For the <Volterra integration operator>
$$
(Af)(x)=\int_0^x f(t)\,dt,
$$
the <adjoint operator> is
$$
(A^*g)(x)=\int_x^1g(t)\,dt.
$$
Since $1/\sigma_n=(2n-1)\pi/2$, direct integration gives
$$
Au_n(x)
=\sqrt2\int_0^x\cos(t/\sigma_n)\,dt
=\sigma_n\sqrt2\sin(x/\sigma_n)
=\sigma_nv_n(x).
$$
Similarly,
$$
A^*v_n(x)
=\sqrt2\int_x^1\sin(t/\sigma_n)\,dt
=\sigma_n\sqrt2
\left(\cos(x/\sigma_n)-\cos(1/\sigma_n)\right)
=\sigma_nu_n(x),
$$
because $\cos((2n-1)\pi/2)=0$. The half-integer sine and cosine families are <orthonormal basis>[orthonormal bases] of $L^2([0,1])$, so $(\sigma_n,u_n,v_n)$ is a <singular system of a compact operator>.

The Tikhonov solution for noisy data $g^{(\delta)}$ is therefore
$$
\boxed{
f_\alpha(x)=
\sum_{n=1}^{\infty}
\frac{\sigma_n}{\sigma_n^2+\alpha}
\langle g^{(\delta)},v_n\rangle u_n(x)}.
$$
Writing out the <inner product> and the singular functions makes this explicit:
$$
\boxed{
f_\alpha(x)=2\sum_{n=1}^{\infty}
\frac{\sigma_n}{\sigma_n^2+\alpha}
\cos\left(\frac{x}{\sigma_n}\right)
\int_0^1g^{(\delta)}(t)
\sin\left(\frac{t}{\sigma_n}\right)\,dt,
\qquad
\sigma_n=\frac{2}{(2n-1)\pi}.}
$$
The factors $\sigma_n/(\sigma_n^2+\alpha)$ suppress the unstable reciprocal growth $1/\sigma_n$ at high index.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 3 / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-335/3/iii/solution