Is L a Self-Adjoint Operator with Non-Negative Eigenvalues?

[brkomir]

Homework Statement

We have a linear differential operator $Ly=-y^{''}$ working on all $y$ that can be derived at least twice on $[-\pi ,\pi ]$ and also note that $y(-\pi )=y(\pi )$ and $y^{'}(-\pi )=y^{'}(\pi )$.
a) Is $0$ eigenvalue for $L$?
b) Is $L$ symmetric? (I think the right English expression would be self-adjoint)
c) Find all positive eigenvalues and corresponding eigenfunctions. What is the dimension of eigenspace?
d) Does $L$ have any negative eigenvalues?

Homework Equations

The Attempt at a Solution

a) No idea how to do this one. Here's my version:

$Ly=-y^{''}=\lambda y$ which gives me $y^{''}+\lambda y=0$. Solution of this differential equation is obviously $y(x)=Acos(\sqrt{\lambda }x)+Bsin(\sqrt{\lambda }x)$

Using this we find out that the boundary conditions for $A\neq 0$ and $B\neq 0$ are fulfilled only when $\lambda =0$. Therefore the answer to question a) is YES.

b)

$L$ is symmetric if $[PW(y,x)]\mid _a^b =0$. Note that originally the DE is $P(z)y^{''}+Q(z)y^{'}+R(z)y=0$ and where $W$ is Wronskian determinant.

Obviously $P=-1$.

$[PW(y,x)]\mid _a^b =[-\begin{vmatrix} y & z\\ y^{'} & z^{'} \end{vmatrix}]\mid _a^b$

After short calculus and knowing that $a=-\pi$ and $b=\pi$ we find out that the equation above is equal to $0$ and therefore the linear operator $L$ is symmetric.

c)

$Ly=-y^{''}=\lambda^2 y$ where I used notation $\lambda ^2$ for eigenvalue instead of $\lambda$ just because I don't want to write square roots every time.

Anyhow, solution to this DE is $y(x)=Acos(\lambda x)+Bsin(\lambda x)$ using boundary conditions:

$y(\pi )-y(-\pi )=2Bsin(\lambda \pi )=0$ and of course for non trivial solutions $B\neq 0$ than:

$\lambda =n$ for $n\in \mathbb{Z}^{+}$ where i suspect these are all the positive eigenvalues.

And accordingly, I assume that $y(x)=Acos(\lambda x)+Bsin(\lambda x)$ are eigenfunctions where for each $\lambda$ the dimension of the eigenspace increases by $1$, therefore the dimension of eigenspace is $n$.

(or do I have to consider $\lambda =0$ here too?)

d) Hmmm, if $\lambda <0$ than the differential equation $y^{''}+\lambda y=0$ changes to $y^{''}-\lambda y=0$.

Now $y$ that solves the equation above is $y(x)=Ae^{\sqrt{\lambda }x}+Be^{-\sqrt{\lambda }x}$.

Again, taking in mind that $y(-\pi )=y(\pi )$ leaves me with

$(B-A)(e^{-\sqrt{\lambda }x\pi }-e^{\sqrt{\lambda }x\pi })=0$ which is only true if $(B-A)=0$.

Therefore the answer is NO, $L$ does not have negative eigenvalues.

 

[pasmith]

Homework Statement

We have a linear differential operator $Ly=-y^{''}$ working on all $y$ that can be derived at least twice on $[-\pi ,\pi ]$ and also note that $y(-\pi )=y(\pi )$ and $y^{'}(-\pi )=y^{'}(\pi )$.
a) Is $0$ eigenvalue for $L$?
b) Is $L$ symmetric? (I think the right English expression would be self-adjoint)
c) Find all positive eigenvalues and corresponding eigenfunctions. What is the dimension of eigenspace?
d) Does $L$ have any negative eigenvalues?

Homework Equations

The Attempt at a Solution

a) No idea how to do this one. Here's my version:

$Ly=-y^{''}=\lambda y$ which gives me $y^{''}+\lambda y=0$. Solution of this differential equation is obviously $y(x)=Acos(\sqrt{\lambda }x)+Bsin(\sqrt{\lambda }x)$

Using this we find out that the boundary conditions for $A\neq 0$ and $B\neq 0$ are fulfilled only when $\lambda =0$. Therefore the answer to question a) is YES.

The correct observation is that setting $\lambda = 0$ gives you $y(x) = A \cos 0 + B \sin 0 = A$. But why not just solve $$Ly = -y'' = 0$$ directly to get $y(x) = Cx + D$, where $y(-\pi) = y(\pi)$ requires $C = 0$ but $D$ can be anything?

b)

$L$ is symmetric if $[PW(y,x)]\mid _a^b =0$. Note that originally the DE is $P(z)y^{''}+Q(z)y^{'}+R(z)y=0$ and where $W$ is Wronskian determinant.

Obviously $P=-1$.

$[PW(y,x)]\mid _a^b =[-\begin{vmatrix} y & z\\ y^{'} & z^{'} \end{vmatrix}]\mid _a^b$

After short calculus and knowing that $a=-\pi$ and $b=\pi$ we find out that the equation above is equal to $0$ and therefore the linear operator $L$ is symmetric.

