Boundary conditions for a ODE from Volterra integral equation

Question

I am modeling a cyclic voltammetry experiment. The mathematical model for it boils down to a Volterra integral equation of the first kind, with an unknown function $\chi(z)$ and known constants $\xi$ and $\theta$:

$$\int_0^{\sigma t}\frac{\chi(z)\mathrm{d}z}{(\sigma t-z)^{1/2}} = \frac{1}{1+\xi\theta e^{-\sigma t}}.$$

Is this a well-posed integral equation? I am troubled by the fact that the right-hand side does not vanish as $\sigma t$ tends to zero.

To solve this integral equation, I am differentiating both sides with respect to $\sigma t$ to obtain an ordinary differential equation:

$$\frac{\chi(z)}{\sqrt{z}}+2\sqrt{z}\chi^\prime(z) = \frac{-\xi\theta e^{-z}}{(1+\xi\theta e^{-z})^2}.$$

What boundary condition should I assume to solve this differential equation?

Answer 1

Your integral equation has the form $$ \int_0^{\tau}\frac{\chi(z)}{\sqrt{\tau-z}}\,dz=g(\tau), \tag{1} $$ where $g(\tau)=\frac{1}{1+\xi\theta e^{-\tau}}$ and $\tau=\sigma t$. The trick to solve it is to divide both sides of $(1)$ by $\sqrt{x-\tau}$ and integrate from $0$ to $x$: $$ \int_0^x\int_0^{\tau}\frac{\chi(z)}{\sqrt{(x-\tau)(\tau-z})}\,dz\,d\tau =\int_0^x\frac{g(\tau)}{\sqrt{x-\tau}}\,d\tau. \tag{2} $$ Changing the order of integration in the LHS of $(2)$, we have $$ \int_0^x\int_z^x\frac{\chi(z)}{\sqrt{(x-\tau)(\tau-z})}\,d\tau\,dz =\int_0^x\frac{g(\tau)}{\sqrt{x-\tau}}\,d\tau. \tag{3} $$ Performing the integral over $\tau$ in the LHS of $(3)$, we get$^{(*)}$ $$ \pi\int_0^x\chi(z)\,dz=\int_0^x\frac{g(\tau)}{\sqrt{x-\tau}}\,d\tau. \tag{4} $$ Finally, using the Fundamental Theorem of Calculus, we obtain $$ \chi(x)=\frac{1}{\pi}\frac{d}{dx}\int_0^x\frac{g(\tau)}{\sqrt{x-\tau}}\,d\tau \tag{5} $$

---

$^{(*)}$ See Proof of this integration shortcut: $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$ or How to evaluate $\int_{U}^{\alpha} \frac{dE}{\sqrt{(\alpha-E)(E-U)}}$.