Sturm–Liouville theory, linear differential equation

Question

We have Sturm–Liouville problem $$\left\{\begin{matrix} {y}''(t)+4{y}'+3y=\lambda y\\ y(0)= y'(\ln(l))=0 \end{matrix}\right. 1<t<l$$ I've found roots of characteristic polynomial of differential equation $$\tau = -2\pm\sqrt{1+\lambda }$$ So there are three three different solutions depending on $\lambda$. $$1)\ \lambda = -1:\ y(t)= C_{1}e^{-2t}+C_{2}te^{-2t}$$ $$2)\ \lambda<-1: \ y(t)=C_{1}e^{-2t}\cos(\sqrt{-1-\lambda}t)+C_{2}e^{-2t}\sin(\sqrt{-1-\lambda}t)$$ $$3)\ \lambda>-1: \ y(t)=C_{1}e^{(-2+\sqrt{1+\lambda})t}+C_{2}e^{(-2-\sqrt{1+\lambda})t}$$ I don't have problmes with cases 1 and 2, but there is problem with case 3.

When we substitute the initial values, we get system of to equations: $$\left\{\begin{matrix} y(0)=C_{1}+C_{2} = 0 \rightarrow C_{1}=-C_{2} \\ y(\ln(l))=(-2+\sqrt{1+\lambda})C_{1}e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})C_{2}e^{(-2-\sqrt{1+\lambda})\ln(l)}=0\end{matrix}\right.$$ Then we get one equation: $$(2-\sqrt{1+\lambda})C_{2}e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})C_{2}e^{(-2-\sqrt{1+\lambda})\ln(l)}=0$$ $$C_{2}\big((2-\sqrt{1+\lambda})e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})e^{(-2-\sqrt{1+\lambda})\ln(l)}\big)=0$$ And as I know, this equation have solution only when $C_{2}=0 \rightarrow C_{1}=0$

So, the equation $$(2-\sqrt{1+\lambda})e^{(-2+\sqrt{1+\lambda})\ln(l)}+(-2-\sqrt{1+\lambda})e^{(-2-\sqrt{1+\lambda})\ln(l)}=0$$ Must not have solutions in real numbers, but I don't know how to prove that

Answer 1

Let $f(t) = e^{-2t} \sinh{t}$. We have $f'(t) = e^{-2t} (-2 \sinh{t} + \cosh{t})$. Take $$ t = \operatorname{arcosh}{\frac{2}{\sqrt{3}}}. $$ We calculate $$ \sinh{t} = \sqrt{\cosh^2{t} - 1} = \sqrt{\frac{4}{3} - 1} = \frac{1}{\sqrt{3}}. $$ So, if I am not mistaken, $f(t)$ is a solution of $$ \left\{\begin{matrix} {y}''+4{y}'+3y= 0 \\ y(0)= y'(\operatorname{arcosh}{\frac{2}{\sqrt{3}}})=0 \end{matrix}\right. $$ with $\lambda = 0 > -1$.

Answer 2

Not quite. We want to solve $$ (-2-\sqrt{1+\lambda})e^{(-2-\sqrt{1+\lambda})t_0} - (-2 + \sqrt{1+\lambda})e^{(-2+\sqrt{1+\lambda})t_0} = 0 $$

where $t_0 = \ln l$ Multiply through by $e^{2t_0}$ and let $\mu = \sqrt{1+\lambda}$ for brevity, we obtain

$$ (-2-\mu)e^{-t_0\mu} - (-2+\mu)e^{t_0\mu} = 0 $$

$$ 2(e^{t_0\mu} - e^{-t_0\mu}) - \mu (e^{t_0\mu} + e^{-t_0\mu}) = 0 $$

or $$ \frac{\mu}{2} = \frac{e^{t_0\mu}-e^{-t_0\mu}}{e^{t_0\mu} + e^{-t_0\mu}} = \tanh (t_0\mu) $$

There is actually one solution if $t_0 > \frac12$ or $l > \sqrt{e}$. A closed form doesn't exist unfortunately, but you can see it graphically.