Find Eigenvalues for $$y''+\lambda y=0$$

[mathmari]

Hey!

$$y''+\lambda y =0$$
$$y(0)=0$$
$$y'(0)=\frac{y'(1)}{2}$$

I have to show that the eigenvalues are complex and are given by the relation $\cos{\sqrt{\lambda}}=2$ except from one that is real.

The characteristic equation is $m^2+\lambda =0 \Rightarrow m= \pm \sqrt{ \lambda}$$*$ $\lambda <0$:
$y(x)=c_1 e^{ \sqrt{ \lambda}x}+c_2 e^{- \sqrt{\lambda}x}$
$y(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_2=-c_1$
$y(x)=c_1(e^{ \sqrt{ \lambda}x}- e^{- \sqrt{\lambda}x})$
$y'(x)= \sqrt{\lambda} c_1(e^{ \sqrt{ \lambda}x}- e^{- \sqrt{\lambda}x})$
$y'(0)=\frac{y'(1)}{2} \Rightarrow \sqrt{\lambda} c_1=\frac{\sqrt{\lambda} c_1(e^{ \sqrt{ \lambda}}- e^{- \sqrt{\lambda}})}{2} \Rightarrow 2=e^{ \sqrt{ \lambda}}- e^{- \sqrt{\lambda}} \Rightarrow e^{2 \sqrt{\lambda}}-2 e^{\sqrt{\lambda}}+1=0 \Rightarrow (e^{\sqrt{\lambda}}-1)^2=0 \Rightarrow e^{\sqrt{\lambda}}=1 \Rightarrow \lambda =0$
That can not be true since I have supposed that $\lambda <0$.$*$ $\lambda =0$:
$y''=0 \Rightarrow y(x)=c_1 x+c_2$
$y(0)=0 \Rightarrow c_2=0$
$y(x)=c_1 x \Rightarrow y'(x)=c_1$
$y'(0)=\frac{y'(1)}{2} \Rightarrow c_1=\frac{c_1}{2} \Rightarrow c_1=0$
It's the trivial solution.$*$ $\lambda >0$:
$y(x)=c_1 \cos{(\sqrt{\lambda}x)}+c_2 \sin{(\sqrt{\lambda}x)}$
$y(0)=0 \Rightarrow c_1=0$
$y(x)=c_2 \sin{(\sqrt{\lambda}x)} \Rightarrow y'(x)=\sqrt{\lambda} c_2 \cos{(\sqrt{\lambda}x)}$
$y'(0)=\frac{y'(1)}{2} \Rightarrow \sqrt{\lambda} c_2 =\frac{\sqrt{\lambda} c_2 \cos{(\sqrt{\lambda})}}{2} \Rightarrow \cos{(\sqrt{\lambda})}=2$
So the eigenvalues are given by the relation $\cos{(\sqrt{\lambda})}=2$.
Is this correct?? (Wondering)
But are the eigenvalues that are given by the relation $\cos{(\sqrt{\lambda})}=2$ complex?
And how can I find the one real eigenvalue?

 

[I like Serena]

Is this correct?? (Wondering)

Hi! ;)

Yep. All correct.
Except the $m= \pm \sqrt{ \lambda}$, which should be $m= \pm \sqrt{- \lambda}$.
So there should also be some more minuses in front of $\lambda$'s further on.

But are the eigenvalues that are given by the relation $\cos{(\sqrt{\lambda})}=2$ complex?

What is the range of the cosine?
Do you know a way to write the cosine with complex numbers?

And how can I find the one real eigenvalue?

You already have it.
You called it the trivial solution. (Wasntme)

 

[mathmari]

Yep. All correct.

Great! (Smile)

Except the $m= \pm \sqrt{ \lambda}$, which should be $m= \pm \sqrt{- \lambda}$.

Oh yes... (Wasntme)

What is the range of the cosine?
Do you know a way to write the cosine with complex numbers?

