Fredholm Integral Equation of the Second Kind in $L_2[0,1]$

Question

Given space $L_2[0,1]$ and the equation $$\displaystyle x(t) + \lambda \int_{0}^{1}(\frac{1}{2} - |t-s|)x(s) ds = \cos(\pi*t)$$

And I want to find a solution to the equation for all values $\lambda \in \mathbb{C}$

My attempt is:

1. Find first derivative: $\displaystyle x'(t) + \lambda \int_{0}^{t}x(s) ds -\lambda \int_{t}^{1}x(s) ds = -\pi*\sin(\pi*t)$
2. Second derivative: $\displaystyle x''(t) + 2\lambda x(t)= -\pi^2*\cos(\pi*t)$
3. And then I want to get solutions depending on values of $\lambda$. For that I need to find general and particular solutions. $$x(t) = \frac{C_1*e^{-2\lambda t}}{\lambda} + \frac{2\pi\lambda sin(\pi*t)}{4\lambda^2 +\pi^2} + \frac{\pi^2 cos(\pi*t)}{4\lambda^2 +\pi^2}$$

And I get stuck there since the results I get doesn't match with solution. Solution is

if $\lambda \neq - \frac{\pi^2}{2}(2n+1)^2$ where $n \in \mathbb{Z}$ then solution exists and is unique $x(t) = \frac{\pi^2}{\pi^2 + 2\lambda}cos(\pi t)$

if $\lambda = - \frac{\pi^2}{2}(2n+1)^2$ where $n \in \mathbb{Z}$, $n \neq 0$ then solution isn't unique $x(t) = ae^{\sqrt{2\lambda t}} + be^{-\sqrt{2\lambda t}} + \frac{\pi^2}{\pi^2 + 2\lambda}cos(\pi t)$

Can someone point out the errors in my reasoning. Thanks for any help!

Answer 1

Let's first consider the simplest case: if $\lambda=0$, the solution to the integral equation $$ x(t) + \lambda \int_{0}^{1}\left(\frac{1}{2} - |t-s|\right)x(s)\, ds = \cos(\pi t) \tag{1} $$ is obviously $x(t)=\cos(\pi t)$. Let's now suppose that $\lambda\neq 0$. Differentiating $(1)$ twice with respect to $t$, we obtain the differential equation$^{(*)}$ $$ x''(t)-2\lambda x(t)=-\pi^2\cos(\pi t). \tag{2} $$ Defining $k:=\sqrt{2\lambda}$, we can write the general solution to $(2)$ as$^{(\dagger)}$ $$ x(t)=C_1\cosh k\left(t-\frac{1}{2}\right)+C_2\sinh k\left(t-\frac{1}{2}\right) +\begin{cases} \frac{\pi^2}{\pi^2+k^2}\,\cos(\pi t)&\text{if $k^2\neq-\pi^2$}, \\ -\frac{\pi}{2}t\sin(\pi t)&\text{if $k^2=-\pi^2$}. \end{cases} \tag{3} $$ It's important to notice that not every solution to the differential equation $(2)$ is a solution to the integral equation $(1)$. For instance, a solution to the latter satisfies the following conditions at $t=0$ and $t=1$: \begin{align} &x(0) + \lambda \int_{0}^{1}\left(\frac{1}{2} - |0-s|\right)x(s)\, ds = \cos(0) \implies x(0)+\lambda \int_{0}^{1}\left(\frac{1}{2} - s\right)x(s)\, ds = 1, \tag{4} \\ &x(1) + \lambda \int_{0}^{1}\left(\frac{1}{2} - |1-s|\right)x(s)\, ds = \cos(\pi) \implies x(1)+\lambda \int_{0}^{1}\left(s-\frac{1}{2}\right)x(s)\, ds = -1. \tag{5} \end{align} Equations $(4)$ and $(5)$ imply that $x(1)=-x(0)$, which then implies that $C_1=0$.

Let's now plug the solution $(3)$ (with $C_1=0$) into Eq. $(5)$. For this, we shall need the following results: \begin{align} &\int_0^1\left(s-\frac{1}{2}\right)\sinh k\left(s-\frac{1}{2}\right)ds= \frac{1}{k}\,\cosh\left(\frac{k}{2}\right)-\frac{2}{k^2}\,\sinh\left(\frac{k} {2}\right), \tag{6} \\ &\int_0^1\left(s-\frac{1}{2}\right)\cos(\pi s)\,ds=-\frac{2}{\pi^2}, \tag{7} \\ &\int_0^1\left(s-\frac{1}{2}\right)s\sin(\pi s)\,ds=\frac{\pi^2-8}{2\pi^3}. \tag{8} \end{align} For $k^2\neq-\pi^2$ we obtain $$ C_2\sinh\left(\frac{k}{2}\right)-\frac{\pi^2}{\pi^2+k^2}+\frac{k^2}{2}\left[C_2 \left(\frac{1}{k}\,\cosh\left(\frac{k}{2}\right)-\frac{2}{k^2}\,\sinh\left(\frac{k} {2}\right)\right)-\frac{2}{\pi^2+k^2}\right]=-1 $$ $$ \implies C_2k\cosh\left(\frac{k}{2}\right)=0. \tag{9} $$ Eq. $(9)$ is satisfied for any $C_2$ if $\frac{k}{2}=\left(n+\frac{1}{2}\right)\pi i$, or $2\lambda=k^2=-(2n+1)^2\pi^2$ $(n\in\mathbb{N}_{>0})$, but $C_2=0$ otherwise.

On the other hand, if $k^2=-\pi^2$, then equations $(3)$ and $(5)$ yield $$ C_2\sinh\left(\frac{i\pi}{2}\right)-\frac{\pi^2}{2}\left[C_2 \left(\frac{1}{i\pi}\,\cosh\left(\frac{i\pi}{2}\right)+\frac{2}{\pi^2}\,\sinh\left(\frac{i\pi}{2}\right)\right)-\frac{\pi}{2}\left(\frac{\pi^2-8}{2\pi^3}\right)\right]=-1 \\ $$ $$ \implies 0\cdot C_2 +\frac{\pi^2}{8}=0, \tag{10} $$ which has no solution.

In summary: the solution to the integral equation $(1)$

• does not exist if $2\lambda=-\pi^2$;
• exists, but is not unique, if $2\lambda= -(2n+1)^2\pi^2$ $(n\in\mathbb{N}_{>0})$; it is given by $x(t)=C\sinh\left(\sqrt{2\lambda}\left(t-\frac{1}{2}\right)\right)+\frac{\pi^2}{\pi^2+2\lambda}\,\cos(\pi t)$, where $C$ is an arbitrary constant;
• exists and is unique otherwise; it is given by $x(t)=\frac{\pi^2}{\pi^2+2\lambda}\,\cos(\pi t)$.

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$^{(*)}$ Notice that \begin{align} \frac{d^2}{dt^2}\int_0^1\left(\frac{1}{2} - |t-s|\right)x(s)\, ds &= -\int_0^1\frac{\partial^2}{\partial t^2}|t-s|x(s)\, ds \\ &=-\int_0^1\frac{\partial}{\partial t}\text{sgn}(t-s)x(s)\,ds \\ &=-\int_0^1 2\delta(t-s)x(s)\,ds \\ &=-2x(t).\phantom{\int} \end{align}

$^{(\dagger)}$ The reason I wrote the argument of the first two terms of $(3)$ as $k\left(t-\frac{1}{2}\right)$, instead of the usual $kt$, will become clear soon.