Solving an integral equation with a simple trick

Oscar Nieves
4 min readMar 29

In a paper published in 2011 by Waldemar Klobus (Motion on a vertical loop with friction) an interesting integral equation (more precisely, a Volterra integral equation of the 2nd kind) appears:

which describes the motion of an object around a circular loop as shown in Fig.1 below:

Image from Motion on a vertical loop with friction

and μ is the coefficient of friction, with respect to the tangential frictional force that the object experiences. V denotes the speed of the object normalized by a constant factor (which the author defines as the square root of gR, where g is the gravitational acceleration and R is the loop’s radius). Although the solution to Equation 1 is presented in the paper, the actual procedure for solving it is not. In this article, I will go through a step-by-step derivation of such a solution. By the way, V(0) is a constant which is derived in the paper based on some critical conditions, but in this article I will mainly focus on the general solution to Equation 1.

This integral equation can be solved by a simple trick, which is to use differentiation first. For example, if we take the 1st derivative of Eq. 1 we obtain

which is more easily written as:

Equation 2 is now a 1st-order differential equation for V² which is linear and has constant coefficients, therefore it has an exact closed-form solution. Using the integrating factor method, in which we integrate the function next to the V² term on the LHS and then put it in an exponential as follows:

we then multiply both sides of Eq. 2 by this factor:

Because of the chain-rule of differentiation, the LHS can be written in the equivalent form:

so Equation 2 now becomes:

Next, we proceed to integrate Eq. 3 with respect to phi but we do so from 0 to phi, hence we choose a “dummy integration variable” t:

The LHS of Eq. 4 is readily integrated as:

The RHS of Eq. 4 can be found from the standard integral techniques such as integration by parts:

so we then find that the RHS of Eq. 4 becomes:

Putting this result together with our result for the LHS, we can finally write:

and re-arranging gives:

and finally, making the substitutions:

we obtain the solution in the form presented in the paper:

What is interesting about this problem is that we obtained a rather straight-forward (though laborious) exact solution to the integral equation by differentiating it first. Sometimes, it is simple steps like these which can make a world of difference when solving mathematical problems.

Oscar Nieves

I write stories about applied math, physics and engineering.