The range of cosine is $[-1,1]$
Do you mean this way: $\displaystyle \frac{e^{-ix}+e^{ix}}{2}$ ?

You already have it.
You called it the trivial solution. (Wasntme)

Aha! So the real eigenvalue is $\lambda =0$ and the corresponding eigenfunction is $y=0$, isn't it?

 

[I like Serena]

The range of cosine is $[-1,1]$
Do you mean this way: $\displaystyle \frac{e^{-ix}+e^{ix}}{2}$ ?

Yep.
So you see it cannot just be real.

Aha! So the real eigenvalue is $\lambda =0$ and the corresponding eigenfunction is $y=0$, isn't it?

Correct! (Nod)

 

[mathmari]

So you see it cannot just be real.

I got stuck... How do we know that $\lambda$ is complex?

$$\cos{(\sqrt{\lambda})}=2 \Rightarrow \frac{e^{-i \sqrt{\lambda}}+e^{i \sqrt{\lambda}}}{2}=2 \Rightarrow e^{-i \sqrt{\lambda}}+e^{i \sqrt{\lambda}}=4 \Rightarrow e^{2 i \sqrt{\lambda}}-4e^{i \sqrt{\lambda}}+1=0$$

We set $e^{i \sqrt{\lambda}}=x$, so:
$x^2-4x+1=0 \Rightarrow x=2 \pm \sqrt{3}$

So $e^{i \sqrt{\lambda}}=2+ \sqrt{3}$ or $e^{i \sqrt{\lambda}}=2 - \sqrt{3}$

So since $2 \pm \sqrt{3}$ is a real number, $e^{i \sqrt{\lambda}}$ should be also a real number, so $\lambda$ should be complex?

 

[I like Serena]

I got stuck... How do we know that $\lambda$ is complex?

$$\cos{(\sqrt{\lambda})}=2 \Rightarrow \frac{e^{-i \sqrt{\lambda}}+e^{i \sqrt{\lambda}}}{2}=2 \Rightarrow e^{-i \sqrt{\lambda}}+e^{i \sqrt{\lambda}}=4 \Rightarrow e^{2 i \sqrt{\lambda}}-4e^{i \sqrt{\lambda}}+1=0$$

We set $e^{i \sqrt{\lambda}}=x$, so:
$x^2-4x+1=0 \Rightarrow x=2 \pm \sqrt{3}$

So $e^{i \sqrt{\lambda}}=2+ \sqrt{3}$ or $e^{i \sqrt{\lambda}}=2 - \sqrt{3}$

So since $2 \pm \sqrt{3}$ is a real number, $e^{i \sqrt{\lambda}}$ should be also a real number, so $\lambda$ should be complex?

Yep! (Nod)

Actually, from what you have now, we can only say that $\sqrt{\lambda}$ must be complex.
$\lambda$ can still be real.

Btw, with the $\lambda$ as defined in your problem statement, everywhere where you mention $\lambda$ it should really be $-\lambda$.

 

[mathmari]

Yep! (Nod)

Actually, from what you have now, we can only say that $\sqrt{\lambda}$ must be complex.
$\lambda$ can still be real.

Btw, with the $\lambda$ as defined in your problem statement, everywhere where you mention $\lambda$ it should really be $-\lambda$.

Ok.. so $\sqrt{-\lambda}$ should be complex.. How can we conclude that $\lambda$ should be complex?

 

[I like Serena]

Ok.. so $\sqrt{-\lambda}$ should be complex.. How can we conclude that $\lambda$ should be complex?

