91Ó°ÊÓ

We are given the Sturm-Liouville problem: \[\mathrm{y''}+\lambda\mathrm{y}=0 \quad \mathrm{y}'(0)=0, \quad \mathrm{y}(\pi)=0.\]

To solve the homogeneous equation, we start by solving the characteristic equation: \[r^2+\lambda=0.\] The roots are: \[r_{1,2}=\pm\sqrt{-\lambda}.\] The general solution of the homogeneous equation is given by: \[\mathrm{y(x)}=c_1\cos(\sqrt{-\lambda}x) + c_2\sin(\sqrt{-\lambda}x).\]

Now let's apply the boundary conditions to the general solution. First, we have: \[\mathrm{y}'(0)=0.\] Taking the derivative of the general solution and plugging in \(x=0\), we get: \[\mathrm{y'(0)}=-c_1\sqrt{-\lambda}\sin(\sqrt{-\lambda}0)+c_2\sqrt{-\lambda}\cos(\sqrt{-\lambda}0)=c_2\sqrt{-\lambda}=0.\] Since \(\sqrt{-\lambda}\) cannot be zero, we conclude that \(c_2=0\). Therefore, our solution simplifies to: \[\mathrm{y(x)}=c_1\cos(\sqrt{-\lambda}x).\] The second boundary condition is: \[\mathrm{y(\pi)}=0.\] Substituting this condition and simplifying, we obtain: \[c_1\cos(\sqrt{-\lambda}\pi)=0.\] This implies that \(\cos(\sqrt{-\lambda}\pi)=0\). Thus, the eigenvalues \(\lambda\) must satisfy: \[\sqrt{-\lambda}\pi = (2n-1)\frac{\pi}{2}, \quad n\in\mathbb{N}.\] This gives us the eigenvalues: \[\lambda_n=-(2n-1)^2\frac{1}{4}.\]

For each eigenvalue \(\lambda_n\), we have a corresponding eigenfunction: \[y_n(x)=C_n\cos\left((2n-1)\frac{\pi}{2}x\right), \quad n\in\mathbb{N}.\] Since there is no linear combination of these eigenfunctions that results in the zero function, they are linearly independent.

Finally, to prove that the set of eigenfunctions is orthogonal, we need to show that the inner product between any two distinct eigenfunctions is zero. The inner product of two functions \(f(x)\) and \(g(x)\) is defined by: \[\langle f,g\rangle = \int_{0}^{\pi} f(x)g(x)\mathrm{dx}.\] Now, let's take the inner product of two eigenfunctions \(y_m(x)\) and \(y_n(x)\) with \(m \neq n\): \[\langle y_m, y_n\rangle = \int_{0}^{\pi} C_mC_n\cos\left((2m-1)\frac{\pi}{2}x\right)\cos\left((2n-1)\frac{\pi}{2}x\right) \mathrm{dx}.\] Using the trigonometric identity \(\cos A\cos B=\frac{1}{2}[\cos(A+B)+\cos(A-B)]\), we obtain: \[\begin{aligned} \langle y_m, y_n\rangle &= C_mC_n\frac{1}{2}\int_{0}^{\pi} \left[\cos\left(\left((2m-1)+(2n-1)\right)\frac{\pi}{2}x\right)+\cos\left(\left((2m-1)-(2n-1)\right)\frac{\pi}{2}x\right)\right] \mathrm{dx}. \end{aligned}\] We know that the integral of \(\cos(kx)\) over the interval \([0,\pi]\) is zero for nonzero integer values of \(k\). Both \((2m-1)+(2n-1)\) and \((2m-1)-(2n-1)\) are nonzero integers, so the value of the above integral is zero. This shows that the set of eigenfunctions is orthogonal.