Sturm Liouville with periodic boundary conditions

Background and motivation: I'm given the boundary value problem: $$y''(x)+2y(x)=-f(x)$$ subject $y(0)=y(2\pi)$ and $y \, '(0)=y \, '(2\pi)$.

EDIT: These were not given to be zero !! Maybe this helps...

The text (Nagle Saff and Snider, end of Chapter 11 technical writing exercise) asks us to construct the Green's function for the problem. At the moment, I'm a bit stumped because there is no $\lambda$ in the given problem. Let me elaborate, if we were given: $$(py')'+qy+\lambda r y= 0$$ where $p,p',q$ and $r$ were continuous, real-valued, periodic functions with period $2\pi$ then I think I'd be able to get started. I know the usual solutions then only fit the given boundary conditions for particular choices of $\lambda$. So, my initial observation is that $p=1$ is certainly continuous and periodic so we can set $p=1$.

• Question: what should I see as $q$ and $r$ for the problem stated at the start of this post? How can we massage the given problem into the standard form of Sturm Liouville?

I suppose it is important to note we must choose $r>0$ as it serves as the weight function in the inner product which is paired with the eigenspace of solutions for this problem.

Added: here is a picture of the problem from the text:

• @Amzoti I was curious, did you see my question, it got buried pretty fast after I posted it. I think you might know the answer... Commented Apr 25, 2014 at 13:31
• It seems to me that $q=0$, $r=1$, $\lambda=2$ would do the trick? However, are you sure the boundary conditions are correct? You're asking for solutions that are sinusoidal, that go to zero on the boundary, and whose derivatives are zero on the boundary. No combination of sines and cosines can achieve this. Commented Apr 27, 2014 at 16:56
• Also I can't seem to find a Chapter 11 in the eighth edition. What edition are you using? Commented Apr 27, 2014 at 17:01
• This problem, as stated, simply does not admit a solution. No function that solves $y''+2y=\delta(x)$ can also have $y(0)=y'(0)=y(2\pi)=y'(2\pi)=0$, which you can show by direct calculation. Maybe this is the point of the exercise? To explore why no solutions exist? Commented Apr 28, 2014 at 14:26
• Or is it supposed to be two problems, one with $y(0)=y(2\pi)=0$, and another one with $y'(0)=y'(2\pi)=0$? Commented Apr 28, 2014 at 14:29