Continue from $e^{i\sqrt{-\lambda}}=2 \pm \sqrt 3$... (Wink)

 

[mathmari]

Continue from $e^{i\sqrt{-\lambda}}=2 \pm \sqrt 3$... (Wink)

(We had found that the eigenvalues are given by the relation $\cos{(\sqrt{-\lambda})}=2$, $\cos{(\sqrt{-\lambda})}= \cos{(\sqrt{\lambda})}$, so they are given by the relation $\cos{(\sqrt{\lambda})}=2 \Rightarrow \dots \Rightarrow e^{i \sqrt{\lambda}}=2 \pm \sqrt{3}$)

$$e^{i \sqrt{\lambda}}=2 \pm \sqrt{3} \Rightarrow i \sqrt{\lambda}=ln(2 \pm \sqrt{3}) \Rightarrow \sqrt{\lambda}=-i ln(2 \pm \sqrt{3}) \Rightarrow \lambda =ln(2 \pm \sqrt{3})$$

Isn't this a real value?
Or have I done something wrong?

 

[I like Serena]

(We had found that the eigenvalues are given by the relation $\cos{(\sqrt{-\lambda})}=2$, $\cos{(\sqrt{-\lambda})}= \cos{(\sqrt{\lambda})}$, so they are given by the relation $\cos{(\sqrt{\lambda})}=2 \Rightarrow \dots \Rightarrow e^{i \sqrt{\lambda}}=2 \pm \sqrt{3}$)

$$e^{i \sqrt{\lambda}}=2 \pm \sqrt{3} \Rightarrow i \sqrt{\lambda}=ln(2 \pm \sqrt{3}) \Rightarrow \sqrt{\lambda}=-i ln(2 \pm \sqrt{3}) \Rightarrow \lambda =ln(2 \pm \sqrt{3})$$

Isn't this a real value?
Or have I done something wrong?

I'm afraid that $\cos{(\sqrt{-\lambda})} \ne \cos{(\sqrt{\lambda})}$.
Suppose you substitute $\lambda=-4$.
Will they be equal then?

Furthermore, when working with complex numbers, many functions become multivalued:
$$e^{i \sqrt{-\lambda}}= 2 \pm \sqrt{3}$$
$$e^{i \sqrt{-\lambda}}= e^{\ln(2 \pm \sqrt{3}) + 2\pi n i}$$
$$i \sqrt{-\lambda}= \ln(2 \pm \sqrt{3}) + 2\pi n i$$

 

[mathmari]

I'm afraid that $\cos{(\sqrt{-\lambda})} \ne \cos{(\sqrt{\lambda})}$.
Suppose you substitute $\lambda=-4$.
Will they be equal then?

No.. (Shake)But isn't it as followed?

The characteristic equation is $m^2+\lambda =0 \Rightarrow m= \pm \sqrt{ - \lambda}= \pm \sqrt{i^2 \lambda}= \pm i \sqrt{\lambda}$

$*$ $\lambda >0$:
$y(x)=c_1 \cos{(\sqrt{\lambda}x)}+c_2 \sin{(\sqrt{\lambda}x)}$
$y(0)=0 \Rightarrow c_1=0$
$y(x)=c_2 \sin{(\sqrt{\lambda}x)} \Rightarrow y'(x)=\sqrt{\lambda} c_2 \cos{(\sqrt{\lambda}x)}$
$y'(0)=\frac{y'(1)}{2} \Rightarrow \sqrt{\lambda} c_2 =\frac{\sqrt{\lambda} c_2 \cos{(\sqrt{\lambda})}}{2} \Rightarrow \cos{(\sqrt{\lambda})}=2$
So the eigenvalues are given by the relation $\cos{(\sqrt{\lambda})}=2$.

Furthermore, when working with complex numbers, many functions become multivalued:
$$e^{i \sqrt{-\lambda}}= 2 \pm \sqrt{3}$$
$$e^{i \sqrt{-\lambda}}= e^{\ln(2 \pm \sqrt{3}) + 2\pi n i}$$
$$i \sqrt{-\lambda}= \ln(2 \pm \sqrt{3}) + 2\pi n i$$

Ahaa! I understand! (Wink)

 

[I like Serena]

No.. (Shake)But isn't it as followed?

