Find all functions such that $f'(t) = f(t) + \int_0^1 f(\tau)\,d\tau.$

Question

From Spivak

Find all functions such that $f'(t) = f(t) + \int_0^1 f(\tau)\,d\tau.$

My approach: differentiate both sides to get $f''(t) = f'(t)$, giving $f'(t) = Ce^t$, implying $f(t) = Ce^t + D$. Plugging this into the diff EQ gives $D=\dfrac{1-e}{2}C$, and indeed $f(t)=Ce^t + \dfrac{1-e}{2}C$ satisfies the equation.

Anyone find any errors in this analysis?

Answer 1

This is a kind of Fredholm integro-differetial equation. Just as the comment of @127.0.9.6, you can not ensure $f''$ is exist. In other words, we can only claim that $f\in C^1$ or $f\in H^1$.

Assume $\int_0^1f(\tau)d\tau=C$ then the equation can be written as $f'(t)=f(t)+C$. This is a simple 1st order linear ODE. It have a general solution that $f(t)=C_1e^t-C$, which $C_1$ is a constant. Next, integrate each sides of this solution, we have: \begin{equation} \int_0^1f(t)dt=C_1\int_0^1 e^tdt-C\\ C=C_1(e-1)-C\\ C=\frac{C_1(e-1)}{2} \end{equation} So the solution of the equation is $f(t)=C_1e^t-\frac{C_1(e-1)}{2}$. If some initial condition have been given, the constant $C_1$ can be determined.

Our solutions are same but I do not use the 2nd derivative of $f(t)$.

Answer 2

I find the presence of integral puzzling; let's just write for now $f'(t)=f(t)+A$ where $A$ is a constant. Since this is assumed to hold for every $t$, we know that $f$ is differentiable. But then $f'(t)$, being $f(t)+A$, is differentiable too. So, $f$ is twice differentiable.

(This is a very simple example of bootstrapping, a way of iteratively improving regularity of solutions. It could be continued to show $f\in C^\infty$, if needed.)

Then your argument goes through as written: $f''(t)=f'(t)$, hence $f'$ is an exponential, and then you figure out the constants. (At the last step, the fact that $A$ is some integral of $f$ becomes relevant.)