Hello!


I got stuck at the following exercise.


Knowing that:
"The eigenvalue problem Ly=(py')'+qy, a <= x <= b is a Sturm-Liouville problem when it satisfies the boundary conditions:
p(a)W(u(a),v*(a))=p(b)W(u(b),v*(b))"


W is the Wronskian
u,v are solutions of the eigenvalue problem
v* is the complex conjugate of v




I have to show that the eigenvalue problem y''+λy=0, with boundary conditions y(0)=0, y'(0)=y'(1) is not a Sturm -Liouville problem.


This is what I've done so far:


Let u, v* solutions of the eigenvalue problem y''+λy=0, then:
u(0)=0, u'(0)=u'(1) and v*(0)=0, v* '(0)=v* '(1).


W(u(0),v*(0))=u(0)v* '(0)-u'(0)v*(0)=0


W(u(1),v*(1))=u(1)v* '(1)-u'(1)v*(1)=u(1) v* '(0)-u'(0)v*(1)

How can I continue? How can I show that this is not equal to 0, so that the two Wronskian are not equal?

To show that the above boundary value problem is not a Sturm - Liouville problem, observe that the general solution to y'' + hy = 0 is:

y = A cos x sqrt(h) + B sin x sqrt(h). As y(0) = 0, for h /= 0, this forces A = 0. Thus y = B sin x sqrt(h) can be the only nontrivial solutions.

As y'(0) = y'(1), B sqrt(h) = B sqrt(h) cos sqrt(h). <==> 1 = cos sqrt(h) as B sqrt(h) /= 0 because y = B sin x sqrt(h) is a nontrivial solution.

Thus, sqrt(h) = 2Pi*k, where k is a positive integer. ==> h = 4(Pi)^2 * k^2 are eigenvalues.

Now consider h = 0. Then the general solution is y = Ax + B. As y(0) = 0, B=0. Thus, y = Ax ==> y'(0) = y'(1) = A. Therefore, 0 is also an
eigenvalue.

An essential property of Sturm - Liouville problems is that eigenfunctions corresponding to distinct eigenvalues are orthogonal. In this particular case, if
f(x) and g(x) correspond to distinct eigenvalues, then

1 /
| f(x)g(x) dx = 0.
0/


Consider f(x) = x, which corresponds to h = 0, and g(x) = sin 2Pi*x, which corresponds to h = 4Pi^2.

1 / |1
Then, | x sin (2Pi*x) dx = -(1/(2Pi)) x cos (2Pi*x) + (1/(4Pi^2)) sin(2Pi*x) | = -1/(2Pi) /= 0.
0/ |0

Hence, we have two eigenfunctions corresponding to two distinct eigenvalues that are not orthogonal. Thus we must conclude that your boundary value
problem is not a Sturm - Liouville problem.
Q.E.D

-CWH :)

To show that the above boundary value problem is not a Sturm - Liouville problem, observe that the general solution to y'' + hy = 0 is:

y = A cos x sqrt(h) + B sin x sqrt(h). As y(0) = 0, for h /= 0, this forces A = 0. Thus y = B sin x sqrt(h) can be the only nontrivial solutions.

As y'(0) = y'(1), B sqrt(h) = B sqrt(h) cos sqrt(h). <==> 1 = cos sqrt(h) as B sqrt(h) /= 0 because y = B sin x sqrt(h) is a nontrivial solution.

Thus, sqrt(h) = 2Pi*k, where k is a positive integer. ==> h = 4(Pi)^2 * k^2 are eigenvalues.

Now consider h = 0. Then the general solution is y = Ax + B. As y(0) = 0, B=0. Thus, y = Ax ==> y'(0) = y'(1) = A. Therefore, 0 is also an
eigenvalue.

An essential property of Sturm - Liouville problems is that eigenfunctions corresponding to distinct eigenvalues are orthogonal. In this particular case, if
f(x) and g(x) correspond to distinct eigenvalues, then

1 /
| f(x)g(x) dx = 0.
0/


Consider f(x) = x, which corresponds to h = 0, and g(x) = sin 2Pi*x, which corresponds to h = 4Pi^2.

1 / |1
Then, | x sin (2Pi*x) dx = -(1/(2Pi)) x cos (2Pi*x) + (1/(4Pi^2)) sin(2Pi*x) | = -1/(2Pi) /= 0.
0/ |0

Hence, we have two eigenfunctions corresponding to two distinct eigenvalues that are not orthogonal. Thus we must conclude that your boundary value
problem is not a Sturm - Liouville problem.
Q.E.D

-CWH :)

But at the second subquestion of this exercise I have to show that the eigenfunctions are orthogonal. How can that be that we have two eigenfunctions corresponding to two distinct eigenvalues that are not orthogonal??