The characteristic equation is $m^2+\lambda =0 \Rightarrow m= \pm \sqrt{ - \lambda}= \pm \sqrt{i^2 \lambda}= \pm i \sqrt{\lambda}$

$*$ $\lambda >0$:
$y(x)=c_1 \cos{(\sqrt{\lambda}x)}+c_2 \sin{(\sqrt{\lambda}x)}$
$y(0)=0 \Rightarrow c_1=0$
$y(x)=c_2 \sin{(\sqrt{\lambda}x)} \Rightarrow y'(x)=\sqrt{\lambda} c_2 \cos{(\sqrt{\lambda}x)}$
$y'(0)=\frac{y'(1)}{2} \Rightarrow \sqrt{\lambda} c_2 =\frac{\sqrt{\lambda} c_2 \cos{(\sqrt{\lambda})}}{2} \Rightarrow \cos{(\sqrt{\lambda})}=2$
So the eigenvalues are given by the relation $\cos{(\sqrt{\lambda})}=2$.

Oh yeah.
I didn't stop to think that the $i$ got slurped up. (Blush)

Ahaa! I understand!

Good! (Smile)

 

[mathmari]

Oh yeah.
I didn't stop to think that the $i$ got slurped up. (Blush)

Good! (Smile)

And to show that this eigenvalue problem is not a Sturm-Liouville problem using the Wronskian, do I have to take $u(x)=\sin{(\sqrt{\lambda}x)}$ and $v(x)=v^*(x)=0$?

 

[I like Serena]

And to show that this eigenvalue problem is not a Sturm-Liouville problem using the Wronskian, do I have to take $u(x)=\sin{(\sqrt{\lambda}x)}$ and $v(x)=v^*(x)=0$?

Did you try? (Thinking)

 

[mathmari]

Did you try? (Thinking)

I tried it now..Using these $u$ and $v^*$ the Wronskian at $0$ is equal with the Wronskian at $1$, isn't it?

 

[I like Serena]

I tried it now..Using these $u$ and $v^*$ the Wronskian at $0$ is equal with the Wronskian at $1$, isn't it?

Yes. A solution that is zero everywhere tends to do that. (Smirk)

 

[mathmari]

Yes. A solution that is zero everywhere tends to do that. (Smirk)

Ok..So, to show that the Wronskians are different do I have to take other $u$ and $v^*$? (Wondering) Which ones?

 

[I like Serena]

Ok..So, to show that the Wronskians are different do I have to take other $u$ and $v^*$? (Wondering) Which ones?

Since we have:
$$i\sqrt \lambda = \ln(2 \pm \sqrt 3) + 2\pi n i$$
$$\sqrt \lambda = - i \ln(2 \pm \sqrt 3) + 2\pi n$$
Perhaps we can use:
$$u(x) = \sin(i x \ln(2 + \sqrt 3)), \quad v(x) = \sin(i x \ln(2 - \sqrt 3))$$
(Wondering)

 

[mathmari]

Since we have:
$$i\sqrt \lambda = \ln(2 \pm \sqrt 3) + 2\pi n i$$
$$\sqrt \lambda = - i \ln(2 \pm \sqrt 3) + 2\pi n$$
Perhaps we can use:
$$u(x) = \sin(i x \ln(2 + \sqrt 3)), \quad v(x) = \sin(i x \ln(2 - \sqrt 3))$$
(Wondering)

I tried the following:

$u(x)=\sin{(i \ln(2+\sqrt{3})x)}= i \sinh{(x \ln(2+\sqrt{3}))}$
$v(x)=\sin{(i \ln(2-\sqrt{3})x)}= i \sinh{(x \ln(2-\sqrt{3}))} \Rightarrow v^*(x)=-i \sinh{(x \ln(2-\sqrt{3}))}$$W(u(0),v^*(0))=u(0) v^{*'}(0)-u'(0)v^*(0)=0$

$W(u(1),v^*(1))=u(1) v^{*'}(1)-u'(1)v^*(1)=2[u(1)v^{*'}(0)-u'(0)v^*(1)]=2[\ln(2 -\sqrt{3}) \sinh{(\ln(2 + \sqrt{3}))}-\ln(2+\sqrt{3}) \sinh{(\ln(2 - \sqrt{3}))}]=0$But then $W(u(0),v^*(0))=W(u(1),v^*(1))$..

