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Find the solution to the Integral Equation:

$\phi(x)=x+\int _0^x \sin (x-t)\phi(t) dt $

The solution is given as $x+\dfrac{x^3}{3!}$.

My try:

$K(x,t)=\sin (x-t)=K_1(x,t)$;

$K_n(x,t)=\int _t^x K(x,z) K_{n-1}( z,t) dz$;

I got $K_2(x,t)=\sin(-x-t)-\sin (x-3t)-(x-t)\cos (x-t)$.

So it's getting difficult.I am appearing for an exam where the time given will be small for this 1 mark question.

Is there any easy way to do this?

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3 Answers

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Differentiate to get $\phi'(x) = 1+ \int_0^x \cos(x-t) \phi(t) dt$ and once more to get $\phi''(x) = \phi(x) - \int_0^x \sin(x-t) \phi(t) dt$. Hence the equation is $0= x - \phi''(x)$.

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You are looking for the solution over $[0, \infty)$ of

$$\phi(x)=x+\phi(x)*\sin(x)$$

Take the Fourier/Laplace transform and solve.

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Take Laplace of this Integral Equation: $${\cal L}(\phi)={\cal L}(x)+{\cal L}(\sin x){\cal L}(\phi) $$ $${\cal L}(\phi)=\frac{\frac{1}{s^2}}{1-\frac{1}{s^2+1}}=\frac{s^2+1}{s^4}=\frac{1}{s^2}+\frac{1}{s^4}$$ so $$\phi(x)=x+\dfrac{x^3}{3!}$$

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