Calculating SVD for some integral operators I am currently trying to analytically find SVD for operators $T_1, T_2: L^2\big([0;1]\big) \to L^2\big([0;1]\big)$ using definition, where:

*

*$(T_1f)(x) = \int_0^x f(y) \ dy$


*$(T_2f)(x) = \int_0^1 K(x,y) f(y) \ dy$, where $K(x,y) = \begin{cases} x(y-1), x < y \\ y(x-1), x \ge y \end{cases} $
These are compact operators. We can easily find their adjoint:

*

*$(T_1^\dagger f)(x) = \int_x^1 f(y) \ dy$


*$(T_2^\dagger f)(x) = \int_0^1 K(y,x) f(y) \ dy$
Now I have to obtain eigenvalues and eigenvectors of $T_1^\dagger T_1$ and $T_2^\dagger T_2$.
In first case we have
$\int_t^1 \int_0^x f(y) \ dy \ dx= \lambda f(t) $
$\int_1^t\int_0^x f(y) \ dy \ dx = -\lambda f(t) $.
Now can I take $f \in C\big([0;1] \big)$. Then I have $F(t) = \int_1^t\int_0^x f(y) \ dy \ dx$ and $F \in C^2\big([0;1] \big).$ I can rewrite my problem as
$ \begin{cases} F''(t) + \frac{1}{\lambda}F(t) = 0 \\ F(1) = F'(0) = 0 \end{cases}$. But am I able to obtain solution for $f \in L^2\big([0;1] \big)$ (using e.g. density $ C\big([0;1] \big)$ in $L^2\big([0;1] \big)$)?
Second equation is
$\int_0^1 K(x,t) \int_0^1 K(x,y) f(y) \ dy \ dx= \lambda f(t)$.
When I write out integral on left side of equation I will obtain
\begin{equation*}\int_0^1 K(x,t) \Bigg( \int_x^1 x(y-1) f(y) \ dy + \int_0^x y(x-1) f(y) \ dy \Bigg) \ dx = \lambda f(t)\end{equation*}
\begin{equation*}\int_0^1 K(x,t) \Bigg( x \int_0^1 y f(y) \ dy - x \int_x^1 f(y) \ dy - \int_0^x y f(y) \ dy \Bigg) \ dx = \lambda f(t)\end{equation*}
Now outer integral can be rewritten in the same way. What I get is equation with nine double integrals on the left.  Here to be fair I am not sure how reduce problem to differential form or maybe how to write out those integrals to obtain  simpler problem.
I would be grateful for any hints with this two operators.
 A: The first operator is not self-adjoint, so it's left and right singular vectors are different. To find these, solve the eigenvalue problems for $T_1T_1^{\dagger},T_1^{\dagger}T_1$. You can write those down explicitly as done above:
$$T_1T_1^{\dagger}: ~~ \sigma^2 f''(t)+f(t)=0~,~ ~~f(0)=f'(1)=0$$
$$T_1^{\dagger}T_1: ~~ \sigma^2 g''(t)+g(t)=0~,~ ~~f'(0)=f(1)=0$$
These can be easily solved to yield the discrete sets
$$(\sigma_n, f_n(t), g_n(t))=\left(\frac{1}{\pi(n+1/2) }~,~ \sin(n+1/2)\pi t~,~ \cos(n+1/2)\pi t\right)~,~ n\in \mathbb{N}^+$$
The other operator is a bit trickier but you can make your life easier by noticing that $T_2^\dagger=T_2$. In this case, the left and right singular value vectors are the same, and the set of singular values can be found by solving the eigenvalue problem $T_2 f=\lambda f$. The operator $T_2$ can be rewritten as
$$T_2 f=x\int_{0}^1yf(y)dy-\int_{0}^x dy\int_{y}^1dtf(t)$$
Taking two derivatives as we did before we discover that
$$T_2 f=\lambda f \Rightarrow\lambda f''(t)-f(t)=0~~,~~f(0)=f(1)=0$$
We get
$$(\lambda_n, f_n)=\left(-\frac{1}{n^2\pi^2}, \sin n\pi t\right)$$