Is this correct? (Wondering) Or have I done something wrong?

 

[I like Serena]

I tried the following:

$u(x)=\sin{(i \ln(2+\sqrt{3})x)}= i \sinh{(x \ln(2+\sqrt{3}))}$
$v(x)=\sin{(i \ln(2-\sqrt{3})x)}= i \sinh{(x \ln(2-\sqrt{3}))} \Rightarrow v^*(x)=-i \sinh{(x \ln(2-\sqrt{3}))}$$W(u(0),v^*(0))=u(0) v^{*'}(0)-u'(0)v^*(0)=0$

$W(u(1),v^*(1))=u(1) v^{*'}(1)-u'(1)v^*(1)=2[u(1)v^{*'}(0)-u'(0)v^*(1)]=2[\ln(2 -\sqrt{3}) \sinh{(\ln(2 + \sqrt{3}))}-\ln(2+\sqrt{3}) \sinh{(\ln(2 - \sqrt{3}))}]=0$But then $W(u(0),v^*(0))=W(u(1),v^*(1))$..

Is this correct? (Wondering) Or have I done something wrong?

What did you do with the $\cosh$ that comes out of the derivative? (Wondering)

 

[mathmari]

What did you do with the $\cosh$ that comes out of the derivative? (Wondering)

We take the derivative at the point $0$, so $\cosh{(0)}=1$.
Is this wrong?

 

[I like Serena]

We take the derivative at the point $0$, so $\cosh{(0)}=1$.
Is this wrong?

Looks good...
It seems the Wronskian condition is satisfied.
What is your reason to think that it wouldn't be? (Wondering)

 

[mathmari]

Looks good...
It seems the Wronskian condition is satisfied.
What is your reason to think that it wouldn't be? (Wondering)

Because I have to show that this eigenvalue problem is not a Sturm-Liouville problem.. (Worried) (Sweating)

 

[mathmari]

I got stuck... (Worried)
What can I do to show that the Wronskian at $0$ is not equal with the Wronskian at $1$?

 

[mathmari]

I have an idea.. (flower) Could you tell me if it's right?

Normally we are searching for eigenvalues where the corresponding eigenfunctions are non-trivial. So do we search maybe the eigenvalues only at the case $\lambda >0$?
Then from the relation $e^{i \sqrt{\lambda}}=2 \pm \sqrt{3}$ we have the following:
$e^{i \sqrt{\lambda}}= e^{\ln(2 \pm \sqrt{3}) + 2\pi n i}$
$i \sqrt{\lambda}= \ln(2 \pm \sqrt{3}) + 2\pi n i$
$\sqrt \lambda = - i \ln(2 \pm \sqrt 3) + 2\pi n$

So the complex eigenvalues are $ \lambda = (- i \ln(2 \pm \sqrt 3) + 2\pi n)^2, n \neq 0$ and the real eigenvalue is $\lambda = (- i \ln(2 \pm \sqrt 3) )^2=-(\ln(2 \pm \sqrt 3) )^2$. Is this correct?

 

[I like Serena]

I have an idea.. (flower) Could you tell me if it's right?

Normally we are searching for eigenvalues where the corresponding eigenfunctions are non-trivial. So do we search maybe the eigenvalues only at the case $\lambda >0$?
Then from the relation $e^{i \sqrt{\lambda}}=2 \pm \sqrt{3}$ we have the following:
$e^{i \sqrt{\lambda}}= e^{\ln(2 \pm \sqrt{3}) + 2\pi n i}$
$i \sqrt{\lambda}= \ln(2 \pm \sqrt{3}) + 2\pi n i$
$\sqrt \lambda = - i \ln(2 \pm \sqrt 3) + 2\pi n$

So the complex eigenvalues are $ \lambda = (- i \ln(2 \pm \sqrt 3) + 2\pi n)^2, n \neq 0$ and the real eigenvalue is $\lambda = (- i \ln(2 \pm \sqrt 3) )^2=-(\ln(2 \pm \sqrt 3) )^2$. Is this correct?

