For an equation of the form, $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{y}(t) = \int_0^t \mathbf{K}(t-\tau)\mathbf{y}(\tau)d\tau,$$ Can it be shown that in the long time, $t\to\infty$, limit, the solution is given by $$ \mathbf{y}(t) = \exp(\hat{\mathbf{K}}(0)t)\cdot\mathbf{y}(0), $$ where $\hat{\mathbf{K}}(s)$ is the Laplace transform of $\mathbf{K}(t)$?
If anyone can provide a simple answer or knows of a useful reference where a result like this for the asymptotic limit is shown, that would be very helpful. Thanks!
I should say no. The Laplace transform of the given equation is $$ s{\hat y}(s)-y(0)={\hat K}(s)y(s) $$ that gives $$ {\hat y}(s)=\frac{y(0)}{s-{\hat K}(s)}. $$ Then, when you will invert the transform, the poles, obtained by the equation $s-{\hat K}(s)=0$, will determine the asymptotic behaviour. Especially, the leading one is the greatest with $Re(s)>0$ while, if you have $Re(s)<0$ for all the poles, it will be the smallest one. If $Re(s)=0$ for any pole, you will get just oscillating behaviour. Therefore, the choice ${\hat K}(0)$ does not seem to have any special meaning in this context.