Sturm Liouville problem (2 questions about the solution)

Question

These are my conditions:

$$f'' +λf = 0 \tag{1}$$ $$f'(0) - f(0) = 0 \tag{2}$$ $$f(1) = 0 \tag{3}$$ $$0<x<1 \tag{4}$$

I'm finding the eigenfunctions and eigenvalues.

We start with $λ = 0$:

$$f = ax+b \Rightarrow f'=a; a-b = 0 \Rightarrow 2a = 0 \Rightarrow a=b=0 \Rightarrow f = 0 $$

$λ = 0$ is not an eigenvalue.

Question: Since we want values of $λ$ where the solution of $(1)$ is non-zero, and we know $f=0$, is that why $λ=0$ is not an eigenvalue?

For $λ > 0$:

$$f = c_1\cdot \cos(\sqrt{λ})x+c_2\cdot \sin(\sqrt{λ})x$$

$$f' =-c_1(\sqrt{λ})\sin(\sqrt{λ}) x+c_2(\sqrt{λ})\cos(\sqrt{λ})x$$

Question: We know $(2)$, so if we subtract $f'$ and $f$ from each other, after substituting $x = 0$, we apparently have:

$$f'(0) - f(0) = c_2 \sqrt{λ}\cdot \cos(\sqrt{λ}) -c_1 = 0$$ Why is that? We have a multiple $x = 0$ for every term in both $f$ and $f''$. How do we get this result?

Thanks so much.

Answer 1

Let me offer a different, more general approach, where you don't end up with extra multiplicative terms that can trick you into finding incorrect eigenvalues.

Every non-trivial solution of $f''+\lambda f=0$ that satisfies $f(1)=0$ must satisfy $f'(1)\ne 0$, which means that it can be normalized so that $f(1)=0$, $f'(1)=1$. This unique solution is $$ f_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}}. $$ The added normalization insures that $\lambda\mapsto f_{\lambda}(x)$ has an everywhere convergence power series in $\lambda$ for all $x \in [0,1]$. This is true in general for fixed endpoint conditions. And it means that the special case $\lambda=0$ is handled with L'Hopital's rule: $$ f_{0}(x) = \lim_{\lambda\rightarrow 0}f_{\lambda}(x)=(x-1). $$ It is easily checked that $f''+0f=0$, with $f_0(1)=0$ and $f_0'(1)=1$, just as for all the other $f_{\lambda}$.

A general $\lambda$ is an eigenvalue iff $\lambda$ is a zero of the following power series obtained from the series expansions of $\cos\sqrt{\lambda}$ and $\sin\sqrt{\lambda}$: $$ 0 = f_{\lambda}'(0)-f_{\lambda}(0) = \cos(\sqrt{\lambda})+\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}} = \sum_{n=0}^{\infty}\frac{(2n+2)}{(2n+1)!}(-\lambda)^n . $$ All eigenvalues must be real, and you can see that the series is strictly positive for $\lambda < 0$ because all the terms in the series are strictly positive for such $\lambda$; clearly $\lambda=0$ is not a solution. Hence the eigenvalues must be strictly positive. For $\lambda > 0$, the equation is transcendental: $$ \tan(\sqrt{\lambda}) = -\sqrt{\lambda},\;\;\; \lambda > 0. $$ You can plot $\tan(x)$ vs. $-x$ to see that these graphs intersect at a sequence $$0 < x_1 < x_2 < x_3 < \cdots,$$ where asymptotically $x_n \sim (2n-1)\frac{\pi}{2}$ holds for large $n$. Then $\lambda_n = x_n^2$.

Answer 2

When solving these sorts of questions you need to check what happens when:

$\lambda$ = 0, $\lambda$ > 0 and $\lambda$ < 0

Our operator here is defined as $L(y)=y'' =\lambda u$

When $\lambda$ = 0:

$$y''=0 \rightarrow y(x) = Ax+B $$

Plugging in our boundary values: $$y'(0)−y(0)=0 \rightarrow A-B = 0 \rightarrow A=B$$

Now we can rewrite our equation as:

$$Ax+A=0$$

$$y(1)= 0 \rightarrow A+A=0 =2A=0 \rightarrow A=0 $$

Hence we arrive at the trivial solution, which doesn't help us find the correct eigenfunctions.

Now if $\lambda >0$ we get a solution composed of exponents of the form

$$y(x)= Ae^\sqrt{\lambda x}+Be^{-{\sqrt\lambda x}} $$

I'll let you plug in the BVs and determined whether we can find our eigenfunctions from here.

Finally if $\lambda < 0$ we get our solution composed of sin and cos

$$ y(x) = Acos(\lambda x) + Bsin(\lambda x)$$

Let's plug in our BVs:

$$f(1)=0 \rightarrow Acos(\lambda) + Bsin(\lambda) = 0 $$

and :

$$f'(x) = -\lambda Asin(\lambda x) + \lambda Bcos(\lambda x) = 0$$

$$f'(0) = \lambda B $$

$$f(0) = A$$

So we get $$\lambda B - A = 0 \rightarrow \lambda B = A$$

Hopefully I haven't made any mistakes up until here, but the rest involves plugging in what we've found, and finding the possible solutions for $\lambda$ based.