I like ideas. ;)

Once we look at complex eigenvalues, conditions like $\lambda > 0$ and $\lambda < 0$ tend to lose their meaning.
Consider $e^{\lambda x}$. This is an exponential function corresponding to $\cosh/\sinh$ yes?
But if $\lambda$ is imaginary, it corresponds to $\cos/\sin$, or a combination.

So yes, $\lambda = -(\ln(2 \pm \sqrt 3) )^2$ are real eigenvalues corresponding to imaginary eigenfunctions.
And $\lambda = 0$ is also a real eigenvalue, corresponding to a real eigenfunction (the only one).
All others are imaginary eigenvalues with imaginary eigenfunctions.

 

[mathmari]

I like ideas. ;)

Once we look at complex eigenvalues, conditions like $\lambda > 0$ and $\lambda < 0$ tend to lose their meaning.
Consider $e^{\lambda x}$. This is an exponential function corresponding to $\cosh/\sinh$ yes?
But if $\lambda$ is imaginary, it corresponds to $\cos/\sin$, or a combination.

So yes, $\lambda = -(\ln(2 \pm \sqrt 3) )^2$ are real eigenvalues corresponding to imaginary eigenfunctions.
And $\lambda = 0$ is also a real eigenvalue, corresponding to a real eigenfunction (the only one).
All others are imaginary eigenvalues with imaginary eigenfunctions.

I think I have a mistake when I calculated the eigenvalues at the case $\lambda <0$. It should be as followed:
$\lambda=-k, k>0$
$y(x)=c_1 e^{ \sqrt{ k}x}+c_2 e^{- \sqrt{k}x}$
$y(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_2=-c_1$
$y(x)=c_1(e^{ \sqrt{ k}x}- e^{- \sqrt{k}x})$
$y'(x)= \sqrt{k} c_1(e^{ \sqrt{ k}x}+e^{- \sqrt{k}x})$
$y'(0)=\frac{y'(1)}{2} \Rightarrow 2 \sqrt{k} c_1=\frac{\sqrt{k} c_1(e^{ \sqrt{ k}}+ e^{- \sqrt{k}})}{2} \Rightarrow 4=e^{ \sqrt{ k}}+e^{- \sqrt{k}} \Rightarrow e^{2 \sqrt{k}}-4 e^{\sqrt{k}}+1=0 \Rightarrow e^{ \sqrt{k}}=2 \pm \sqrt{3} \Rightarrow e^{ \sqrt{- \lambda}}=2 \pm \sqrt{3}$The exercise asks besides from the complex eigenvalues, the ONE real eigenvalue. Which is meant? $\lambda = -(\ln(2 \pm \sqrt 3) )^2$ or $\lambda = 0$ ? (Wondering)Taking $u(x)=\sin{(i \ln{(2+ \sqrt{3})}x)}=i \sinh{( \ln{(2+\sqrt{3})}x)}$ and $v(x)=v^*(x)=e^{x {\ln{(2+\sqrt{3})}}}-e^{-x {\ln{(2+\sqrt{3})}}}=2 \sinh{({\ln{(2+\sqrt{3})}})x)}$ we have the following:

$W(u(0), v^*(0))=0$

$W(u(1), v^*(1))=u(1) v^{*'}(0)-u'(1)v^*(1)=2[u(1)v^{*'}-u'(0)v^*(1)]=2[i \sinh{(\ln{(2+\sqrt{3})})} 2 \ln{(2+\sqrt{3})}-i \ln{(2+\sqrt{3})} 2 \sinh{({\ln{(2+\sqrt{3}})})}]=\dots=0$

If I have done correct the calculations, the Wronksians are again the same at the these two points.. (Worried)

 

[I like Serena]

