For the unknown y : [0, 1] → ℝ, consider the following two-point boundary value problem:

\(\rm \left\{\begin{aligned}\rm y^{\prime \prime}(x)+2 y(x) & =0 \quad \text { for } \rm x ∈(0,1), \\ \rm y(0) & =\rm y(1)=0 .\end{aligned}\right.\).

It is given that the above boundary value problem corresponds to the following integral equation:

y(x) = 2\(\displaystyle\int_0^1\) K(x, t) y(t) dt for x ∈ [0, 1].

Which of the following is the kernel K(x, t)?

Concept:

Leibnitz rule for differentiation:

\(\frac{\partial }{\partial x}\int_{a(x)}^{b(x)}f(x,t)dt\) = \(\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}dt+f(b(x),x)\frac{\partial b}{\partial x}-f(a(x),x)\frac{\partial a}{\partial x}\) 

Explanation:

(1): 

K(x, t) = \(\begin{cases} \rm t(1-x) & \text { for } \rm t<x \\ \rm x(1-t) & \text { for } \rm t>x\end{cases}\) 

So 

y(x) = 2\(\displaystyle\int_0^1\) K(x, t) y(t) dt

⇒ y(x) = 2 \(\displaystyle\int_0^x\)t(1 - x)ydt + 2 \(\displaystyle\int_x^1\)x(1 - t)ydt....(i)

⇒ y' = 2\(\displaystyle\int_0^x\)(-t)y(t)dt + x(1 - x)y(x) - 0 + 2\(\displaystyle\int_x^1\)1(1-t)y(t)dt + 0 -x(1 - x)y(x).1

⇒ y' = 2\(\displaystyle\int_0^x\)(-t)y(t)dt + 2\(\displaystyle\int_x^1\)(1 - t)y(t)dt

⇒ y'' = 2[0 - xy(x).1 - 0] + 2[0 + 0 -(1 - x)y(x)] 

⇒ y'' = -2y(x)

⇒ y''(x) + 2y(x) = 0

 Also by (i)

y(0) = 2 \(\displaystyle\int_0^0\)t(1 - 0)ydt + 2 \(\displaystyle\int_0^1\)0(1 - t)ydt = 0 and  

y(1) = 2 \(\displaystyle\int_0^1\)t(1 - 1)ydt + 2 \(\displaystyle\int_1^1\)1(1 - t)ydt = 0

hence \(\rm \left\{\begin{aligned}\rm y^{\prime \prime}(x)+2 y(x) & =0 \quad \text { for } \rm x ∈(0,1), \\ \rm y(0) & =\rm y(1)=0 .\end{aligned}\right.\) satisfies

Therefore option (1) is correct