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Convert $$y(x)=\lambda \int_{-1}^1 (1-|x-t|)y(t)dt$$ to differential equation with boundary conditions.

I know how to deal with this kind of problems but not when the kernel is a modulus function. I started solving it in the typical manner:

$$y'(x)=\lambda \int_{-1}^1 \frac{t-x}{|t-x|} y(t) dt.$$ It sort of rings the bell that $\frac{t-x}{|t-x|}$ could be $-1$ or $1$ depending on the values $x$ and $t$ on $[-1,1]$. Is there a way to determine those values? If I am right, it would be easy to solve this equation.

I also found another similar question (Transforming the integral equation $u(x) + \frac{\lambda}{2}\int_{0}^{1}|x - s|u(s)ds = ax + b$ into its equivalent differential equation) so I tried one of the methods in the answers:

$$\begin{align} \int_{-1}^1 (1-|x-t|) y(t) dt = \int_{-1}^{-x} (1-|x-t|) y(t) dt + \int_{-x}^0 (1-|x-t|) y(t) dt \\\\+ \int_{0}^x (1-|x-t|) y(t) dt + \int_{x}^1 (1-|x-t|) y(t) dt \end{align}$$

Then I'm really not sure what happens with the signs:

$$\begin{align} \int_{-1}^1 (1-|x-t|) y(t) dt = \int_{-1}^{-x} (1+x-t) y(t) dt + \int_{-x}^0 (1-x+t) y(t) dt \\\\+ \int_{0}^x (1-x+t) y(t) dt + \int_{x}^1 (1+x-t) y(t) dt \end{align}$$

Then by Leibnitz's Rule it should (?) follow that $$\frac{d^2}{dx^2} \int_{-1}^1 (1-|x-t|)u(t)dt=-4u(x)?$$ I feel like I just messed up with the signs...

Karagum
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1 Answers1

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If $x\le-1$, then $$y(x)=\lambda \int_{-1}^1 (1-|x-t|)y(t)dt=\lambda \int_{-1}^1 (1-(t-x))y(t)dt=\lambda \int_{-1}^1 (1-t+x)y(t)dt$$ and hence $$ y'(x)=\lambda \int_{-1}^1 y(t)dt. $$ If $|x|<1$, then $$y(x)=\lambda \int_{-1}^x (1-|x-t|)y(t)dt+\lambda \int_x^1 (1-(t-x))y(t)dt=\lambda \int_{-1}^x(1-x+t)y(t)dt+\lambda \int_x^1 (1-t+x)y(t)dt $$ and hence $$ y'(x)=\lambda y(x)-\lambda\int_{-1}^xy(t)dt-\lambda y(x)+\lambda\int_x^1y(t)dt=-\lambda\int_{-1}^xy(t)dt+\lambda\int_x^1y(t)dt. $$ So $$ y''(x)=-2\lambda y(x). $$ If $x\ge1$, then $$y(x)=\lambda \int_{-1}^1 (1-(x-t))y(t)dt=\lambda \int_{-1}^1 (1-x+t)y(t)dt$$ and hence $$ y'(x)=-\lambda \int_{-1}^1 y(t)dt. $$ Thus the equation is $$ y''=-2\lambda y $$ with boundary condition $$ y'(-1)=\lambda\int_{-1}^1y(t)dt, y'(1)=-\lambda\int_{-1}^1y(t)dt. $$

xpaul
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