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Consider the integro-differential equation $$ \frac{df(t)}{dt} = - \int_0^t ds \ g(s) f(t-s) $$ subject to the initial condition $f(0)=0$ and where $g$ is a known function.

My Question: Is the solution to the above always $f(t)=0$?

The way I've thought about solving this is to take the Laplace transform with $$ F(q) := \int_0^\infty dt \; e^{-q t} f(t) \ , \\ G(q) := \int_0^\infty dt \; e^{-q t} g(t) \ , $$ so that the integro-differential equation becomes $$ q F(q) - f(0^{+}) = - G(q) F(q) \ . $$ Since the initial condition is assumed to be $f(0^{+})=0$ we get $$ F(q) [ q + G(q) ] = 0 \ . $$ This means that so long as $q+G(q)\neq 0$ we can say that $F(q)=0$ and so $f(t)=0$.

However, I am worried about the case when $q+G(q)= 0$. Does this ruin the solution $f(t)=0$ in some cases?

QuantumEyedea
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    For what it's worth, if $G(q) = -q$ then $g(t) = -\delta^\prime (t)$. – bob.sacamento May 31 '19 at 19:57
  • Interesting....when I plug $g(t) = - \delta'(t)$ I find that the integral $-\int_0^t ds; g(t) f(t-s)$ evaluates to exactly $\dot{f}(t)$. So in this case we have a no information with an equation $\dot{f}(t)=\dot{f}(t)$. What is the meaning of this? – QuantumEyedea May 31 '19 at 20:02
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    Well, at least it's consistent! :-) That is an interesting result, though. I think it means that, in that case, the RHS is identically equal to the LHS and $f(t)$ could be anything. Actually, that would be right. Look again at your LT. If the quantity in brackets is zero, then there is no restriction on what $F(q)$ could be and, therefore, no restriction on $f(t)$. – bob.sacamento May 31 '19 at 21:51
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    One more way to look at it is that the convolution of a function with $-\delta^\prime$ is identically the same as the derivative of that function. – bob.sacamento May 31 '19 at 21:59

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