$L$ is self-adjoint with respect to an inner product, presumably $\langle f,g \rangle = \int_{-\pi}^{\pi} f(x)g(x)\,dx$ if we're dealing only with real-valued twice-differentiable functions, if and only if $$\langle Lf, g \rangle = \langle f, Lg \rangle$$ for all relevant $f$ and $g$. So you need to show that for all twice-differentiable $f$ and $g$ which satisfy the boundary conditions, you have $$\int_{-\pi}^{\pi} -f''(x) g(x) \,dx = \int_{-\pi}^{\pi} -f(x) g''(x) \,dx.$$

c)

$Ly=-y^{''}=\lambda^2 y$ where I used notation $\lambda ^2$ for eigenvalue instead of $\lambda$ just because I don't want to write square roots every time.

Best then to set $\lambda = k^2$ for $k > 0$.

Anyhow, solution to this DE is $y(x)=Acos(\lambda x)+Bsin(\lambda x)$ using boundary conditions:

$y(\pi )-y(-\pi )=2Bsin(\lambda \pi )=0$ and of course for non trivial solutions $B\neq 0$ than:

$\lambda =n$ for $n\in \mathbb{Z}^{+}$ where i suspect these are all the positive eigenvalues.

And accordingly, I assume that $y(x)=Acos(\lambda x)+Bsin(\lambda x)$ are eigenfunctions where for each $\lambda$ the dimension of the eigenspace increases by $1$, therefore the dimension of eigenspace is $n$.

The eigenvalue is now $\lambda^2$ rather than $\lambda$, so the eigenvalues are $\lambda_n^2 = n^2 > 0$ for $n \in \mathbb{Z}^{+}$.

Each eigenspace is two-dimensional: the linearly independent eigenfunctions corresponding to the eigenvalue $n^2$ are $\cos nx$ and $\sin nx$.

(or do I have to consider $\lambda =0$ here too?)

d) Hmmm, if $\lambda <0$ than the differential equation $y^{''}+\lambda y=0$ changes to $y^{''}-\lambda y=0$.

If you want to do that, then you need to write $y'' + \lambda y = y'' - |\lambda|y = 0$. It would be easier to define $\lambda = -k^2$ for $k > 0$.

Now $y$ that solves the equation above is $y(x)=Ae^{\sqrt{\lambda }x}+Be^{-\sqrt{\lambda }x}$.

Again, taking in mind that $y(-\pi )=y(\pi )$ leaves me with

$(B-A)(e^{-\sqrt{\lambda }x\pi }-e^{\sqrt{\lambda }x\pi })=0$ which is only true if $(B-A)=0$.

Therefore the answer is NO, $L$ does not have negative eigenvalues.

So far you've only shown that you must have $B = A$; you have yet to show that $B = A = 0$.

 

[brkomir]

The correct observation is that setting $\lambda = 0$ gives you $y(x) = A \cos 0 + B \sin 0 = A$. But why not just solve $$Ly = -y'' = 0$$ directly to get $y(x) = Cx + D$, where $y(-\pi) = y(\pi)$ requires $C = 0$ but $D$ can be anything?

That is a lot easier, I agree. So the conclusion is that if $\lambda =0$ we can find a function that satisfies boundary conditions. In this case $y=D$ is a constant function, therefore $\lambda$ can also be $0$.

b) Is $L$ self-adjoint?

$<f,g>=\int _{-\pi }^{\pi }f(x)g(x)dx$ for real functions.

$<Lf,g>=<f,Lg>$

$\int _{-\pi }^{\pi }-f^{''}gdx=\int _{-\pi }^{\pi }-fg^{''}dx$

$\int _{-\pi }^{\pi }(fg^{''}-f^{''}g)dx=0$ That here is still a question. We are trying to prove that the LHS of the equation is 0.

$\int _{-\pi }^{\pi }(fg^{''}-f^{''}g)dx=\int _{-\pi }^{\pi }[(g^{'}f)^{'}-f^{'}g^{'}-(f^{'}g)^{'}+f^{'}g^{'}]dx=[g^{'}f-f^{'}g]\mid _{-\pi }^{\pi }$

$[g^{'}f-f^{'}g]\mid _{-\pi }^{\pi }=0$ for given boundary conditions. Nice, so yes, $L$ is self-adjoint.

c)

Best then to set $\lambda = k^2$ for $k > 0$.

Ok, I will do that.

The eigenvalue is now $\lambda^2$ rather than $\lambda$, so the eigenvalues are $\lambda_n^2 = n^2 > 0$ for $n \in \mathbb{Z}^{+}$.

Each eigenspace is two-dimensional: the linearly independent eigenfunctions corresponding to the eigenvalue $n^2$ are $\cos nx$ and $\sin nx$.

So for $n$ eigenfunctions the eigenspace is $2n$ dimensional? I assume that is because $sin$ and $cos$ are already linearly independent, therefore for each $\lambda \neq 0$ I get two additional linearly independent functions.

But, is it really $\lambda _n^2=n^2$ ?? Because, if I am not mistaken for $\lambda = k^2$ for $k > 0$ I get

$2Bsin(k\pi )=0$ therefore $k=n$ and if anything, than $\lambda _n=k^2=n^2$. Or not?d)

So far you've only shown that you must have $B = A$; you have yet to show that $B = A = 0$.

The second part of course comes from second condition, which gives me $(A+B)(e^{\sqrt{\lambda }\pi }-e^{-\sqrt{\lambda }\pi })=0$

Now we have both conditions saying that $A=B$ and $A=-B$ which is only possible if $A=B=0$