How to convert Integral Equations into Differential Equations?

We’ve covered the basics of integral equations and learned how to derive them from ordinary differential equations. We even solved a linear integral equation using the trial method. Now comes the interesting part—going in reverse.

Before we dive deeper into solving integral equations, you need to master the reverse conversion: turning integral equations back into differential equations. Trust me, this skill becomes essential when you’re working through complex problems.

If you’re just joining this series, here’s what we’ve covered so far:

• Part I: Basics of Integral Equations
• Part II: Square Integrable Functions, Norms, and Trial Method
• Part III: Converting Differential Equations into Integral Equations

Converting Integral Equations to Differential Equations

This process is the exact opposite of what we did in Part III. Remember how we converted boundary value differential equations into integral equations? Now we’re reversing that operation.

Here’s the pattern I’ve observed after working through dozens of these conversions:

• Volterra Integral Equations → Initial Value Problems (IVPs)
• Fredholm Integral Equations → Boundary Value Problems (BVPs)

The technique is straightforward once you understand the core principle. In the forward conversion, we continuously integrated the differentials within given boundary values. For the reverse process, we continuously differentiate the integral equation and extract conditions by substituting the constant integration limits.

Let me walk you through two worked examples that demonstrate both cases.

Problem 1: Volterra Integral Equation → Initial Value Problem

Problem: Convert the following Volterra integral equation into an initial value problem:

$$y(x) = – \int_{0}^x (x-t) y(t) \, dt$$

Quick context: This is the same integral equation we obtained when converting the initial value problem \( y” + y = 0 \) with conditions \( y(0) = 0 \) and \( y'(0) = 0 \). See Problem 1 of Part III for the forward conversion.

Solution

We start with our integral equation:

$$y(x) = – \int_{0}^x (x-t) y(t) \, dt \quad \ldots (1)$$

Step 1: First differentiation. Differentiate equation (1) with respect to \( x \) using the Leibniz rule:

$$y'(x) = -\frac{d}{dx} \int_{0}^x (x-t) y(t) \, dt$$

When we differentiate under the integral sign, the term \( (x-t) \) differentiates to \( 1 \) with respect to \( x \), while the upper limit contribution vanishes since \( (x-x) = 0 \). This gives us:

$$y'(x) = -\int_{0}^x y(t) \, dt \quad \ldots (2)$$

Step 2: Second differentiation. Differentiate equation (2) with respect to \( x \):

$$y”(x) = -\frac{d}{dx} \int_{0}^x y(t) \, dt$$

By the Fundamental Theorem of Calculus, differentiating an integral with respect to its upper limit yields the integrand evaluated at that limit:

$$y”(x) = -y(x) \quad \ldots (3′)$$

Rearranging gives us our differential equation:

$$y”(x) + y(x) = 0 \quad \ldots (3)$$

Step 3: Extract initial conditions. Substitute the lower limit \( x = 0 \) into equations (1) and (2):

From equation (1):

$$y(0) = – \int_{0}^0 (0-t) y(t) \, dt = 0 \quad \ldots (4)$$

From equation (2):

$$y'(0) = -\int_{0}^0 y(t) \, dt = 0 \quad \ldots (5)$$

Final Result: The differential equation (3) with initial conditions (4) and (5) form the complete initial value problem:

$$y” + y = 0, \quad y(0) = 0, \quad y'(0) = 0 \quad \Box$$

Problem 2: Fredholm Integral Equation → Boundary Value Problem

Problem: Convert the following Fredholm integral equation into a boundary value problem:

$$y(x) = \lambda \int_{0}^{l} K(x,t) y(t) \, dt$$

where the kernel \( K(x,t) \) is defined piecewise as:

$$K(x,t) = \frac{t(l-x)}{l} \quad \text{for } 0 < t < x$$

$$K(x,t) = \frac{x(l-t)}{l} \quad \text{for } x < t < l$$

Reference: This kernel appeared in Example 2 of Part III, where we derived this integral equation from a boundary value problem.

Solution

The given integral equation is:

$$y(x) = \lambda \int_{0}^{l} K(x,t) y(t) \, dt \quad \ldots (1)$$

Since the kernel is piecewise-defined, we split the integral at \( x \):

$$y(x) = \lambda \left( \int_{0}^{x} \frac{(l-x)t}{l} y(t) \, dt + \int_{x}^{l} \frac{x(l-t)}{l} y(t) \, dt \right) \quad \ldots (2)$$

Step 1: First differentiation. Differentiate equation (2) with respect to \( x \). This requires careful application of the Leibniz rule since \( x \) appears both in the limits and inside the integrands:

$$y'(x) = -\frac{\lambda}{l} \int_{0}^x t \, y(t) \, dt + \frac{\lambda}{l} \int_{x}^l (l-t) y(t) \, dt \quad \ldots (3)$$

The boundary terms from Leibniz cancel due to the kernel’s continuity at \( x = t \).

Step 2: Second differentiation. Differentiate equation (3) with respect to \( x \):

$$y”(x) = -\frac{\lambda}{l} \cdot x \cdot y(x) – \frac{\lambda}{l} \cdot (l-x) \cdot y(x)$$

$$y”(x) = -\frac{\lambda}{l} y(x) \left[ x + (l-x) \right] = -\frac{\lambda}{l} y(x) \cdot l$$

$$y”(x) = -\lambda y(x)$$

Rearranging gives us the differential equation:

$$y”(x) + \lambda y(x) = 0 \quad \ldots (4)$$

Step 3: Extract boundary conditions. Substitute both integration limits into equation (2):

At \( x = 0 \): The first integral vanishes (limits 0 to 0), and in the second integral, the factor \( x = 0 \) makes the integrand zero:

$$y(0) = 0 \quad \ldots (5)$$

At \( x = l \): The second integral vanishes (limits \( l \) to \( l \)), and in the first integral, the factor \( (l-x) = 0 \) makes the integrand zero:

$$y(l) = 0 \quad \ldots (6)$$

Final Result: The ODE (4) with boundary conditions (5) and (6) constitutes the boundary value problem:

$$y” + \lambda y = 0, \quad y(0) = 0, \quad y(l) = 0 \quad \Box$$

Key Takeaways

After working through these conversions, here’s what you should remember:

1. Differentiation is the reverse of integration—repeatedly differentiate the integral equation until the integral disappears.
2. Volterra → IVP: Use the lower limit to extract initial conditions.
3. Fredholm → BVP: Use both limits to extract boundary conditions.
4. The Leibniz rule is essential when the variable appears both in the limits and under the integral sign.
5. Piecewise kernels require splitting the integral and careful bookkeeping during differentiation.

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