Compact operators and Fredholm's theory

Question

Can you help me to find all real numbers $\beta \in \mathbb {R}$ for which the equation $x(t) + \int_0^1(1+\alpha ts)x(s)ds = \beta + t^2$ is solvable in space $L_2[0,1]$ for any real number $\alpha \in \mathbb {R} $? I have no idea.

Answer 1

Though the given integral equation certainly "smells" like Fredholm, such concerns as that and compactness of operators don't seem to be needed to find solutions. Furthermore, the inter-relationship 'twixt $\alpha$ and $\beta$ is slightly different than proposed in the text of the problem. See below.

Given that

$x(t) + \displaystyle \int_0^1(1+\alpha ts)x(s) \; ds = \beta + t^2, \tag 1$

we may by linearity of integration write

$x(t) + \displaystyle \int_0^1 x(s) \; ds + \alpha t \int_0^1 sx(s) \; ds = \beta + t^2, \tag 2$

which indicates that $x(t)$ is a quadratic polynomial in $t$, that is,

$x(t) = \beta + t^2 - \displaystyle \int_0^1 x(s) \; ds - \alpha t \int_0^1 sx(s) \; ds; \tag 3$

thus, $x(t)$ takes the general form

$x(t) = at^2 + bt + c; \tag 4$

it is then easy to see that the integrals occurring in (2) must be

$\displaystyle \int_0^1 x(s) \; ds = \int_0^1 (as^2 + bs + c) \;ds = \dfrac{1}{3}a + \dfrac{1}{2} b + c, \tag 5$

and

$\displaystyle \int_0^1 sx(s) \; ds = \int_0^1 (as^3 + bs^2 + cs) \; ds = \dfrac{1}{4}a + \dfrac{1}{3}b + \dfrac{1}{2}c; \tag 6$

we combine these three equations into (2):

$at^2 + bt + c + \dfrac{1}{3}a + \dfrac{1}{2} b + c + \alpha t(\dfrac{1}{4}a + \dfrac{1}{3}b + \dfrac{1}{2}c) = \beta + t^2; \tag 7$

gathering and comparing coefficients of the different powers of $t$ we immediately find that

$a = 1, \tag 8$

$\dfrac{1}{3}a + \dfrac{1}{2} b + 2c = \beta, \tag 9$

$b + \alpha (\dfrac{1}{4}a + \dfrac{1}{3}b + \dfrac{1}{2}c) = 0; \tag{10}$

we may simplify via (8):

$ \dfrac{1}{2} b + 2c = \beta - \dfrac{1}{3}, \tag{11}$

$\dfrac{3 + \alpha}{3}b + \dfrac{\alpha}{2}c = -\dfrac{\alpha}{4}; \tag{12}$

these two equations form a linear system for $b$ and $c$ which we write as

$\begin{bmatrix} \dfrac{1}{2} & 2 \\ \dfrac{3 + \alpha}{3} & \dfrac{\alpha}{2} \end{bmatrix} \begin{pmatrix} b \\ c \end{pmatrix} = \begin{pmatrix} \beta - \dfrac{1}{3} \\ -\dfrac{\alpha}{4} \end{pmatrix}; \tag{13}$

the determinant of the matrix on the left is

$\dfrac{\alpha}{4} - \dfrac{2\alpha}{3} - 2 = -\dfrac{5\alpha}{12} - 2, \tag{14}$

which takes the value $0$ precisely when

$-\dfrac{5\alpha}{12} - 2 = 0 \Longleftrightarrow \alpha = -\dfrac{24}{5}; \tag{15}$

for all other $\alpha$, when

$\alpha \ne -\dfrac{24}{5}, \tag{16}$

there is a unique solution $(b, c)^T$ to (13) for any $\beta$ and hence a unique $x(t)$ of the form (4) satisfying (1). In the event that (15) binds, (13) reduces to

$\begin{bmatrix} \dfrac{1}{2} & 2 \\ -\dfrac{3}{5} & -\dfrac{12}{5} \end{bmatrix} \begin{pmatrix} b \\ c \end{pmatrix} = \begin{pmatrix} \beta - \dfrac{1}{3} \\ -\dfrac{6}{5} \end{pmatrix}; \tag{17}$

we see as expected then that the columns of the matrix are proportional:

$\begin{pmatrix} 2 \\ -\dfrac{12}{5} \end{pmatrix} = 4\begin{pmatrix} \dfrac{1}{2} \\ -\dfrac{3}{5} \end{pmatrix}, \tag{18}$

and since (17) with the aid of (18) may be written

$(b + 4c) \begin{pmatrix} \dfrac{1}{2} \\ -\dfrac{3}{5} \end{pmatrix} = b \begin{pmatrix} \dfrac{1}{2} \\ -\dfrac{3}{5} \end{pmatrix} + c\begin{pmatrix} 2 \\ -\dfrac{12}{5} \end{pmatrix} = \begin{pmatrix} \beta - \dfrac{1}{3} \\ -\dfrac{6}{5} \end{pmatrix}, \tag{19}$

we find upon comparing the second components that

$b + 4c = 2, \tag{20}$

whence

$\beta - \dfrac{1}{3} = 1 \Longleftrightarrow \beta = \dfrac{4}{3}, \tag{21}$

the only possible value of $\beta$ when $\alpha = -24 / 5$. In this case, since the columns of the coefficient matrix are linearly dependent, we cannot further determine $b$ and $c$ beyond (20); thus we may pick either one, say $c$, arbitrarily and then we have

$b = 2 - 4c, \tag{22}$

which will of course yield an infinite family of solutions

$x(t) = t^2 + (2 - 4c)t + c \tag{23}$

to our initial equation (1).

We summarize: under (16), we are free to choose any $\beta$ and obtain a unique $x(t)$; when $\alpha = -24 / 5$, there is a unique $\beta$ forced upon us, but we obtain an infinite one-parameter family of solutions parametrized by $b$ or $c$.

Answer 2

Hint: $\int_0^1 (1 + \alpha t s) x(s)\; ds$ is a polynomial in $t$.