# Role of the interval for defining inner product and boundary conditions in Sturm Liouville problems.

I have been learning about the Sturm Liouville method and am quite confused as to the role of boundary conditions and interval over which we take the inner product. In particular, it seems to me that the interval for which the Sturm Lioville operator is self adjoint and Hermitian depends on the operator and the functions on which it acts.

On the other hand we can choose the physical interval over which we wish to solve oir problem. It may or may noy coincide with the interval limits for the operator to be Hermitian.

When i have attempted questions, it seems that boundary conditions are given for the problem (which are applied to the eigenfunctions since I have only dealt with homogeneous BCs) and this interval is also then used for the inner product! Why? What guarantee is there that this interval allows the operator to be Hermitian or self-adjoint, and why does this matter anyway?

Sturm and Liouville formulated a general case of the ODE coming out of separation of variabales for the Heat equation and other equations involving the Laplacian in various orthogonal coordinate systems such as cylindrical and spherical coordinates. They found that separation of variables produced an eigenvalue/eigenfunction problem of the form $$\frac{d}{dx}\left(p(x)\frac{df}{dx}\right)-q(x)f(x)+\lambda w(x)f(x) = 0.$$ A regular problem is posed on a finite interval $[a,b]$ where $p,q,w$ are continuous on $[a,b]$ and where $p,w$ are non-vanishing and non-negative. Separated endpoint conditions are typically formulated as $$\cos\alpha f(a)+\sin\alpha f'(a) = 0 \\ \cos\beta f(b)+\sin\beta f'(b) = 0.$$ Angles $\alpha,\beta$ are introduced in order to normalized the equation and to ensure that the coefficients of $f$ and $f'$ are real and not both zero at $a,b$. Fourier had demonstrated "integral orthogonality" of several different cases, and Sturm, Liouville were able to show the same for the above case. Explicitly, if $f_1,f_2$ are solutions of the same problem with corresponding parameters $\lambda_1,\lambda_2$ where $\lambda_1\ne\lambda_2$, then it was found that one has the integral orthogonality condition $$\int_{a}^{b}f_1(x)f_2(x)w(x)dx = 0.$$ This is important because Fourier needed to expand a function in a sum of the following type in order to solve his PDEs: $$f = \sum_{n=1}^{\infty}\alpha_n f_n$$ And this he did using this new notion of orthogonality \begin{align} \int_a^b f(x)f_m(x)w(x)dx & = \sum_{n=1}^{\infty}\alpha_n\int_a^b f_n(x)f_m(x)w(x)dx \\ & = \alpha_m \int_{a}^{b}f_m(x)^2w(x) dx \end{align} This gave an explicit formula for unknown constants $\alpha_m$ in the form $$\alpha_m = \frac{\int_{a}^{b}f(x)f_m(x)w(x)dx}{\int_{a}^{b}f_m(x)^2w(x)dx}$$ Fourier's problem, in modern terminology, is formulated on an interval, with endpoint conditions, as an eigenvalue problem. The eigenfunctions are mutually orthogonal with respect to this integral expression on $[a,b]$, giving rise to orthogonal function expansions with respect to this weighted inner product. The symmetric form of the Sturm-Liouville ODE is critical to orthogonality. The inner product is defined on the same interval in order to obtain the correct orthogonality conditions of Fourier, and the endpoint conditions must be chosen in order to have a symmetric problem, which results in orthogonality.