I'm trying to show the continuity of the solution to this simple integral
equation (or the corresponding ODE):
$$
x(t)=x(0)+at-b\int_0^t x(s)\wedge u(s) \, ds,\quad t\in[0,T]
$$
where $x(s)\wedge u(s)=\min(x(s),u(s))$ and $u$ is a piecewise constant (right-continious)
function, i.e.,
$$
u(s)=u_k,\quad t_{k}\leq s I haven't been able to find any general results. I'd be grateful if
you could point me to such results or give me a hint in proving the
statement. Perhaps bounding $\left|x(t)-x(t_1)\right|$ and taking
$t\rightarrow t_1$?
Proving continuity of the solution to a simple integral equation
0
$\begingroup$
real-analysis
ordinary-differential-equations
continuity
1 Answers
2
The integral is continuous on $t$ and so any solution (say measurable) of the integral equation will be continuous (the right-hand side is continuous and so the left-hand side is continuous).
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0Thanks. Is it trivial that the integral is continuous? Or needs some argument like "because it's derivative exists almost everywhere"? – 2017-01-27
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0We know that $\int_s^t \text{integrable function}\to0$ when $t\to s$, so if we take say bounded measurable functions, one could say that it is "trivial" without any damage. – 2017-01-27