I think I have a mistake when I calculated the eigenvalues at the case $\lambda <0$. It should be as followed:
$\lambda=-k, k>0$
$y(x)=c_1 e^{ \sqrt{ k}x}+c_2 e^{- \sqrt{k}x}$
$y(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_2=-c_1$
$y(x)=c_1(e^{ \sqrt{ k}x}- e^{- \sqrt{k}x})$
$y'(x)= \sqrt{k} c_1(e^{ \sqrt{ k}x}+e^{- \sqrt{k}x})$
$y'(0)=\frac{y'(1)}{2} \Rightarrow 2 \sqrt{k} c_1=\frac{\sqrt{k} c_1(e^{ \sqrt{ k}}+ e^{- \sqrt{k}})}{2} \Rightarrow 4=e^{ \sqrt{ k}}+e^{- \sqrt{k}} \Rightarrow e^{2 \sqrt{k}}-4 e^{\sqrt{k}}+1=0 \Rightarrow e^{ \sqrt{k}}=2 \pm \sqrt{3} \Rightarrow e^{ \sqrt{- \lambda}}=2 \pm \sqrt{3}$

Right!
I believe that your other (real) solutions would actually not be valid, since you had the condition $\lambda > 0$, while it turned out that $\lambda < 0$ (or imaginary).

Anyway, you did not really have to distinguish those 2 conditions - the resulting formulas are the same.

The exercise asks besides from the complex eigenvalues, the ONE real eigenvalue. Which is meant? $\lambda = -(\ln(2 \pm \sqrt 3) )^2$ or $\lambda = 0$ ? (Wondering)

Just to be sure, I verified that those eigenvalues really lead to a corresponding solution, and they do. So I do not think there is a calculation error.
Actually, W|A comes up with the following eigenfunction for $\lambda = -\ln^2(2+\sqrt 3)$:
$$y = (2+\sqrt 3)^x - (2+\sqrt 3)^{-x}$$
This is a real function!
And indeed it satisfies the boundary conditions.

We get the expected similar function for $\lambda = -\ln^2(2-\sqrt 3)$.
And $\lambda = 0$ is also an eigenvalue.

So I really count 3 real eigenvalues.

Taking $u(x)=\sin{(i \ln{(2+ \sqrt{3})}x)}=i \sinh{( \ln{(2+\sqrt{3})}x)}$ and $v(x)=v^*(x)=e^{x {\ln{(2+\sqrt{3})}}}-e^{-x {\ln{(2+\sqrt{3})}}}=2 \sinh{({\ln{(2+\sqrt{3})}})x)}$ we have the following:

$W(u(0), v^*(0))=0$

$W(u(1), v^*(1))=u(1) v^{*'}(0)-u'(1)v^*(1)=2[u(1)v^{*'}-u'(0)v^*(1)]=2[i \sinh{(\ln{(2+\sqrt{3})})} 2 \ln{(2+\sqrt{3})}-i \ln{(2+\sqrt{3})} 2 \sinh{({\ln{(2+\sqrt{3}})})}]=\dots=0$

If I have done correct the calculations, the Wronksians are again the same at the these two points.. (Worried)

I don't see either how they can be different.
Sorry. (Doh)

 

[mathmari]

I believe that your other (real) solutions would actually not be valid, since you had the condition $\lambda > 0$, while it turned out that $\lambda < 0$ (or imaginary).

Yes, you're right! (Nod)

Just to be sure, I verified that those eigenvalues really lead to a corresponding solution, and they do. So I do not think there is a calculation error.
Actually, W|A comes up with the following eigenfunction for $\lambda = -\ln^2(2+\sqrt 3)$:
$$y = (2+\sqrt 3)^x - (2+\sqrt 3)^{-x}$$
This is a real function!
And indeed it satisfies the boundary conditions.

We get the expected similar function for $\lambda = -\ln^2(2-\sqrt 3)$.
And $\lambda = 0$ is also an eigenvalue.

So I really count 3 real eigenvalues.

