# Sturm-Liouville eigenvalue problem

#### Silverflow

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?

#### Opalg

MHF Hall of Honor
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 $$\displaystyle \alpha>0$$.

• Silverflow

#### Silverflow

That makes a lot more sense. Thank you very much!