How to solve an ODE in Sturm-Liouville form
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I am attempting to solve the following ODE:
$frac{d}{dz} bigg[F'(z) bigg(frac{z-1}{z}bigg)^2bigg] = frac{2(z-1)F(z)}{z^4}$
with the conditions that both $F(z)$ and $F'(z)$ approach 0 as $z$ approaches $pm infty$.
I have tried integrating both sides over different domains but nothing has worked so far.
differential-equations
asked Nov 14 at 2:38
1123581321
10918
add a comment |
up vote
0
down vote
favorite
I am attempting to solve the following ODE:
$frac{d}{dz} bigg[F'(z) bigg(frac{z-1}{z}bigg)^2bigg] = frac{2(z-1)F(z)}{z^4}$
with the conditions that both $F(z)$ and $F'(z)$ approach 0 as $z$ approaches $pm infty$.
I have tried integrating both sides over different domains but nothing has worked so far.
differential-equations
asked Nov 14 at 2:38
1123581321
10918
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am attempting to solve the following ODE:
$frac{d}{dz} bigg[F'(z) bigg(frac{z-1}{z}bigg)^2bigg] = frac{2(z-1)F(z)}{z^4}$
with the conditions that both $F(z)$ and $F'(z)$ approach 0 as $z$ approaches $pm infty$.
I have tried integrating both sides over different domains but nothing has worked so far.
differential-equations
asked Nov 14 at 2:38
1123581321
10918
I am attempting to solve the following ODE:
$frac{d}{dz} bigg[F'(z) bigg(frac{z-1}{z}bigg)^2bigg] = frac{2(z-1)F(z)}{z^4}$
with the conditions that both $F(z)$ and $F'(z)$ approach 0 as $z$ approaches $pm infty$.
I have tried integrating both sides over different domains but nothing has worked so far.
differential-equations
differential-equations
asked Nov 14 at 2:38
1123581321
10918
asked Nov 14 at 2:38
1123581321
10918
asked Nov 14 at 2:38
1123581321
10918
asked Nov 14 at 2:38
1123581321
10918
asked Nov 14 at 2:38
1123581321
10918
10918
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Your given equation has the structure $$(F'(z)u(z)^2))'=u(z)v(z)F(z)$$ with $u(z)=frac{z-1}z=1-frac1z$ and $v(z)=frac{2}{z^3}$. This expands under the product rule to (wherever $u(z)ne 0$, $zne 0$) $$F''(z)u(z)+2F'(z)u'(z)=v(z)F(z).$$ Now you will also find the terms on the left side in the second derivative of a product,
$$
(F(z)u(z))''=F''(z)u(z)+2F'(z)u'(z)+u''(z)F(z)=(u''(z)+v(z))F(z)
$$
and with $u''(z)=-frac{2}{z^3}$ one gets $u''(z)+v(z)=0$ so that
$$
F(z)frac{z-1}z=Az+B.
$$
But $$F(z)=frac{Az^2+Bz}{z-1}$$ has no non-trivial solutions with $F(pminfty)=0$, so that for the given conditions $F(z)=0$ is the only solution.
answered Nov 14 at 9:37
LutzL
53.4k41953
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
Yes of course. Thank you.
– LutzL
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your given equation has the structure $$(F'(z)u(z)^2))'=u(z)v(z)F(z)$$ with $u(z)=frac{z-1}z=1-frac1z$ and $v(z)=frac{2}{z^3}$. This expands under the product rule to (wherever $u(z)ne 0$, $zne 0$) $$F''(z)u(z)+2F'(z)u'(z)=v(z)F(z).$$ Now you will also find the terms on the left side in the second derivative of a product,
$$
(F(z)u(z))''=F''(z)u(z)+2F'(z)u'(z)+u''(z)F(z)=(u''(z)+v(z))F(z)
$$
and with $u''(z)=-frac{2}{z^3}$ one gets $u''(z)+v(z)=0$ so that
$$
F(z)frac{z-1}z=Az+B.
$$
But $$F(z)=frac{Az^2+Bz}{z-1}$$ has no non-trivial solutions with $F(pminfty)=0$, so that for the given conditions $F(z)=0$ is the only solution.
answered Nov 14 at 9:37
LutzL
53.4k41953
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
Yes of course. Thank you.
