IEVS3

Formulate an integral equation from the BVP: $u'' + p(x)u' + q(x)u=f(x),\,\,x>a,\,\,u(a)=u_0,\,\,u'(a)=u_1$

Solve for $u''\,$ .

$u'' = -p(x)u' - q(x)u + f(x)\,$

Integrate both sides from $a$ to $x$ with respect to a new dummy variable $y$ .

$\int_a^x u''(y)\,dy = -\int_a^x p(y)u'(y)\,dy - \int_a^x \left[q(y)u(y)-f(y)\right]\,dy\,$

$u'(x) - u'(a) = -\left[ p(y)u(y)\big|_{y=a}^x - \int_a^x p'(y)u(y)dy \right] - \int_a^x \left[ q(y)u(y) - f(y)\right]\,dy \,$

$u'(x) = u_1 - \left[ p(x)u(x)-p(a)u(a)\right] + \int_a^x u(y)\left[ p'(y)-q(y)\right] + f(y)\,dy \,$

Integrate again. Notice how the limits of integration change.

$u(x) - u(a) = \int_a^x \left[u_1 + p(a)u_0 - p(y)u(y)\right]dy + \int_a^x \int_a^s u(y)\left[ (p'(y)-q(y))\right]+f(y)\,dy\,ds \,$

Fact:

$\int_a^x \int_a^s f(y)\,dy\,ds = \int_a^x f(y)(x-y)\,dy\,$

So the last result becomes

$u(x) = u_0 + \int_a^x u_1 + p(a)u_0 - p(y)u(y) + \left\{ u(y)\left[p'(y)-q(y)\right] + f(y)\right\}(y-x) dy \,$

$u(x) = u_0 + \int_a^x u(y)\left[- p(y)+p'(y)-q(y)\right](y-x)dy + \int_a^x u_1 + p(a)u_0 + f(y)(y-x)dy \,$

This equation is of the form $u(x) = \int_a^x k(x,y)u(y)\,dy + F(x)\,$ , which is a Volterra equation of the second kind.