IEVS3

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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.

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