IE13

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Transform the BVP to an integral equation:\frac{d^2y}{dx^2} + y = x, y(0) = 0, y'(1) = 0


First, integrate both sides from x to 1 (so that the boundary condition y'(1) = 0 can be applied). This gives
y'(1) - y'(x) + \int_x^1 y(u)\,du = \int_x^1 u\,du = \frac{u^2}{2}\Big|_x^1 = \frac{1}{2} - \frac{x^2}{2}

Applying the condition y'(1) = 0 and multiplying both sides by (-1) results in

y'(x) - \int_x^1 y(u)\,du = \frac{x^2}{2} - \frac{1}{2}

Next, integrate both sides from 0 to x:

\big(y(x) - y(0)\big) - \int_0^x \int_z^1 y(u)\,du\,dz = \int_0^x \frac{u^2}{2} - \frac{1}{2}\,du = \left(\frac{u^3}{6} - \frac{u}{2}\right)\Big|_0^x = \frac{x^3}{6} - \frac{x}{2}

or, applying the condition y(0) = 0,

y(x) - \int_0^x \int_z^1 y(u)\,du\,dz = \frac{x^3}{6} - \frac{x}{2}

The double integral may be simplified if we invert the order of integration. To do this, consider the region of integration for the double integral:
Image:Integ reg1.gif

This same region, when the order of integration is reversed, looks like this:
Image:Integ reg2.gif


We see that the region must be split into two pieces, and a separate integral written for each piece:

\int_0^x \int_z^1 y(u)\,du\,dz = \int_0^x \int_0^u dz\,\, y(u)\,du + \int_x^1 \int_0^x dz\,\, y(u)\,du

or after doing the trivial z-integration,

\int_0^x \int_z^1 y(u)\,du\,dz = \int_0^x u\,y(u)\,du + \int_x^1 x\,y(u)\,du

The desired integral equation, then, is

y(x) - \int_0^x u\,y(u)\,du - \int_x^1 x\,y(u)\,du = \frac{x^3}{6} - \frac{x}{2}

Such an integral equation is often written in the form

y(x) - \int_0^1 k(x,u)\, y(u)\,du = \frac{x^3}{6} - \frac{x}{2}

where

k(x,u) = \begin{cases} u,\, 0 < u < x \\ x,\, x < u < 1 \end{cases}

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