Proving that $\int_0^l (\cos \beta_nx-\frac2{\beta_nl}\sin \beta_nx)(\cos \beta_mx-\frac2{\beta_ml}\sin \beta_mx)dx=0$

Question

We have known that $\beta_nl\sin (\beta_nl)=2(1-\cos (\beta_nl))$ and $sin (\beta_nl)=\frac12\beta_nl(1+\cos (\beta_nl))$.

Consider natural number m, n which satisfied above condition (m≠n)

Proved that $\int_0^l (\cos \beta_nx-\frac2{\beta_nl}\sin \beta_nx)(\cos \beta_mx-\frac2{\beta_ml}\sin \beta_mx)dx=0$

This is a very small part of the complex problem in my excercise, I have tried integrated directly and simplified the formula with above condition. However, I can only get the sum of the last three term was zero. In other words, I could not find a way to prove $\int_0^l \cos \beta_nx\cos \beta_mxdx=0$. Does somebody know how to prove the property?

Thanks a lot!!

EDIT: This is my work so far. A little complicated. $\delta_{nm}$ represents the integral after direct integration.

Answer 1

Preliminary commentary :

If your initial constraints are written under the form :

$$\begin{cases}a \sin a=2(1-\cos a)\\ \sin a=\frac12a(1+\cos a)\end{cases}$$

Using classical trigonometric formulas, this system boils down to

$$\begin{cases}\frac{a}{2}\cos \frac{a}{2} = \sin \frac{a}{2} \iff \tan b = b\\ \sin \frac{a}{2}=\frac{a}{2} \cos \frac{a}{2} \iff \tan b = b\end{cases}$$

with $b=\frac{a}{2}=\frac12\beta_n \ell$.

So, your two first equations contain the same information, i.e. that the

$$r_n:=\frac12\beta_n l$$

are the roots of equation $$\tan x = x$$ which has an infinite number of solutions (with $x_n\approx (2n+1)\pi/2$ for large $n$ : draw a graphical representation...).

which amounts to say that :

$$\begin{cases} \cos(\frac12\beta_n \ell)-\frac{2}{\beta_n \ell}\sin(\frac12\beta_n \ell)&=&0\\ \cos(\frac12\beta_m \ell)-\frac{2}{\beta_m \ell}\sin(\frac12\beta_m \ell)&=&0 \end{cases}$$

Let us now prove that :

$$I_{mn}:=\int_0^{\color{red}{\ell/2}} \underbrace{(\cos \beta_nx-\frac2{\beta_nl}\sin \beta_nx)}_{e_n(x)}\underbrace{(\cos \beta_mx-\frac2{\beta_ml}\sin \beta_mx)}_{e_m(x)}dx=0$$

[take care to the $\color{red}{\ell/2}$ !]

Please note that

$$e''_m(x)=-(\beta_m)^2e_m(x) \ \text{and} \ e''_n(x)=-(\beta_n)^2e_n(x)$$

A classical technique in such questions is double integration by parts formula :

$$\int_a^{b} u v'' dx = \int_a^{b} u'' v dx + \begin{cases}+ u(b) v'(b) \\- u(a) v'(a)\\ - u'(b) v(b) \\ + u'(a) v(a)\end{cases}$$

Let us take in the previous formula

$u(x)=e_n(x), \ \ v(x)=e_m(x), \ \ with \ \ a=0,b=\ell/2,$

we obtain the following expression (in which *** are for unimportant factors)

$$\int_0^{\ell/2} e_n e''_m dx = \int_0^{\ell/2} e''_n e_m dx +\begin{cases} + \underbrace{(\cos(\frac12\beta_n \ell)-\frac{2}{\beta_n \ell}\sin(\frac12\beta_n \ell))}_{= 0} \ \times \ (***) \\ - (-2/\ell) \\ + (***) \ \times \ \underbrace{(\cos(\frac12\beta_m \ell)-\frac{2}{\beta_m \ell}\sin(\frac12\beta_m \ell))}_{= 0}\\ -2/\ell \end{cases}$$

The previous identity can be written :

$$\underbrace{\int_0^{\ell/2} e_n (-\beta_m^2)e_m dx} = \int_0^{\ell/2} (-\beta_n^2)e_n e_m dx + 0$$

As $\beta_n \ne \beta_m$, this proves that

$$\int_0^{\ell/2} e_n e_m dx = 0$$

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Important remark : This issue can be turned, in the setting of Sturm-Liouville theory, into an eigenvalues/eigenfunctions problem where the $r_n$s (as defined above) are eigenvalues of a certain self-adjoint operator associated with a system of orthogonal eigenfunctions, the $e_n$s, which is a natural way to interprete your integral equal to zero.

("self-adjoint" qualificative refers naturaly to the property $\int u''v=\int uv''$ we have obtained above).