# Why boundary conditions in Sturm-Liouville problem are homogeneous?

Boundary conditions in Sturm-Liouville problem looks like this: $$\alpha_1 y(a)+\alpha_2 y'(a)=0$$ $$\beta_1y(b)+\beta_2 y'(b)=0$$ The ordinary boundary conditions for boundary-value problem looks: $$\alpha_1 y(a)+\alpha_2 y'(a)=\gamma_1$$ $$\beta_1y(b)+\beta_2 y'(b)=\gamma_2$$ Why in Sturm-Liouville conditions $\gamma_1$ and $\gamma_2$ both zeros? Does it have some hidden (or physical) sense?

## 2 Answers

If you want a solution with $\gamma_1\ne 0$ and/or $\gamma_2\ne 0$, then you can subtract a function from your solution that satisfies the non-zero endpoint conditions, and you have effectively converted the problem to an inhomogeneous problem with homogeneous endpoint conditions, which can be solved using separation of variables.

• Thanks a lot for answer, but I am mostly interested in the origin of zero conditions, in fact why in Sturm-Liouville problem we have $\gamma_1 = \gamma_2 = 0$ – danielleontiev Jan 15 '18 at 22:57
• @danielleontiev : Everything started with Fourier. He invented separation of variables, and used it to solve heat problems. He also created the Heat Equation. Homogeneous conditions of the type you have stated would be applied to the end of a wire where the radiation of heat would be proportional to the temperature. A non-homogeneous term would represent a source of some type. – DisintegratingByParts Jan 15 '18 at 23:12

Methods of linear algebra may be brought to bear on vector spaces. The space of solutions satisfying homogeneous boundary conditions is a vector space, whereas the space of solutions satisfying inhomogeneous conditions is only an affine space. Moreover, there is an easy way to translate the homogeneous vector space into the affine space: by translation!