I found a differential ordinary equation that uses Diric Delta Function that I don't know how to solve. Thank you if you can help me.
The ODE is like that
$$
z' = \rho \cdot \sqrt{1+z^2} + w \cdot \delta(x-x_0)
$$
And my domain is $\left[a, \ b\right]$ with $a < x_0 < b$. With $\rho$ and $w$ constants.
How can I solve it?
The solution I tried was:
$$
\dfrac{z'}{\rho \cdot \sqrt{1+z^2} + w \cdot \delta(x-x_0)} = 1
$$
$$
I = \int_{a}^{b} \dfrac{z'}{\rho \sqrt{1+z^2}+w\delta} dx = \int_{a}^{b} 1 \ dx = b-a
$$
Now I divide the domain $\left[a, \ b\right]$ in three domains $\left[a, \ x_0 – \varepsilon\right] \cup \left[x_0 – \varepsilon, \ x_0 + \varepsilon\right] \cup \left[x_0 + \varepsilon, \ b\right]$ with $\varepsilon > 0$ small and then make three integrals separately $I_1$, $I_2$ and $I_3$.
$$
I = \lim_{\varepsilon \to 0^{+}} I_{1}(\varepsilon)+I_{2}(\varepsilon)+I_{3}(\varepsilon) = \lim_{\varepsilon \to 0^{+}} \left(\underbrace{\int_{a}^{x_0 – \varepsilon} \square \ dx}_{I_1(\varepsilon)} + \underbrace{\int_{x_0 – \varepsilon}^{x_0 + \varepsilon} \square \ dx}_{I_2(\varepsilon)} + \underbrace{\int_{x_0 + \varepsilon}^{b} \square \ dx}_{I_3(\varepsilon)}\right)
$$
With
\begin{align*}
I_{1}(\varepsilon) & = \int_{a}^{x_0-\varepsilon} \dfrac{z'}{\rho \sqrt{1+z^2}+w\underbrace{\delta}_{0}} dx = \dfrac{1}{\rho}\int_{z(a)}^{z(x_0-\varepsilon)} \dfrac{dz}{\sqrt{1+z^2}} = \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(a)}^{z(x_0-\varepsilon)} \\
I_2(\varepsilon) & = \int_{x_0-\varepsilon}^{x_0+\varepsilon} \dfrac{z'}{\rho \sqrt{1+z^2}+w\delta} dx \overset{\underset{\mathrm{?}}{}}{=} \int_{x_0-\varepsilon}^{x_0+\varepsilon} 1 \ dx = 2\varepsilon \\
I_{3}(\varepsilon) & = \int_{x_0+\varepsilon}^{b} \dfrac{z'}{\rho \sqrt{1+z^2}+w\underbrace{\delta}_{0}} dx = \dfrac{1}{\rho}\int_{z(x_0+\varepsilon)}^{z(b)} \dfrac{dz}{\sqrt{1+z^2}} = \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(x_0+\varepsilon)}^{z(b)}
\end{align*}
And therefore
\begin{align*}
b – a = I & = \lim_{\varepsilon \to 0^{+}} I_1 + I_2 + I_3 \\
& = \lim_{\varepsilon^{+}} \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(a)}^{z(x_0-\varepsilon)} + 2 \varepsilon + \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(x_0+\varepsilon)}^{z(b)} \\
& = \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(a)}^{z(b)} + \lim_{\varepsilon \to 0^{+}} \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(x_0-\varepsilon)}^{z(x_0+\varepsilon)} \\
& \overset{\underset{\mathrm{?}}{}}{=} \dfrac{1}{\rho} \ln \left(z + \sqrt{z^2-1}\right)_{z(a)}^{z(b)}
\end{align*}
I thought about using another function in the place of $\delta$, like
$$
\delta_\varepsilon(x-x_0) =
\begin{cases}
0 \ \ \ \text{if} \ x < x_0 – \varepsilon \\
\frac{1}{2\varepsilon} \ \ \ \text{if} \ x_0 – \varepsilon < x < x_0 + \varepsilon \\
0 \ \ \ \text{if} \ x_0 + \varepsilon < x \\
\end{cases}
$$
or even a continuous function, but it's harder(or impossible) to calculate the integral:
$$
\delta_\varepsilon(x-x_0) =
\dfrac{1}{\varepsilon \sqrt{\pi}} \exp\left(-\dfrac{1}{\varepsilon^2}(x-x_0)^2\right)
$$
Unfortunately, the physical problem (of this ODE) suggests that $z$ has a linear term of Heaviside Function at the point $x_0$.
Best Answer
If you divide by $\frac{1}{\sqrt{1+z^2}}$ and integrate you get $$\int_a^x \frac{z'(w)dw}{\sqrt{1+z(w)^2}}=\rho(x-a)+\frac{w}{\sqrt{1+z^2(x_0)}}\theta(x-x_0)$$ Where $\theta(x)$ is the Heaviside step function. With the substitution $y=z(w)$ in the first integral the antiderivative is elementary and is $\sinh^{-1}y$, so that the solution becomes $$z(x)= \sinh \bigg [\sinh^{-1}z(a)+\rho(x-a)+\frac{w}{\sqrt{1+z^2(x_0)}}\theta(x-x_0)\bigg]$$ Admittedly, this is a pretty weird solution though given the two initial conditions; do you have any physical reason to know the value of $z$ at $x=a$ and $x=x_0$?