# A Standard Integral Equation

Consider the integral equation

$$\phi(x) = x + \lambda\int_0^1 \phi(s)\,ds$$

Integrating with respect to $$x$$ from $$x=0$$ to $$x=1$$:

$$\int_0^1 \phi(x)\,dx = \int_0^1x\,dx + \lambda \int_0^1\Big[\int_0^1\phi(s)\,ds\Big]\,dx$$

which is equivalent to

$$\int_0^1 \phi(x)\,dx = \frac{1}{2} + \lambda \int_0^1\phi(s)\,ds$$

How can I go from here in order to solve the problem for the homogeneous case and find the corresponding characteristic values and associated rank?

• What is $\lambda$? What do you mean by "solve the problem"? I don't see what "the problem" is supposed to mean. Of which object do you want to find the characteristic values and ranks? Have you checked your definition of $\phi$? – James Mar 25 at 17:50
• My apologies, $\lambda$ is an arbitrary constant. In essence I want to obtain an expression of $\phi(x)$ which does not contain a function of $s$, which the initial integral equation has. – LightningStrike Mar 25 at 17:54

Relabelling the dummy variable $$x\mapsto s$$ on the LHS of your final equation, $$\int_0^1\phi(s)\,ds-\lambda\int_0^1\phi(s)\,ds=\frac12\\\implies \int_0^1\phi(s)\,ds=\frac1{2(1-\lambda)}$$

Thus $$\phi(x)=x+\frac\lambda{2(1-\lambda)}$$

Note $$\int_{0}^{1}{\phi(s)ds}$$ is a constant say $$a$$. Your functional equation (FE) can be rewritten as: $$\phi(x)=x+a\lambda$$ Putting into FE yields:

$$x+a\lambda=x+\lambda\int_{0}^{1}{(s+a\lambda )ds }\iff a\lambda=\lambda\big(\frac{1}{2}+\lambda a\big)$$ If $$\lambda=0$$ then $$\phi(x)=x$$

if $$\lambda\ne 1$$ $$a=\frac{1}{2}+\lambda a\iff ( 1-\lambda)a=\frac{1}{2}\iff a=\frac{1}{2-2\lambda}$$ and then $$\phi(x)=x+\frac{\lambda}{2-2\lambda}$$

If $$\lambda=1$$ there won’t besuch $$\phi$$.

If you are after finding $$\phi(x)$$, one approach that comes to mind is to assume it is smooth enough to have a normally convergent (so we can interchange series summation and integration) Taylor expansion on $$[0, 1]$$: $$\phi(x) = \sum_{n \geq 0} a_{n} x^{n}.$$ Substituting it into your equation, we get: $$\sum_{n \geq 0} a_{n} x^{n} = x + \lambda \sum_{n \geq 0}{a_{n} \over n+1}.$$ Matching up the coefficients of the difference powers of $$x$$, we get: $$a_{n} = 0 \quad \mbox{ for } n \geq 2,$$ $$a_{1} = 1,$$ and $$a_{0} = \lambda \left(a_{0} + {a_{1} \over 2}\right).$$ This gives a relationship between $$a_{0}$$ and $$\lambda$$.

Note that since $$\lambda\int_0^1 \phi(s)\,ds$$ is a constant (with respect to $$x$$), then we can write$$\phi(x)=x+a$$and by substitution we conclude that $$x+a=x+\lambda\int _{0}^{1}x+adx\implies\\a=\lambda({1\over 2}+a)\implies\\a={\lambda\over 2-2\lambda}$$ and we obtain$$\phi(x)=x+{\lambda\over 2-2\lambda}\quad,\quad \lambda\ne 1$$The case $$\lambda=1$$ leads to no solution.