The Green's function is the solution when $f(x)=\delta(x-x_s)$, where $x_s$ is some kind of point source position that forces the system. Let's suppose that $x_s\in(0,2\pi)$. For $x\neq x_s$, the delta function is zero, and so we solve the homogeneous equation $$\left| \begin{array}{cc} y'' + 2y = 0, & x<x_s\\ y'' + 2y = 0, & x>x_s \end{array} \right.$$ And we'll worry about what happens right at $x=x_s$ in a bit. The homogeneous equations presented are solved by sines and cosines $$y(x) = \left\{ \begin{array}{cc} A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x), & x<x_s\\ C\cos(\sqrt{2}x)+D\sin(\sqrt{2}x), & x>x_s\\ \end{array} \right.$$ Now the first boundary condition is that $y(0)=y(2\pi)$. For the left boundary, we use the left part of the piecewise $y$ above, and for the right boundary, we use the right part, so this reads $$A = C\cos(2\sqrt{2}\pi)+D\sin(2\sqrt{2}\pi)$$ Doing the same with the derivative conditions gives $$B = D \cos \left(2 \sqrt{2} \pi \right)-C \sin \left(2 \sqrt{2} \pi \right)$$ Now we need conditions to match the delta function. We expect the solution to be continuous at $x=x_s$, so $$A\cos(\sqrt{2}x_s)+B\sin(\sqrt{2}x_s) = C\cos(\sqrt{2}x_s)+D\sin(\sqrt{2}x_s)$$ And then there is the appropriate "jump" condition on the derivative at $x=x_s$. We need $y$ to be discontinuous enough so that taking two derivatives will result in a negative delta function. This means that there must be a negative unit step discontinuity in the derivative: $$-\sqrt{2} A \sin \left(\sqrt{2} x_s\right)+\sqrt{2} B \cos\left(\sqrt{2} x_s\right)-1=\sqrt{2}D\cos\left(\sqrt{2} x_s\right)-\sqrt{2} C \sin \left(\sqrt{2} x_s\right)$$ The above are four linear equation in the four unknowns $(A,B,C,D)$, which we can formulate as $$\left[ \begin{array}{cccc} 1 & 0 & -\cos(2\sqrt{2}\pi) & -\sin(2\sqrt{2}\pi) \\ 0 & 1 & \sin \left(2 \sqrt{2} \pi \right) & -\cos \left(2 \sqrt{2} \pi \right)\\ \cos(\sqrt{2}x_s)&\sin(\sqrt{2}x_s)&-\cos(\sqrt{2}x_s)&-\sin(\sqrt{2}x_s) \\ -\sqrt{2} \sin \left(\sqrt{2} x_s\right)&\sqrt{2} \cos\left(\sqrt{2} x_s\right)&\sqrt{2}\sin \left(\sqrt{2} x_s\right)&-\sqrt{2}\cos\left(\sqrt{2} x_s\right) \end{array} \right] \left[ \begin{array}{c} A\\B\\C\\D \end{array} \right] = \left[ \begin{array}{c} 0\\0\\0\\1 \end{array} \right]$$ Solving that system after a good bit of trigonometric simplifications gives $$A=-\frac{ \cos \left(\sqrt{2} (x_s-\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}$$ $$B=\frac{ \sin \left(\sqrt{2} (\pi -x_s)\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}$$ $$C=-\frac{ \cos \left(\sqrt{2} (x_s+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}$$ $$D=-\frac{ \sin \left(\sqrt{2} (x_s+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}$$ Plugging into the piecewise form proposed for $y$ and doing more trigonometric simplifying, $$y(x) = \left\{ \begin{array}{cc} -\frac{ \cos \left(\sqrt{2} (x-x_s+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}, & x<x_s\\ -\frac{ \cos \left(\sqrt{2} (x_s-x+\pi )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}, & x>x_s\\ \end{array} \right.$$ Notice these are the same except the role of $x$ and $x_s$ are switched between the two expressions. This allows us to write a more compact expression $$y(x)=-\frac{ \cos \left(\sqrt{2} (\pi-|x-x_s| )\right)}{2 \sqrt{2}\sin(\sqrt{2}\pi)}$$ which is the Green's function for this problem.
I will write $a$ for $\sqrt{2}$ for simplicity. The general solution of the homogenous equation $y''+2y=0$ has the form $$y(x)=C\cos(ax)+D\sin(ax)$$ Using the variation of parameters method to find a particular solution of the non homogenous problem, we have to determine $C$ and $D$ with \eqalign{C'\cos(ax)+D'\sin(ax)&=0\cr -C'\sin(ax)+D'\cos(ax)&=-\frac{1}{a}f(x) } This yields $$C'=\frac{1}{a}\sin(ax)f(x),\quad D'=-\frac{1}{a}\cos(ax)f(x)$$ Thus, the general solution of $y''+2y+f=0$ is given by \eqalign{ y(x)&=c\cos(ax)+d\sin(ax)+\frac{\cos(ax)}{a}\int_0^x\sin(at)f(t)dt-\frac{\sin(ax)}{a}\int_0^x\cos(at)f(t)dt\\ &=c\cos(ax)+d\sin(ax)-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt} and $$y'(x)=-ac\sin(ax)+ad\cos(ax)- \int_0^x\cos(ax-at)f(t)dt$$ Now, the conditions $y(0)=y(2\pi)$ and $y'(0)=y'(2\pi)$ give us two equations:
\eqalign{c &=c\cos(2a\pi)+d\sin(2a\pi )-\frac{1}{a}\int_0^{2\pi}\sin(2a\pi-at)f(t)dt\cr ad&= -ac\sin(2a\pi)+ad\cos(2a\pi)- \int_0^{2\pi}\cos(2a\pi-at)f(t)dt} This can be arranged as follows \eqalign{c\sin(a\pi)-d\cos(a\pi) &=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi}\sin(2a\pi-at)f(t)dt\cr c\cos(a\pi)+d\sin(a\pi)&= - \frac{1}{2a\sin(a\pi)}\int_0^{2\pi}\cos(2a\pi-at)f(t)dt} Solving for $c$ and $d$ we obtain \eqalign{c&=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi}(\cos(2a\pi-at)\cos(a\pi)+\sin(a\pi)\sin(2a\pi-at))f(t)dt\cr &=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi} \cos(a\pi-at)f(t)dt\cr d&=-\frac{1}{2a\sin(a\pi)}\int_0^{2\pi}(\cos(2a\pi-at)\sin(a\pi)-\cos(a\pi)\sin(2a\pi-at))f(t)dt\cr &=\frac{1}{2a\sin(a\pi)}\int_0^{2\pi} \sin(a\pi-at)f(t)dt\cr } Finally \eqalign{ y(x)& =\frac{-1}{2a\sin(a\pi)}\int_0^{2\pi} \big(\cos(ax)\cos(a\pi-at)-\sin(ax)\sin(a\pi-at)\big)f(t)dt\\ &\phantom{=}-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt\\ &=\frac{-1}{2a\sin(a\pi)}\int_0^{2\pi}\cos(a\pi+ax-at)f(t)dt-\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt } This expression of $y$ can be simplified as follows: \eqalign{ y(x)& =\frac{-\cot(a\pi)}{2a }\int_0^{2\pi} \cos(ax-at)f(t)dt\cr &\phantom{=}+\frac{1}{2a }\int_0^{2\pi} \sin(ax-at)f(t)dt -\frac{1}{a}\int_0^x\sin(ax-at)f(t)dt\cr &=\frac{-\cot(a\pi)}{2a }\int_0^{2\pi} \cos(ax-at)f(t)dt+\frac{1}{2a }\int_x^{2\pi} \sin(ax-at)f(t)dt\cr &\phantom{=} -\frac{1}{2a }\int_0^{x} \sin(ax-at)f(t)dt \cr &=\frac{-\cot(a\pi)}{2a }\int_0^{2\pi} \cos(ax-at)f(t)dt-\frac{1}{2a }\int_0^{2\pi} \sin(a|x-t|)f(t)dt\cr &=\frac{-1}{2a\sin(a\pi)}\int_0^{2\pi} \cos(a\pi-a|x-t|)f(t)dt } Therefore, $$y(x)=\int_0^{2\pi}G(x,t)f(t)dt$$ with $$G(x,t)=\frac{-\cos(\sqrt{2}(\pi-|x-t|))}{2\sqrt{2}\sin(\sqrt{2}\pi)}.$$ which is the desired conclusion.$\qquad\square$.