91Ó°ÊÓ

For the given Sturm-Liouville problem, the differential equation and boundary conditions are: $$ y^{\prime \prime}+\lambda y =0, \\ y(0) =0, \quad y(\pi)+y^{\prime}(z)=0 . $$ Eigenfunctions are solutions of the differential equations with specific eigenvalues \(λ\). The general solution to this differential equation can be written as: $$ y(x) = A\sin(\sqrt{\lambda}x) + B\cos(\sqrt{\lambda}x) $$ Applying boundary conditions: 1. \(y(0) = 0\): \(B\cos(\sqrt{\lambda}(0)) = 0\) which implies \(B=0\). 2. \(y(\pi) + y^{\prime}(z) = 0\) Let's apply the second boundary condition on the simplified general solution.

The simplified general solution is: $$ y(x) = A\sin(\sqrt{\lambda}x) $$ And its derivative is: $$ y^{\prime}(x) = A\sqrt{\lambda}\cos(\sqrt{\lambda}x) $$ Now, apply the boundary condition \(y(\pi) + y^{\prime}(z) = 0\): $$ A\sin(\sqrt{\lambda}\pi) + A\sqrt{\lambda}\cos(z) = 0 $$ We know \(A\) cannot be zero, so we divide by \(A\): $$ \sin(\sqrt{\lambda}\pi) + \sqrt{\lambda}\cos(z) = 0 $$ From this equation, we can determine the eigenvalues and eigenfunctions.

Rewrite the equation from the previous step: $$ \sin(\sqrt{\lambda}\pi) = -\sqrt{\lambda}\cos(z) $$ We notice that the left side is a sine function and the right side is a negative cosine function multiplied by an unknown constant. To satisfy the equality, eigenvalues can be found when \(\cos(z) = 0\), which occurs when \(\lambda_n = \left(n-\frac{1}{2}\right)^2\) for all integer values of \(n\). For each eigenvalue, the corresponding eigenfunction is: $$ y_n(x) = A_n\sin\left(\sqrt{\lambda_n}x\right) = A_n\sin\left(\left(n-\frac{1}{2}\right)x\right) $$ Now that we have the eigenfunctions, the next step is finding the coefficients \(A_n\).

The expansion of the function \(f(x) = \sin x\) in terms of eigenfunctions is given by: $$ f(x) = \sum_{n=1}^{\infty} A_n\sin\left(\left(n-\frac{1}{2}\right)x\right) $$ To find \(A_n\), multiply both sides by \(\sin\left(\left(m-\frac{1}{2}\right)x\right)\), integrate from \(0\) to \(m\), and use orthogonal properties: $$ \int_0^m \sin x \sin\left(\left(m-\frac{1}{2}\right)x\right) dx = \int_0^m \sum_{n=1}^{\infty} A_n\sin\left(\left(n-\frac{1}{2}\right)x\right) \sin\left(\left(m-\frac{1}{2}\right)x\right) dx $$ Due to orthogonal properties, the right side of the equation simplifies to: $$ \frac{A_m \pi}{2} = \int_0^m \sin x \sin\left(\left(m-\frac{1}{2}\right)x\right) dx $$ Solve for \(A_m\): $$ A_m = \frac{2}{\pi}\int_0^m \sin x \sin\left(\left(m-\frac{1}{2}\right)x\right) dx $$ Now, we have everything we need to express \(f(x)=\sin x\) as a series of eigenfunctions.

Using the coefficients \(A_m\) and eigenfunctions \(y_n(x)\), the given function \(f(x) = \sin x\) can be written as: $$ \sin x = \sum_{n=1}^{\infty} \frac{2}{\pi}\left(\int_0^m \sin x \sin\left(\left(n-\frac{1}{2}\right)x\right) dx\right)\sin\left(\left(n-\frac{1}{2}\right)x\right) $$ This is the final expanded form of the function \(f(x) = \sin x\).