
– WA Don
Dec 18 '18 at 12:27
$begingroup$
Thanks so much, and the last term simplified to zero if either of $ alpha $ or $ beta $ is zero and if neither is, it becomes $ int_{partial Omega} p frac{beta}{alpha} left| frac{partial y}{partial n}right|^2 ds = int_{partialOmega} p frac{alpha}{beta} left| y right|^2 ds $ which is where I get stuck. I am wondering if it is not true without restrictions on $alpha, beta$ - for instance both must have the same sign when non-zero.
$endgroup$
– WA Don
Dec 18 '18 at 12:27
add a comment |
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$begingroup$
For any non-trivial solution $(λ,y)$ you get $$ newcommand{vol}{{rm vol}} newcommand{area}{{rm area}} λ=frac{ int_Ωp(x)|∇y(x)|^2,dvol(x) +int_Ωq(x)|y(x)|^2,dvol(x) - int_{∂Ω}y(x)p(x)langle ∇y(x), n(x)rangle,darea(x) }{ int_Ωρ(x)|y(x)|^2,dvol(x) } $$ so essentially you "only" have to treat the last term in the numerator.
$endgroup$
– LutzL
Dec 17 '18 at 11:45
$begingroup$
Thanks so much, and the last term simplified to zero if either of $ alpha $ or $ beta $ is zero and if neither is, it becomes $ int_{partial Omega} p frac{beta}{alpha} left| frac{partial y}{partial n}right|^2 ds = int_{partialOmega} p frac{alpha}{beta} left| y right|^2 ds $ which is where I get stuck. I am wondering if it is not true without restrictions on $alpha, beta$ - for instance both must have the same sign when non-zero.
$endgroup$
– WA Don
Dec 18 '18 at 12:27