orthogonality of relevant integrals involving trigonometric functions

Question

I'm given a question for my Partial Differential Equations course and I see this step and I can't wrap my head around it. I've been drawing graphs for the past hour and I can't understand how to integrate it and how to see how the coefficients interact. So if someone could explain it to me I'd be very grateful.

My lecturer tried to help by saying "In problems involving series of trigonometric functions, you may assume orthogonality over [better: of] relevant intervals".

Answer 1

One way to see it is with the identity $$ 2\sin(A)\sin(B) = \cos(A-B) - \cos(A+B),$$ which itself follows from the angle addition formula $\cos(A\pm B) = \cos(A)\cos(B) \mp \sin(A)\sin(B)$.

If $n=m$, then we are integrating $\frac{1}{2} - \frac{1}{2}\cos\left(\frac{(2m+1)\pi}{a}x\right)$ over $[0,a]$. This is easily seen to be $a/2$. In a similar fashion, for $n≠m$ we see that we are integrating $\frac{1}{2}\cos\left(\frac{(m-n)\pi}{a}x\right) - \frac{1}{2}\cos\left(\frac{(m+n+1)\pi}{a}x\right)$, and both of these integrate to $0$.

Answer 2

The general case is handled quite nicely. Suppose you are looking for solutions $y$ of $$ y''(x)+\lambda y(x) = 0,\;\;\; a \le x \le b \\ Ay(a)+By'(a) = 0 \\ Cy(b)+Dy'(b) = 0. $$ Here $A, B$ are real constants that are not both zero; likewise $C, D$ are real constants that are not both $0$. This generality allows cases such as $y(a)=y(b)=0$ or $y'(a)=y'(b)=0$ or $y(a)=y'(b)=0$, and more general cases. There will always be a discrete set of $\lambda$ for which there are non-trivial solutions (i.e., not identically $0$ solutions,) and they can always be arranged as follows $$ \lambda_1 < \lambda_2 < \lambda_3 < \cdots < \lambda_n < \cdots $$ And $\lambda_n \rightarrow\infty$ as $n\rightarrow\infty$. Suppose $y_n,y_m$ are solutions corresponding to $\lambda_n,\lambda_m$ respectively with $\lambda_n\ne \lambda_m$. Then $-y_n''=\lambda_n y_n$ and $-y_m''=\lambda_m y_m$, and \begin{align} &(\lambda_n-\lambda_m)\int_{a}^{b}y_n(t)y_m(t)dt \\ &=\int_{a}^{b}-y_n''y_m+y_ny_m''dt \\ &=\int_{a}^{b}\frac{d}{dt}(-y_n'y_m+y_ny_m')dt \\ &= y_ny_m'-y_n'y_m|_{a}^{b} \\ &= \left.\left|\begin{matrix}y_n(t) & y_m(t) \\ y_n'(t) & y_m'(t)\end{matrix}\right|\right|_{t=a}^{b} = 0. \;\;\; \tag{$\dagger$} \end{align} The reason that the last terms are $0$ is that the last terms are determinants of matrices that are singular at $t=a$ and at $t=b$ because the endpoint conditions give non-trivial null space vectors: $$ \left[\begin{matrix} A & B \end{matrix}\right] \left[\begin{matrix}y_n(a) & y_m(a) \\ y_n'(a) & y_m'(a)\end{matrix}\right] = 0, \\ \left[\begin{matrix} C & D \end{matrix}\right] \left[\begin{matrix}y_n(b) & y_m(b) \\ y_n'(b) & y_m'(b)\end{matrix}\right] = 0. $$ Then, because $\lambda_n\ne \lambda_m$ was assumed, equation $(\dagger)$ forces $$ \int_{a}^{b}y_n(t)y_m(t)dt = 0,\;\;\; n \ne m. $$ Summary: So, regardless of the types of conditions you choose, you automatically end up with mutually orthogonal solutions on the interval $[a,b]$. You can have conditions such as $$ y(a)=y(b)=0, \\ y'(a)=y(b)=0, \\ y(a)=y'(b) = 0, \\ y(a)+y'(a)=0,\; y(b)+3y'(b)=0. $$ Regardless of the type of linear homogeneous endpoint conditions you choose with real coefficients, you'll always end up with mutually orthogonal trigonometric solutions. (And it should be noted that the the normalized solutions always form a complete orthonormal basis of $L^2[a,b]$.)