Where did $a$ and $b$ come from in this eigenvalue problem?
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I have this eigenvalue problem:
$$-x^2u_xx + 2xu_x - 2u = lambda x^2 u$$
for $0 < x < 1$, with boundary conditions $u_x(0) = 0, u(1) = u_x(1)$.
I'm trying to find a function $M(x)$, such that under the change of variable $u(x) = M(x)w(x)$, it can be transformed to
$$w_xx = - lambda w$$
for $0 < x < 1$.
The solution says this:
With $u = Mw$, the ODE $-x^2 u'' + 2xu' - 2u = lambda x^2u$ becomes
$$-x^2 (M'' -aM' - bM) + 2x(M' w + Mw') - 2Mw = lambda x^2Mw$$
$$-x^2 Mw'' + (2xM - 2x^2 M')w' + (-x^2 M'' + 2xM' - 2M)w = lambda x^2Mw$$
I was wondering if someone could please explain where the $a$ and $b$ came from? Thanks.
differential-equations pde eigenvalues-eigenvectors boundary-value-problem
asked Aug 16 at 22:42
Wyuw
1368
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up vote
2
down vote
favorite
I have this eigenvalue problem:
$$-x^2u_xx + 2xu_x - 2u = lambda x^2 u$$
for $0 < x < 1$, with boundary conditions $u_x(0) = 0, u(1) = u_x(1)$.
I'm trying to find a function $M(x)$, such that under the change of variable $u(x) = M(x)w(x)$, it can be transformed to
$$w_xx = - lambda w$$
for $0 < x < 1$.
The solution says this:
With $u = Mw$, the ODE $-x^2 u'' + 2xu' - 2u = lambda x^2u$ becomes
$$-x^2 (M'' -aM' - bM) + 2x(M' w + Mw') - 2Mw = lambda x^2Mw$$
$$-x^2 Mw'' + (2xM - 2x^2 M')w' + (-x^2 M'' + 2xM' - 2M)w = lambda x^2Mw$$
I was wondering if someone could please explain where the $a$ and $b$ came from? Thanks.
differential-equations pde eigenvalues-eigenvectors boundary-value-problem
asked Aug 16 at 22:42
Wyuw
1368
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have this eigenvalue problem:
$$-x^2u_xx + 2xu_x - 2u = lambda x^2 u$$
for $0 < x < 1$, with boundary conditions $u_x(0) = 0, u(1) = u_x(1)$.
I'm trying to find a function $M(x)$, such that under the change of variable $u(x) = M(x)w(x)$, it can be transformed to
$$w_xx = - lambda w$$
for $0 < x < 1$.
The solution says this:
With $u = Mw$, the ODE $-x^2 u'' + 2xu' - 2u = lambda x^2u$ becomes
$$-x^2 (M'' -aM' - bM) + 2x(M' w + Mw') - 2Mw = lambda x^2Mw$$
$$-x^2 Mw'' + (2xM - 2x^2 M')w' + (-x^2 M'' + 2xM' - 2M)w = lambda x^2Mw$$
I was wondering if someone could please explain where the $a$ and $b$ came from? Thanks.
differential-equations pde eigenvalues-eigenvectors boundary-value-problem
asked Aug 16 at 22:42
Wyuw
1368
I have this eigenvalue problem:
$$-x^2u_xx + 2xu_x - 2u = lambda x^2 u$$
for $0 < x < 1$, with boundary conditions $u_x(0) = 0, u(1) = u_x(1)$.
I'm trying to find a function $M(x)$, such that under the change of variable $u(x) = M(x)w(x)$, it can be transformed to
$$w_xx = - lambda w$$
for $0 < x < 1$.
The solution says this:
With $u = Mw$, the ODE $-x^2 u'' + 2xu' - 2u = lambda x^2u$ becomes
$$-x^2 (M'' -aM' - bM) + 2x(M' w + Mw') - 2Mw = lambda x^2Mw$$
$$-x^2 Mw'' + (2xM - 2x^2 M')w' + (-x^2 M'' + 2xM' - 2M)w = lambda x^2Mw$$
I was wondering if someone could please explain where the $a$ and $b$ came from? Thanks.
differential-equations pde eigenvalues-eigenvectors boundary-value-problem
asked Aug 16 at 22:42
Wyuw
1368
asked Aug 16 at 22:42
Wyuw
1368
asked Aug 16 at 22:42
Wyuw
1368
asked Aug 16 at 22:42
Wyuw
1368
1368
add a comment |Â
add a comment |Â
1 Answer
1
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Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.
answered Aug 16 at 23:28
Rócherz
2,1962518
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
2
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.
answered Aug 16 at 23:28
Rócherz
2,1962518
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
2
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
add a comment |Â
up vote
0
down vote
accepted
Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.
answered Aug 16 at 23:28
Rócherz
2,1962518
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
2
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.
answered Aug 16 at 23:28
Rócherz
2,1962518
Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.
answered Aug 16 at 23:28
Rócherz
2,1962518
answered Aug 16 at 23:28
Rócherz
2,1962518
answered Aug 16 at 23:28
Rócherz
2,1962518
answered Aug 16 at 23:28
Rócherz
2,1962518
2,1962518
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
2
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
add a comment |Â
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
2
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
Hmm, it seems unlikely to be a typo though? Putting in $a$ and $b$ seems deliberate.
– Wyuw
Aug 16 at 23:31
2
2
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Wyuw Why do you think it is deliberate? If it was, shouldn't $a$ and $b$ appear in the original substitution $u = Mw$ and also in the second line?
– Mattos
Aug 16 at 23:50
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
@Mattos Ahh you make a good point! Yes, I think you are correct.
– Wyuw
Aug 16 at 23:53
add a comment |Â
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