# 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?

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