Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an equation : $\ f(x) - \lambda\int\limits_0^1 9xtf(t) dt = ax^2 - 4x^2\ $ in $\ L_2[0,1]\ $ space.

And I want to understand how to solve it, not just obtain an answer.

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

For this particular equation, it is simple. Differentiate both sides of the equation with respect to $x$: $$ f^\prime(x) = 2 (a-4) x + \lambda \int_0^1 9 t f(t) \mathrm{d} t $$ Since the integral is independent of $x$, the solution is a quadratic $f(x) = (a-4) x^2 + b x + c$. Substituting into the original equation we get a linear system of equations for $b$ and $c$: $$ (a-4) x^2 + b x + c - \lambda x \int_0^1 9 t f(t) \mathrm{d} t = (a-4) x^2 $$ Since $\int_0^1 9 f(t) \mathrm{d} t = \frac{9}{4} a+ 3b+ \frac{9}{2} c - 9$, we get the system: $$ c = 0 \qquad b (1-3 \lambda) - \frac{9}{4} \lambda \left( 2 c + a -4\right) = 0 $$ Therefore we obtain: $$ f(x) = (a-4) x \left( x -\frac{9}{4} \frac{\lambda}{3 \lambda - 1} \right) $$

share|improve this answer
    
Can you recomend a good book on these theme? –  Philipp G. Sinicyn Nov 23 '11 at 15:51
1  
@PhilippG.Sinicyn The book "First course on integral equations" by Wazwaz may be a good start. –  Sasha Nov 23 '11 at 16:17
    
How to solve this if $\lambda = \frac{1}{3}$ –  Philipp G. Sinicyn Nov 28 '11 at 10:38
    
@PhilippG.Sinicyn Come back to equations for polynomial coefficients. They would say $c=0$, $a=4$ and would leave $b$ free, which is to say, equation will have no solutions if $a\not=4$, and if $a=4$, then $f(x) = b*x$ would be a solution –  Sasha Nov 28 '11 at 14:05
add comment

If the Kernel (what is in along with the unknown function in the integral) is polynomial, then it is easy to solve. In this case, multiply your equation by x and integrate between 0 and 1, then obtain the value of ∫tf(t)dt and replace in the original equation to get the solution.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.