Transforming the integral equation $u(x) + \frac{\lambda}{2}\int_{0}^{1}|x - s|u(s)ds = ax + b$ into its equivalent differential equation

Question

Let $u \in C^2[0, 1]$ satisfy for some $ \lambda \neq 0$ and $a \neq 0,$ $$u(x) + \frac{\lambda}{2}\int_{0}^{1}|x - s|u(s)ds = ax + b.$$ Then show that u also satisfies $\frac{d^2u}{dx^2} + \lambda u = 0$

I differentiate the integral equation twice w.r.t $x$. On first differentiation, I get $$\frac{du}{dx} + \frac{\lambda}{2}\int_{0}^{1}\frac{x - s}{|x - s|}u(s)ds = a$$

On second differentiation w.r.t., $x$, I get $$\frac{d^2u}{dx^2} + \frac{\lambda}{2} = 0.$$

But how to do the prblem?

Answer 1

Let's split the integral into

$$\begin{align} \int_0^1 |x-s|u(s)ds&=\int_0^x |x-s|u(s)ds+\int_x^1 |x-s|u(s)ds\\\\ &=\int_0^x (x-s)u(s) ds-\int_x^1 (x-s)u(s) ds \end{align}$$

Now, differentiate using Leibnitz's Rule gives

$$\begin{align} \frac{d}{dx}\int_0^1 |x-s|u(s)ds&=\left(\int_0^x \frac{d(x-s)}{dx}\,u(s)\, ds\,+(x-x)u(x)\right)\\\\ &+\left(-\int_x^1 \frac{d(x-s)}{dx}\,u(s)\, ds\,-(-1)(x-x)u(x)\right)\\\\ &=\int_0^x u(s) ds-\int_x^1 u(s) ds \end{align}$$

A second application of Leibnitz's Rule (or here, simply the Fundamental Theorem of Calculus) yields

$$\begin{align} \frac{d^2}{dx^2}\int_0^1 |x-s|u(s)ds&=2u(x) \end{align}$$

Thus,

$$u''+\lambda u=0.$$

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NOTE:

A second way to carry out the analysis relies on use of the theory of generalized functions, and NOT classical analysis. In this context,

$$\frac{d^2|x-s|}{dx^2}=2\delta(x-s)$$

where $\delta$ is the Dirac distribution (aka, the Dirac delta "function"). Then,

$$\begin{align} \frac{d^2}{dx^2}\int_0^1 |x-s|u(s)ds&=\int_0^1\,\frac{d^2|x-s|}{dx^2}u(s)ds\\\\ &=\int_0^1\,2\delta(x-s)\,u(s)ds\\\\ &=2u(x) \end{align}$$

as expected.

Answer 2

Using the identity

$$ \frac{d}{dx} \frac{x-s}{|x-s|} = 2\delta(x - s)$$

Differentiating a second time should give $$ \frac{\lambda}{2}\int_0^1 2\delta(x-s)u(s)\,ds = \lambda u(x) $$

Thus the equation is $$ \frac{d^2u}{dx^2} + \lambda u = 0$$