2
$\begingroup$

Can anyone please point to any references regarding solutions of ODE of the form

$$(a+1_{x=0})f'(x)-f(x)=h(x)-\int_0^\infty h(t)e^{-t}dt \tag{1}$$ with $f(0)=0$, where $h$ is a given function, a is some constant and $1_{x=0}$ is the indicator function. I only need the solution for $x\geq 0$. I see that when the coefficient is some continuous function of $x$, say $b(x)$, that is $$b(x)f'(x)-f(x)=h(x)-\int_0^\infty h(t)e^{-t}dt$$

the solution is given by $$-e^{-\int_0^x\frac{1}{b(u)}du}\int_{x}^\infty\frac{1}{b(y)}\left(h(y)-\int_0^\infty h(t)e^{-t}dt\right)e^{\int_0^y\frac{1}{b(u)}du}dy$$ I am not sure how to proceed when the coefficient is not continuous.

$\endgroup$
2
  • $\begingroup$ Suggestion: take the limit $c\to 0^+$ of the solution to the ODE with $b(x)=a+1_{[0,c]}$. $\endgroup$
    – Gonçalo
    Commented Aug 21, 2023 at 0:04
  • $\begingroup$ Usually the differential equation should be fulfilled on an open intervall. So if your interval is $(0,\infty)$, the indicator function would not actually do something. Can you please clarify on the $x$-domain? $\endgroup$
    – maxmilgram
    Commented Aug 22, 2023 at 11:38

1 Answer 1

3
+25
$\begingroup$

Let's consider the case when $x\neq0$ because the derivative generally makes sense there. And let's call the solution $g(x).$ We have $1_{\{0\}}(x\neq0)=0.$ $$ag'(x)-g(x)=h(x)-\int_0^\infty h(t)e^{-t}dt$$ This can be easily solved like you pointed out. \begin{eqnarray} g'(x)e^{-x/a}-\frac{e^{-x/a}}ag(x)&=&\frac{e^{-x/a}}a\left(h(x)-\int_0^\infty h(t)e^{-t}dt\right)\\ \left(g(x)e^{-x/a}\right)'&=&\frac{e^{-x/a}}a\left(h(x)-\int_0^\infty h(t)e^{-t}dt\right)\\ \left(g(u)e^{-u/a}\right)'&=&\frac{e^{-u/a}}a\left(h(u)-\int_0^\infty h(t)e^{-t}dt\right)\\ \int_b^x\left(g(u)e^{-u/a}\right)'du&=&\int_b^x\frac{e^{-u/a}}a\left(h(u)-\int_0^\infty h(t)e^{-t}dt\right)du\\ g(x)e^{-x/a}-g(b)e^{-b/a}&=&\int_b^x\frac{e^{-u/a}}a\left(h(u)-\int_0^\infty h(t)e^{-t}dt\right)du\\ g(x)&=&g(b)e^{(x-b)/a}+e^{x/a}\int_b^x\frac{e^{-u/a}}a\left(h(u)-\int_0^\infty h(t)e^{-t}dt\right)du \end{eqnarray} We need another boundary condition to be able to calculate values of $g$ because $0$ is not in the domain of $g.$ Having that, we can show $\lim_{x\to0^+}g(x)=0.$ And thus, $g=f$ because $g(x)=f(x)$ for all appropriate $x$ and we have $f(x)$ as desired.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .