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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?

user242028
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  • I provided 2 distinct approaches. Please let me know how I can improve my answer. I just want to give you the best answer I can give you. – Mark Viola May 21 '15 at 18:25
  • Your answer, I got it properly. But I am not able to find mistake in my approach @Dr.MV – user242028 May 21 '15 at 18:47
  • I want to know that whether $\frac{du}{dx} + \frac{\lambda}{2}\int_{0}^{1}\frac{x - s}{|x - s|}u(s)ds = a$ is correct. Secondly If I do like as follows: $\frac{d}{dx} \int_{0}^{1}\frac{x - s}{|x - s|}u(s) = \int_{0}^{1}\frac{\del}{\del x} \frac{x - s}{|x - s|}u(s) = \int_{0}^{1}0. u(s) = 1$ then where does problem occur? – user242028 May 21 '15 at 18:53
  • The problem is that $frac{d}{dx}|x|$ doesn't exist at $x=0$. And $frac{d}{dx}\sgn(x)$ doesn't exist at $x=0$. So, classically speaking, the interchange of operations is not permitted. – Mark Viola May 21 '15 at 19:11

2 Answers2

2

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,

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

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.

Mark Viola
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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 $$ \frac{d^2u}{dx^2} + \lambda u = 0$$

Dylan
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  • This is the non-classical approach and relies on an understanding of the Generalized Function Theory. – Mark Viola May 21 '15 at 18:05
  • I'm just using what I know, since I'm familiar with the delta function. But I like your answer too. – Dylan May 21 '15 at 18:06
  • Sure, both are legitimate. I provided both approaches. But the classical approach is likely the one that others would have seen. – Mark Viola May 21 '15 at 18:16