# Compact operators and Fredholm's theory

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.

• For which the equation does what? Has a solution? – Robert Israel Feb 21 at 4:20
• @RobertIsrael yes, has a solution. – Gera Slanova Feb 21 at 4:29

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 $$c$$.

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