Solving a simple Volterra integral equation
Mar 12, 2022

An integral equation is one that contains an unknown function under the integral sign. When that integral is over an interval varying with $$x$$, it is more specifically classified as a Volterra integral equation.

I came across this interesting, relatively simple Volterra equation in Stewart's Calculus and thought it would be interesting to try out.

$y(x) = 2 + \int_2^x (t - t\cdot y(t))\ \text{d}t.$

The first thing that comes to mind is that we need to get the unknown function $$y(t)$$ out from the integral sign. To do this, we can use the Fundamental Theorem of Calculus to differentiate the equation, cancelling the integral sign. When you use FTC, you replace the $$t$$ inside the integral (which is simply a dummy variable) with $$x$$.

$y'(x) = x - x \cdot y(x).$

Note this is a separable differential equation,

$\frac{\text{d}y}{\text{d}x} = x(1-y(x)) \to \int \frac{\text{d}y}{1-y(x)} = \int x\ \text{d}x.$

Integrating both sides, we get the following.

\begin{align*} -\ln(1-y) &= \frac{x^2}{2} + c\\ &\implies \ln(1-y) = -\frac{x^2}{2} - c\\ &\implies 1-y = e^{-\frac{x^2}{2} -c}\\ &\implies y = 1 - e^{-c}\cdot e^{-\frac{x^2}{2}}\\ &\text{Let } C = e^{-c}\\ &\therefore y = 1-Ce^{-\frac{x^2}{2}} \end{align*}

From this general solution, a particular solution can be found by obtaining an initial condition from the integral. Note that $$y(2) = 2$$, because an integral evaluated at equal upper and lower bounds is just zero.

Solving for the arbitary constant at $$x=2$$,

$2 = 1-Ce^{-2} \implies -C = e^2.$

Therefore, the particular solution to the Volterra integral equation is,

$y = 1 + e^{\frac{4-x^2}{2}}.$

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