Writing y′′+P(x)y′+Q(x)y=g(x) as a Fredholm Integral Equation

[bolbteppa]

Is it possible to convert a general linear second order boundary value ode

y'' + P(x)y' + Q(x)y = g(x), y(a) = y_a, y(b) = y_b

to a Fredholm integral equation, explicitly determining the Kernel in the process, without removing the y' term? (Here is an example of doing it without the y' term) I seem to be getting stuck.

My answer (computed below if necessary) is

y(x) = y_a+[\frac{y_b-y_a}{b-a}+\frac{1}{b-a}\int_a^bp(t)y(t)dt + \frac{1}{b-a}\int_a^b[q(t)-p'(t)](b-t)y(t)dt - \frac{1}{b-a}\int_a^bg(t)(b-t)dt](x-a)-\int_a^xp(t)y(t)dt - \int_a^x[q(t)-p'(t)](x-t)y(t)dt+\int_a^xg(t)(x-t)dt

How do we write this in terms of an integral kernel, and can we call our result a Green function?

Note: Here is an example from Arfken of how it is done for the simple case of y''+\omega^2 y = 0, y(0)=0, y(b)=0:

- Computation:

Integrating

y'' = - P(x)y' - Q(x)y + g(x)

gives us

y&#039;(x) = y&#039;_a -<br /> \int_a^xp(t)y&#039;(t)dt-\int_a^xq(t)y(t)dt+\int_a^xg(t)dt

which, on getting rid of the $y'$ term by I.B.P.,

y&#039;(x) = y&#039;_a +y_ap(a)-p(x)y(x) +<br /> \int_a^xp&#039;(t)y(t)dt-\int_a^xq(t)y(t)dt+\int_a^xg(t)dt

gives us

y&#039;(x) = y&#039;_a +y_ap(a)-p(x)y(x) -<br /> \int_a^x[q(t)-p&#039;(t)]y(t)dt+\int_a^xg(t)dt

Integrating to find $y$ gives

y(x) = y_a+[y&#039;_a +y_ap(a)](x-a)-\int_a^xp(t)y(t)dt -<br /> \int_a^x\int_a^u[q(t)-p&#039;(t)]y(t)dtdu+\int_a^x\int_a^ug(t)dtdu

or


y(x) = y_a+[y&#039;_a +y_ap(a)](x-a)-\int_a^xp(t)y(t)dt -<br /> \int_a^x[q(t)-p&#039;(t)](x-t)y(t)dt+\int_a^xg(t)(x-t)dt

or


y(x) = y_a+[y&#039;_a +y_ap(a)](x-a)-\int_a^xp(t)y(t)dt -<br /> \int_a^x[q(t)-p&#039;(t)](x-t)y(t)dt+\int_a^xg(t)(x-t)dt

Now, to remove the undetermined constant y&#039;_a we can use the B.C.
y(b)=y_b to find

y&#039;_a = \frac{y_b-y_a}{b-a}-y_ap(a)+\frac{1}{b-a}\int_a^bp(t)y(t)dt<br /> + \frac{1}{b-a}\int_a^b[q(t)-p&#039;(t)](b-t)y(t)dt - \frac{1}{b-a}\int_a^bg(t)(b-t)dt

So that the solution is

y(x) = y_a+[\frac{y_b-y_a}{b-a}+\frac{1}{b-a}\int_a^bp(t)y(t)dt +<br /> \frac{1}{b-a}\int_a^b[q(t)-p&#039;(t)](b-t)y(t)dt -<br /> \frac{1}{b-a}\int_a^bg(t)(b-t)dt](x-a)-\int_a^xp(t)y(t)dt -<br /> \int_a^x[q(t)-p&#039;(t)](x-t)y(t)dt+\int_a^xg(t)(x-t)dt

 

[bigfooted]

maybe this will help:

Any second order linear homogeneous ODE of the form az"+bz'+cz=0 can be written as y"-ry=0
using the transformation y=z \exp(-\int\frac{b}{2a}dx) with r=\frac{b^2+2ab&#039;-2ba&#039;-4ac}{4a^2}

so you transform first into something you know how to solve, then transform the answer back (in this case transforming the answer in y back to z by using the inverse transform z=y \exp(\int\frac{b}{2a}))