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I have started answering a fredholm integral equation of the second kind and do not know where to go from here.

The answer has to be written in the form $$ \sum a_jx^{j-1} $$

The Fredholm integral equation is

$$ x^3+\frac16x^2+\frac15x = g(x) + \int_0^1(x^2y+xy^2) f(y) dy$$.

My method so far:

Let: $$C_1 = \int_0^1yf(y)dy$$ and $$C_2 = \int_0^1y^2f(y)dy$$

Then $$ x^3+\frac16x^2+\frac15x = (C_1x^2 +C_2x) + g(x)$$.

Eliminating f(y) to get $$C_1 = (\frac14C_1 + \frac13C_2) + \int_0^1yg(y)dy$$ and $$C_2 = (\frac15C_1 + \frac14C_2) + \int_0^1y^2g(y)dy$$

I don't know where to go from here to get it into the form $$ \sum a_jx^{j-1} $$

Do I put it into matrix form, and solve simultaneously, (not sure how to do this) If I have gotten anything wrong here please let me know. Or if you need any more information I may be able to provide. (Like how I got to a specific equation) Any help will be appreciated

Thank you very much

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$$g(x) = x^3 - \frac{38}{1077} x^2 + \frac{58}{1795} x$$

By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:

The integral from 0 to 1 of $$x^2y(y^3+Ay^2+By) dy = x^2[\frac{y^5}{5} + \frac{Ay^4}{4} + \frac{By^3}{3}] = x^2(\frac{1}{5} + \frac{A}{4} + \frac{B}{3})$$ and the integral from 0 to 1 of $$xy^2(y^3+Ay^2+By) dy = x[y^6/6 + \frac{Ay^5}{5} + \frac{By^4}{4}] = x(\frac{1}{6} + \frac{A}{5} + \frac{B}{4})$$

So the equality becomes: $$x^3 + \frac{1}{6} x^2 + \frac{1}{5} x = (x^3 + Ax^2 + Bx) + x^2(\frac{1}{5} + \frac{A}{4} + \frac{B}{3}) + x(\frac{1}{6} + \frac{A}{5} + \frac{B}{4})$$

Rearranging gives: $$x^2(\frac{-1}{30} - \frac{5A}{4} - \frac{B}{3}) + x(\frac{1}{30} - \frac{A}{5}- \frac{5B}{4}) = 0$$

So, by comparing coefficients; $$\frac{-1}{30} - \frac{5A}{4} - \frac{B}{3} = 0$$ and $$\frac{1}{30} - \frac{A}{5} - \frac{5B}{4} = 0$$

Solving simultaneously gives $$A = - 38/1077 $$ and $$B = \frac{58}{1795} $$

$$ g(x)= x^3-\frac{38}{1077} x^2 + \frac{58}{1795}$$

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  • $\begingroup$ If i was to write it in the form of a sum of functions, how would I write it as $$ \sum a_jx^{j-1} $$. $\endgroup$ – p s Jan 4 at 11:17
  • $\begingroup$ If we let $a_1=\frac{58}{1795}$, $a_2=0$, $a_3=-\frac{38}{1077}$ and $a_4=1$ we have:$$g(x)=\sum_{j=1}^4{a_jx^{j-1}}$$ $\endgroup$ – Peter Foreman Jan 4 at 20:47
  • $\begingroup$ Thank you for your help $\endgroup$ – p s Jan 6 at 10:16
  • $\begingroup$ Thank you for your help. If I was to continue my method would I still get the same answer? I didn't know what to do next in my method to get an answer. $\endgroup$ – p s Jan 6 at 10:25

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