Quite challenge: how to solve this functional equation?

Question

I have asked This question before, but I forget to add some term and factors. When adding those, the difficulties increased significantly.

Consider the following equation:

$$\omega(t) = 1 + \frac{2}{t^2} \cdot \int_0^{t-a} x \cdot \omega(t - x) dx$$

I have tried to use similar method to convert it into ordinary differential equation. But I failed.

When I simultaneously make derivatives of both sides,

• First order: $$\omega'(t) = \frac{2 (t-a) \omega (a)}{t^2}+\int_0^{t-a} \left(\frac{2 x \omega '(t-x)}{t^2}-\frac{4 x \omega (t-x)}{t^3}\right) \, dx$$

• Second order: $$\omega''(t) = -\frac{8 (t-a) \omega (a)}{t^3}+\frac{2 (t-a) \omega '(a)}{t^2}+\frac{2 \omega (a)}{t^2}+\int_0^{t-a} \left(\frac{12 x \omega (t-x)}{t^4}-\frac{8 x \omega '(t-x)}{t^3}+\frac{2 x \omega ''(t-x)}{t^2}\right) \, dx$$

As a result, I believe the solution for the previous question is not suitable for this problem.

Is there anyone have any ideas about this problem?

If converting it into ordinary differential equation is the correct routine, then the key procedure to solve the problem is how to eliminate integral part.

Answer 1

It's best to transform the equation into one that will differentiate more easily before trying to convert to an ODE. In this particular example, we can convert the integral on the RHS to a second antiderivative of $\omega$: $$ \int_0^{t-a}x\omega(t-x)dx = \int_a^t(t-x)\omega(x)dx = \left.(t-x)\omega^{(-1)}(x)\right|_a^t +\int_a^t\omega^{(-1)}(x)dx = \omega^{(-2)}(t) $$ Where we have chosen the antiderivative $\omega^{(-1)}(x)$ such that $\omega^{(-1)}(a) = 0$, and similarly for the second antiderivative $\omega^{(-2)}$. Then we rearrange the original equation a bit to get $$ \omega(t) - \frac{2}{t^2}\omega^{(-2)}(t) = 1. $$ This is a second-order inhomogenous linear ODE for $\omega^{(-2)}(t)$, with initial conditions $\omega^{(-2)}(t) = \omega^{(-1)}(t) = 0$. This can be solved through standard methods for such equations to get $$ \omega^{(-2)}(t) = \frac{t^2}{3}\left[\ln\left(\frac{t}{a}\right)-\frac{1}{3}\right] + \frac{a^3}{9t}. $$ Then differentiate twice to get $\omega$: $$ \omega(t) = \frac{7}{9}+\frac{2a^3}{9t^3} + \frac{2}{3}\ln\left(\frac{t}{a}\right), $$ which you can check satisfies the integral equation.

EDIT: The answer given in this post changed a few times because I made some algebra mistakes calculating the answer. The current one is correct according to WolframAlpha.

Answer 2

First rewrite the expression as $$t^{2}w(t)-t^{2}= 2\int_0^{t-a} x \cdot \omega(t - x) dx,$$ and make the change of variable $y=t-x$, to get $$t^{2}w(t)-t^{2}=2\int_a^{t} (t-y)w(y)dy=2t\int_a^{t} w(y)dy-2\int_a^{t}yw(y)dy.$$

Differentiating this expression yields $$t^{2}w'(t)-2tw(t)-2t=2\int_a^{t} w(y)dy+2tw(t)-2tw(t)=2\int_a^{t} w(y)dy,$$ and differentiating once more gives you an ODE.