Solve $f(t)=1+t-\frac{8}{3}\int ^t_0 (t-\tau)^3f(\tau)d\tau$

Question

Solve the integral equation $$f(t)=1+t-\frac{8}{3}\int ^t_0 (t-\tau)^3f(\tau)d\tau.$$

My attempt: Re-write the given integral equation, $$f(t)=1+t-\frac{8}{3}{t^3*f(t)}$$ Apply Laplace transform $$\mathcal{L}\{f(t)\}=\mathcal{L}\left\{1+t-\frac{8}{3}{t^3*f(t)}\right\}$$ I got $f(t)=\mathcal{L}^{-1}\left\{\frac{p^2+p^3}{p^4+16}\right\}$. How to continue from here?

Answer 1

Here is a way of tackling some inverse Laplace transform problems assuming you are good at manipulating summations $$ \mathcal{L}^{-1}\left[\frac{s^2+s^3}{16+s^4}\right]=\mathcal{L}^{-1}\left[\frac{s^2}{16+s^4}\right]+\mathcal{L}^{-1}\left[\frac{s^3}{16+s^4}\right] $$ expanding about $s$ equals infinity we have $$ \frac{s^2}{16+s^4} = \sum_{n=0}^\infty \frac{(-1)^n 16^n}{s^{4n+2}}\\ \frac{s^3}{16+s^4} = \sum_{n=0}^\infty \frac{(-1)^n 16^n}{s^{4n+1}} $$ using that $\mathcal{L}^{-1}(1/s^n)=t^{n-1}/(n-1)!$ $$ \mathcal{L}^{-1}\left[\frac{s^2}{16+s^4}\right] = \sum_{n=0}^\infty \frac{(-1)^n 16^n t^{4n+1}}{(4n+1)!}=\frac{ch(x)s(x)+c(x)sh(x)}{2\sqrt{2}}\\ \mathcal{L}^{-1}\left[\frac{s^3}{16+s^4}\right] = \sum_{n=0}^\infty \frac{(-1)^n 16^n t^{4n}}{(4n)!}=c(x)ch(x) $$ where $ch=\cosh,sh=\sinh,s=\sin,c=\cos,x=\sqrt{2}t$.

Answer 2

Recall the Cauchy repeated integral formula:

$$\int_0^t (t-\tau)^3 f(\tau){\rm d}\tau =3! \underbrace{\int\int\int\int f(t)}_{g(t)}$$ (abuse of notation, bear with me).

Using the new function $g(t)$, we get a linear ordinary differential equation:

$$g''''(t)+\frac{8}{3}3! g(t)=1+t$$ $$g''''(t)+2^4 g(t)=1+t$$

This one, you can solve with ease. Particular solution is just

$$g_p(t)=\frac{1}{16}(1+t)$$ and the homogeneous solution is, from the characteristic equation $\lambda^4=-2^4$ → $\lambda=\sqrt{2}(\pm 1 \pm i)$, a superposition

$$g_h(t)=e^{\sqrt 2 t}(A\sin\sqrt2 t + B\cos\sqrt2 t)+e^{-\sqrt 2 t}(C\sin\sqrt2 t + D\cos\sqrt2 t)$$

The original function $f(t)$ is simply the fourth derivative of $g_h(t)$ (the particular solution vanishes after differentiation anyway). Because differentiation of exponentials is the same, the solution has the same form:

$$f(t)=e^{\sqrt 2 t}(A\sin\sqrt2 t + B\cos\sqrt2 t)+e^{-\sqrt 2 t}(C\sin\sqrt2 t + D\cos\sqrt2 t)$$ with $A,B,C,D$ set by initial conditions, which you can read from the original integral equation ($f(0)=1$, $f'(0)=1$, $f''(0)=0$, $f'''(0)=0$). You can carry on from here.

Answer 3

Hint. Let $z_k$ for $k=1,2,3,4$ be the roots of $z^4=-16$. Then find complex numbers $A,B,C,D$ such that $$\frac{p^2+p^3}{p^4+16}=\frac{A}{p-z_1}+\frac{B}{p-z_2}+\frac{C}{p-z_3}+\frac{D}{p-z_4}$$ and recall that $$\mathcal{L}^{-1}\left(\frac{1}{p-z_k}\right)(t)=e^{z_kt}.$$