You must have a typographical error in the problem, because the boundary conditions that were given allow y(x)=x to be an eigenfunction corresponding to lambda = 0, and y(x) = sin (2Pi x) to be an eigenfunction corresponding to lambda = 4Pi^2, but y(x) = x and y(x) = sin (2Pi x) are not orthogonal on [0,1].

-CWH

You must have a typographical error in the problem, because the boundary conditions that were given allow y(x)=x to be an eigenfunction corresponding to lambda = 0, and y(x) = sin (2Pi x) to be an eigenfunction corresponding to lambda = 4Pi^2, but y(x) = x and y(x) = sin (2Pi x) are not orthogonal on [0,1].

-CWH

I understand!! Thank you for your answer!! :-)

If we have the eigenvalue problem:
y''+λy=0
y(0)=0
y'(0)=y'(1)/2

the eigenvalues are:
*the ones that are given by the relation cos(square{λ})=2 and
*λ=0 and the corresponding eigenfunction is y(x)=0.

What can I do in this case to show that this eigenvalue problem is not a Sturm-Liouville problem?
When I take the dot product of an eigenfunction, that corresponds to an eigenvalue given by cos(square{λ})=2, and y(x)=0, it would be equal to 0.

I understand!! Thank you for your answer!! :-)

If we have the eigenvalue problem:
y''+λy=0
y(0)=0
y'(0)=y'(1)/2

the eigenvalues are:
*the ones that are given by the relation cos(square{λ})=2 and
*λ=0 and the corresponding eigenfunction is y(x)=0.

What can I do in this case to show that this eigenvalue problem is not a Sturm-Liouville problem?
When I take the dot product of an eigenfunction, that corresponds to an eigenvalue given by cos(square{λ})=2, and y(x)=0, it would be equal to 0.



In this case, lambda = 0 is not an eigenvalue, as the only solution that corresponds to it is the trivial solution. The only real eigenvalue in this case is lambda = -(ln(2+sqrt(3)))^2. A corresponding eigenfunction is
u(x) = sinh(ln(2+sqrt(3)))x.

If omega = 2k(Pi) + I ln(2+sqrt(3)), k an integer, then lambda = omega^2
= (4k^2 (Pi)^2 - (ln(2+sqrt(3)))^2 + I 4k(Pi) ln(2+sqrt(3))) is an eigenvalue with the corresponding eigenfunction y(x) = sin(omega x).

Let omega0 = I ln(2+sqrt(3)) so y0(x) = -i sin(omega0 x) = sinh(ln(2+sqrt(3)))x.
Let omega1 = 2(Pi) + I ln(2+sqrt(3)) and let v(x) = sin(omega1 x).

Then, v*(x) is an eigenfunction corresponding to eigenvalue lambda = omega1*^2.

Clearly, W[u,v*](0) = 0 and W[u,v*](1) = 2sqrt(3) (omega1* + omega0)
= 4(Pi)sqrt(3).

This tells me that the problem is not a Sturm - Liouville problem. This unfortunately also tells me that u and v, which are eigenfunctions corresponding to distinct eigenvalues, are not orthogonal. Hint: Differentiate the Wronskian with respect to x. The boundary condition y(0) = 0 is your problem, as this forces the eigenvalue problem to either be a Sturm - Liouville problem or to have non-orthogonal eigenfunctions corresponding to distinct eigenvalues.

If you are looking for an eigenvalue problem in which eigenfunctions corresponding to distinct eigenvalues are orthogonal that is not a Sturm-Liouville problem, try y'' + hy = 0 with the boundary conditions

y(0) = 2y(1)
y'(0) = (0.5)y'(1).

This will give you what you want, but proving this is a bit tricky, so if you need more help feel free to let me know.

-CWH :)

Ok, I understand!! Thanks a lot!!

I looked again at the first eigenvalue and I got stuck..
The exercise is:
"Show that the eigenvalue problem y''+λy=0, with boundary conditions y(0)=0, y'(0)=y'(1) is not a Sturm -Liouville problem.
Show that for this problem the eigenvalues are real, and the eigenfunctions are orthogonal to each other, but the eigenfunctions don't form a complete set."

To show that the problem is not a Sturm-Liouville problem, we have shown that two eigenfunctions corresponding to two distinct eigenvalues are not orthogonal. But at the second question I have to show that the eigenfunctions are orthogonal. Do I have to show that only the eigenfunctions corresponding to λ>0 are orthogonal?