4

I want to determine the singular value decomposition of the integral operator $$L^2(0,1) \to L^2(0,1) ; f \mapsto Af(\cdot) = \int_0^\cdot f(y)dy.$$

Its adjungate is given by

$$A^*f(\cdot) = \int_\cdot^1 f(y) dy. $$

Hence $A^*Af(x) = \int_x^1 \int_0^y f(z)dzdy.$ Assuming that $f$ is smooth yields

$$\lambda f''(x) = A^*Af''(x) = \int_x^1\int_0^y f''(z)dzdy = -f(x) +f(1) - (1-x)f'(0).$$

Clearly, $\lambda_j = ((j-1/2)\pi)^{-2}$ and $v_j(x)= \sqrt{2}\cos((j-1/2)\pi x)$ are solutions of the above differential equation. Do more solutions exist? Since $A$ is injective the Eigenfunctions should span all of $L^2(0,1)$. We have $v_j(1)=0$ for all $j$, but this is merely a point and should not matter in $L^2(0,1)$. Moreover, the $v_j$ are even for some $j$ and odd for the other.

ViktorStein
  • 5,024
crankk
  • 1,447

1 Answers1

2

As I illustrate below, you can derive boundary conditions on your differential equation in the process, ensuring that the solutions you found are the only ones. Also, since the integration bounds of $A^* A$ involve $x$, so when you differentiate with respect to $x$, you have to take this into account:

In order for $\lambda \ne 0$ to be an eigenvalue of $T:= A^* A$, we have to have $(T x)(\xi) = \lambda x(\xi)$, i.e $$ \int_{\xi}^{1} \int_{0}^{t} x(s) d s dt = \lambda x(\xi) $$ This yields $x(1) = 0$. Differentiating with respect to $\xi$ yields $$ \lambda x'(\xi) = - \int_{0}^{\xi} x(s) ds, $$ implying $x'(0) = 0$ and $\lambda x''(\xi) = -x(\xi)$. We have reduced the eigenvalue problem to a second order ordinary differential equation in $x$: $$ (1) \begin{cases} x''(\xi) + \frac{1}{\lambda} \cdot x(\xi) = 0, \\ x(1) = 0, \\ x'(0) = 0. \end{cases} $$ The solution of $(1)$ is given by $$ x(\xi) = c_1 \cos(t \xi) + c_2 \sin(t \xi), \qquad \text{where } t := \frac{1}{\sqrt{\lambda}}. $$ Plugging in the second boundary condition into $x'(\xi) = c_2 t \cos(t \xi) - c_1 t \sin(t \xi)$ yields $x'(0) = c_2 t$, implying $c_2 = 0$. Plugging in the first boundary condition yields $$ x(1) = c_1 \cos(t) + c_2 \sin(t) = c_1 \cos(t) \overset{!}{=} 0 \implies \frac{1}{\sqrt{\lambda}} = t = \pi \left(n - \frac{1}{2}\right) $$ for all $n \in \mathbb{Z}$.
ViktorStein
  • 5,024