As regards $\lambda = 0$, the corresponding eigenfunction is $y(x)=0$, which is a trivial solution, isn't it? In Wikipedia (Sturm) there is the sentence:
"The value of λ is not specified in the equation; finding the values of λ for which there exists a non-trivial solution of (1) satisfying the boundary conditions is part of the problem called the Sturm–Liouville (S–L) problem."
So I think that we don't take into consideration $\lambda = 0$. (Thinking) Or do we do this only in the case where the problem is a Sturm-Liouville problem?

I don't see either how they can be different.
Sorry. (Doh)

Since I have not found any $u(x)$ and $v^*(x)$ so that the Wronskian at $0$ is different with its value at $1$, I tried to show it otherwise.

For a Sturm-Liouville problem one property is that the eigenvalues are real, so since we have found also complex eigenvalues, the problem is not a Sturm-Liouville problem.

Could we do show this in this way? (Wondering)

 

[I like Serena]

As regards $\lambda = 0$, the corresponding eigenfunction is $y(x)=0$, which is a trivial solution, isn't it? In Wikipedia (Sturm) there is the sentence:
"The value of λ is not specified in the equation; finding the values of λ for which there exists a non-trivial solution of (1) satisfying the boundary conditions is part of the problem called the Sturm–Liouville (S–L) problem."
So I think that we don't take into consideration $\lambda = 0$. (Thinking) Or do we do this only in the case where the problem is a Sturm-Liouville problem?

That sounds very plausible. (Mmm)

Actually, $y=0$ is always a solution, regardless of the value of $\lambda$, so $\lambda=0$ is not really the corresponding eigenvalue.
So it makes sense to leave that one out.

Furthermore, the eigenfunctions for $-\ln^2(2\pm\sqrt 3)$ are identical.
That is because \(\displaystyle \frac 1{2-\sqrt 3} = 2+\sqrt 3\).
So we might consider them as the same.

Since I have not found any $u(x)$ and $v^*(x)$ so that the Wronskian at $0$ is different with its value at $1$, I tried to show it otherwise.

For a Sturm-Liouville problem one property is that the eigenvalues are real, so since we have found also complex eigenvalues, the problem is not a Sturm-Liouville problem.

Could we do show this in this way? (Wondering)

Aha! I see that according to the wiki page that the eigenvalues are real if the Sturm-Liouville problem is regular. That seems to suggest that your Wronskian condition boils down to the same thing.

Anyway, that would be an indication that there is something special with those complex eigenvalues.
Have you tried to calculate the Wronskian for 2 eigenfunctions of which 1 has an imaginary eigenvalue? (Wondering)

 

[mathmari]

That sounds very plausible. (Mmm)

Actually, $y=0$ is always a solution, regardless of the value of $\lambda$, so $\lambda=0$ is not really the corresponding eigenvalue.
So it makes sense to leave that one out.

Furthermore, the eigenfunctions for $-\ln^2(2\pm\sqrt 3)$ are identical.
That is because \(\displaystyle \frac 1{2-\sqrt 3} = 2+\sqrt 3\).
So we might consider them as the same.

Great! (Whew)

Have you tried to calculate the Wronskian for 2 eigenfunctions of which 1 has an imaginary eigenvalue? (Wondering)

So should I try it for $u(x)=\sin{(x(-i \ln{(2+\sqrt{3})}+2 \pi n))}$ and $v(x)=v^*(x)=(2+\sqrt{3})^x-(2+\sqrt{3})^{-x}$ ?

 

[I like Serena]

So should I try it for $u(x)=\sin{(x(-i \ln{(2+\sqrt{3})}+2 \pi n))}$ and $v(x)=v^*(x)=(2+\sqrt{3})^x-(2+\sqrt{3})^{-x}$ ?

Yes... (Sweating)

 

[mathmari]

Yes... (Sweating)

If I have done right the calculations, the Wronskian at $1$ is not equal to $0$! (Wink) (Party) (Dance)

 

[mathmari]

Thank you for your help! (Bow)