1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ For which the equation does what? Has a solution? $\endgroup$ – Robert Israel Feb 21 at 4:20
  • 1
    $\begingroup$ @RobertIsrael yes, has a solution. $\endgroup$ – Gera Slanova Feb 21 at 4:29
0
$\begingroup$

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 $c$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.