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Convert the differential equation $$y''+2y'+(\lambda+1)y=0$$ to Sturm-Liouville form, and obtain the solutions satisfying the boundary conditions $$y(0)=y(\pi)=0.$$

Using the integrating factor $\mu(x)=e^{2x}$, the Sturm-Liouville form of the ODE is $$(e^{2x}y')'+(e^{2x}+\lambda e^{2x})y=0.$$ For the solutions satisfying the boundary conditions, I considered three cases:

If $\lambda=0$, then $y''+2y'+y=0$. This has the general solution $y(x)=Ae^{-x}+Bxe^{-x}$ and upon applying the boundary conditions, leads to a trivial solution.

If $\lambda=w^2<0$, then $y''+2y'+(-w^2+1)y=0$. This has the general solution $y(x)=Ce^{(-1+w)x}+De^{(-1-w)x}$ and upon applying the boundary conditions, leads to a trivial solution.

If $\lambda=w^2>0$, then $y''+2y'+(w^2+1)y=0$ This has the general solution $y(x)=e^{-x}\left(E\cos(wx)+F\sin(wx)\right)$ and upon applying the boundary conditions, a non-trivial solution occurs when $w=n, \ n\in\mathbb{Z^+}$.

Does this mean that the eigenfunctions of this ODE are $\phi_n(x)=e^{-x}\sin(nx)$ with corresponding eigenvalues $\lambda_n=n^2$? I don't fully understand the terms eigenfunctions and eigenvalues.

Furthermore, I wish to show that the eigenfunctions corresponding to distinct eigenvalues are orthogonal: \begin{align} \int_0^\pi w(x)y_n(x)y_m(x) \ dx&=\int_0^\pi e^{2x}(e^{-2x}\sin(xn)\sin(xm)) \ dx \\ &=\frac{1}{2}\left(\frac{sin(\pi(n-m))}{n-m}-\frac{\sin(\pi(n+m))}{n+m}\right) \end{align}

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  • $\begingroup$ Yes, you are correct $\endgroup$ – Dylan Mar 12 at 8:37
  • $\begingroup$ Thank you. But how come when I try to show by direct integration that eigenfunctions corresponding to different eigenvalues are orthogonal? $$\text{e.g.} \ \int_{0}^{\pi} \phi_1(x)\phi_2(x)=0.$$ Clearly this does not work. $\endgroup$ – Stuart-James Burney Mar 12 at 8:58
  • $\begingroup$ They are orthogonal, but it's not straightforward. There's a weighting factor you'll need to find $\endgroup$ – Dylan Mar 12 at 8:59
  • $\begingroup$ Ah, I forgot that. We can use the ODE in Sturm-Liouville form to see that the weighting function is $e^{2x}$. Thus the resulting integral reduces to zero. $\endgroup$ – Stuart-James Burney Mar 12 at 9:00
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Eigenfunctions $y_n=e^{-x}\sin(nx)$ and eigenvalues $\lambda_n=n^2$ is solutions of eigenvalue problem $$Ly=\lambda y, \quad y(0)=y(\pi)=0,$$ here $$Ly=-(y''+2y'+y).$$

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