Sturm-Liouville eigenvalue problem

Oct 2009
54
2
Hi,
I got a question which I'm quite stuck on...
Given the eigenvalue problem,
\(\displaystyle y''+\lambda y=0\)
\(\displaystyle hy(0)+y'(0) = 0, y(1) = 0\)
Show that if \(\displaystyle \lambda = 0\) is an eigenvalue, then \(\displaystyle h\) must have a specific value and find this value. If \(\displaystyle h \geq 1\), show that there is exactly one negative eigenvalue.

I believe I've done the first part, as my working is as follows:
Rearrange the HLDE into self adjoint form: \(\displaystyle -(-y')' = \lambda (-1)y\), and taking that \(\displaystyle \lambda = 0\), means the HLDE becomes \(\displaystyle y'' = 0\). The general solution to this should be of the form \(\displaystyle y(x)= Ax + B\). Using the boundary conditions,
\(\displaystyle hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}\)
\(\displaystyle y(1) = A + B = 0 \Rightarrow A = -B\)
Therefore, \(\displaystyle h = -\frac{(-B)}{B} = 1\)
Does this look correct?

The next part of the question I get stuck on. The general solution for \(\displaystyle -(-y')' = \lambda (-1)y\) is \(\displaystyle y(x)= Ccos (\sqrt{\lambda}x) + Dsin(\sqrt{\lambda}x) \).
Using the boundary conditions
\(\displaystyle hy(0)+y'(0) = h(Ccos (\sqrt{\lambda}0) + Dsin(\sqrt{\lambda}0)) + (-C\sqrt{\lambda}sin (\sqrt{\lambda}0) + D\sqrt{\lambda}cos(\sqrt{\lambda}0))\) \(\displaystyle = h*C +\sqrt{\lambda}D = 0\)
\(\displaystyle y(1) =Ccos (\sqrt{\lambda}) + Dsin(\sqrt{\lambda}) = 0 \)
From here, I not sure where to go... have I do the right thing?

Thanks for your time.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
Hi,
I got a question which I'm quite stuck on...
Given the eigenvalue problem,
\(\displaystyle y''+\lambda y=0\)
\(\displaystyle hy(0)+y'(0) = 0, y(1) = 0\)
Show that if \(\displaystyle \lambda = 0\) is an eigenvalue, then \(\displaystyle h\) must have a specific value and find this value. If \(\displaystyle h \geq 1\), show that there is exactly one negative eigenvalue.

I believe I've done the first part, as my working is as follows:
Rearrange the HLDE into self adjoint form: \(\displaystyle -(-y')' = \lambda (-1)y\), and taking that \(\displaystyle \lambda = 0\), means the HLDE becomes \(\displaystyle y'' = 0\). The general solution to this should be of the form \(\displaystyle y(x)= Ax + B\). Using the boundary conditions,
\(\displaystyle hy(0)+y'(0) = hB + A = 0 \Rightarrow h = -\frac{A}{B}\)
\(\displaystyle y(1) = A + B = 0 \Rightarrow A = -B\)
Therefore, \(\displaystyle h = -\frac{(-B)}{B} = 1\)
Does this look correct? Yes.

The next part of the question I get stuck on. The general solution for \(\displaystyle -(-y')' = \lambda (-1)y\) is \(\displaystyle y(x)= Ccos (\sqrt{\lambda}x) + Dsin(\sqrt{\lambda}x) \).
Using the boundary conditions
\(\displaystyle hy(0)+y'(0) = h(Ccos (\sqrt{\lambda}0) + Dsin(\sqrt{\lambda}0)) + (-C\sqrt{\lambda}sin (\sqrt{\lambda}0) + D\sqrt{\lambda}cos(\sqrt{\lambda}0))\) \(\displaystyle = h*C +\sqrt{\lambda}D = 0\)
\(\displaystyle y(1) =Ccos (\sqrt{\lambda}) + Dsin(\sqrt{\lambda}) = 0 \)
You are looking for solutions with \(\displaystyle \lambda <0\), so it's not a good idea to use functions like \(\displaystyle \cos(\sqrt{\lambda}x)\) and \(\displaystyle \sin(\sqrt{\lambda}x)\). Instead, notice that the equation \(\displaystyle y'' = (-\lambda) y\) has solutions \(\displaystyle y=e^{\pm\alpha x}\), where \(\displaystyle \alpha = \sqrt{-\lambda}\). If \(\displaystyle y = Ae^{\alpha x} + Be^{-\alpha x}\) then the boundary conditions lead to some equation like \(\displaystyle h = \frac{\alpha(e^{\alpha} + e^{-\alpha})}{e^{\alpha} - e^{-\alpha}}\), and you need to show that for \(\displaystyle h\geqslant1\) that has a unique solution for \(\displaystyle \alpha>0\).
 
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