– LutzL
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
add a comment |
up vote
2
down vote
accepted
Your given equation has the structure $$(F'(z)u(z)^2))'=u(z)v(z)F(z)$$ with $u(z)=frac{z-1}z=1-frac1z$ and $v(z)=frac{2}{z^3}$. This expands under the product rule to (wherever $u(z)ne 0$, $zne 0$) $$F''(z)u(z)+2F'(z)u'(z)=v(z)F(z).$$ Now you will also find the terms on the left side in the second derivative of a product,
$$
(F(z)u(z))''=F''(z)u(z)+2F'(z)u'(z)+u''(z)F(z)=(u''(z)+v(z))F(z)
$$
and with $u''(z)=-frac{2}{z^3}$ one gets $u''(z)+v(z)=0$ so that
$$
F(z)frac{z-1}z=Az+B.
$$
But $$F(z)=frac{Az^2+Bz}{z-1}$$ has no non-trivial solutions with $F(pminfty)=0$, so that for the given conditions $F(z)=0$ is the only solution.
answered Nov 14 at 9:37
LutzL
53.4k41953
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
Yes of course. Thank you.
– LutzL
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your given equation has the structure $$(F'(z)u(z)^2))'=u(z)v(z)F(z)$$ with $u(z)=frac{z-1}z=1-frac1z$ and $v(z)=frac{2}{z^3}$. This expands under the product rule to (wherever $u(z)ne 0$, $zne 0$) $$F''(z)u(z)+2F'(z)u'(z)=v(z)F(z).$$ Now you will also find the terms on the left side in the second derivative of a product,
$$
(F(z)u(z))''=F''(z)u(z)+2F'(z)u'(z)+u''(z)F(z)=(u''(z)+v(z))F(z)
$$
and with $u''(z)=-frac{2}{z^3}$ one gets $u''(z)+v(z)=0$ so that
$$
F(z)frac{z-1}z=Az+B.
$$
But $$F(z)=frac{Az^2+Bz}{z-1}$$ has no non-trivial solutions with $F(pminfty)=0$, so that for the given conditions $F(z)=0$ is the only solution.
answered Nov 14 at 9:37
LutzL
53.4k41953
Your given equation has the structure $$(F'(z)u(z)^2))'=u(z)v(z)F(z)$$ with $u(z)=frac{z-1}z=1-frac1z$ and $v(z)=frac{2}{z^3}$. This expands under the product rule to (wherever $u(z)ne 0$, $zne 0$) $$F''(z)u(z)+2F'(z)u'(z)=v(z)F(z).$$ Now you will also find the terms on the left side in the second derivative of a product,
$$
(F(z)u(z))''=F''(z)u(z)+2F'(z)u'(z)+u''(z)F(z)=(u''(z)+v(z))F(z)
$$
and with $u''(z)=-frac{2}{z^3}$ one gets $u''(z)+v(z)=0$ so that
$$
F(z)frac{z-1}z=Az+B.
$$
But $$F(z)=frac{Az^2+Bz}{z-1}$$ has no non-trivial solutions with $F(pminfty)=0$, so that for the given conditions $F(z)=0$ is the only solution.
answered Nov 14 at 9:37
LutzL
53.4k41953
edited 2 days ago
answered Nov 14 at 9:37
LutzL
53.4k41953
answered Nov 14 at 9:37
LutzL
53.4k41953
answered Nov 14 at 9:37
LutzL
53.4k41953
53.4k41953
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
Yes of course. Thank you.
– LutzL
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
add a comment |
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
Yes of course. Thank you.
– LutzL
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
I think the second last line should be $u'' + v = 0$ not $u + v = 0$
– 1123581321
Nov 14 at 10:01
Yes of course. Thank you.
– LutzL
2 days ago
Yes of course. Thank you.
– LutzL
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
@Lutzt. I agree with your conclusion. +1
– JJacquelin
2 days ago
add a comment |
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