# Solution of ODE with discontinuous coefficients

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.

• Suggestion: take the limit $c\to 0^+$ of the solution to the ODE with $b(x)=a+1_{[0,c]}$. Commented Aug 21, 2023 at 0:04
• 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? Commented Aug 22, 2023 at 11:38

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.