Question 11.2.2: Determine the normalized eigenfunctions of the problem y′′ +......
Determine the normalized eigenfunctions of the problem
y^{\prime\prime}+\lambda y=0,\quad y(0)=0,\quad y^{\prime}(1)+y(1)=0. (25)
In Example 1 of Section 11.1, we found that the eigenvalue \lambda_{n} satisfies the equation
\sin{\sqrt{\lambda_{n}}}+{\sqrt{\lambda_{n}}}\cos{\sqrt{\lambda_{n}}}=0 (26)
and that the corresponding eigenfunction is
\phi_{n}(x)=k_{n}\sin\left({\sqrt{\lambda_{n}}}\,x\right), (27)
where k_{n} is arbitrary. We can determine k_{n} from the normalization condition (20).
\int_{0}^{1}r(x)\phi_{n}^{2}(x)d x=1,\quad n=1,2,\,\dots\,. (20)
Since r(x) = 1 in this problem, we have
\int_{0}^{1}r(x)\phi_{n}^{2}(x)d x=k_{n}^{2}\int_{0}^{1}\sin^{2}\Bigl(\sqrt{\lambda_{n}}x\Bigr)d x\\=\left.k_{n}^{2}\int_{0}^{1}\left({\frac{1}{2}}-{\frac{1}{2}}\cos\left(2\sqrt{\lambda_{n}}x\right)\right)d x=k_{n}^{2}\left({\frac{x}{2}}-{\frac{\sin(2\sqrt{\lambda_{n}}x)}{4\sqrt{\lambda_{n}}}}\right)\right|_{0}^{1}\\=k_{n}^{2}\left(\frac{1}{2}-\frac{\sin(2\sqrt{\lambda_{n}})}{4\sqrt{\lambda_{n}}}\right)=k_{n}^{2}\left(\frac{1}{2}-\frac{\sin\sqrt{\lambda_{n}}\cos\sqrt{\lambda_{n}}}{2\sqrt{\lambda_{n}}}\right)\\=k_{n}^{2}\,{\frac{1+\cos^{2}{\sqrt{\lambda_{n}}}}{2}},
where in the last step we have used equation (26). Hence, to normalize the eigenfunctions \phi_{n}, we must choose
k_{n}=\left(\frac{2}{1+\cos^{2}\sqrt{\lambda_{n}}}\right)^{1/2}. (28)
The normalized eigenfunctions of the given problem are
\phi_{n}(x)=\frac{\sqrt{2}\sin\sqrt{\lambda_{n}}x}{\left(1+\cos^{2}\sqrt{\lambda_{n}}\right)^{1/2}};\;\;\;n=1,2,\,\dots\,. (29)