How to change Differential Equations into Integral Equations?

Here’s something that tripped me up when I first learned integral equations: you can convert any differential equation into an integral equation. Just integrate it.

Sounds too simple, right? It is simple. But the details matter. And those details are what separate a clean integral equation from a mess.

In this guide, I’ll walk you through exactly how to do it. We’ll start with first-order equations, move to second-order, and handle both initial value problems and boundary value problems. By the end, you’ll have a clear strategy you can apply to any ODE.

Why Convert a Differential Equation to an Integral Equation?

Before we dive in, let’s address the obvious question: why bother?

Differential equations and integral equations are two representations of the same underlying problem. Converting between them isn’t just an academic exercise. Here’s why it matters:

• Different solution techniques: Some equations that are hard to solve as ODEs become easier as integral equations (and vice versa)
• Numerical methods: Integral equations often have better numerical stability properties
• Existence and uniqueness: Proving solutions exist is sometimes easier in integral form
• Boundary conditions built-in: The integral equation already incorporates the boundary conditions, so you can’t accidentally forget them

Think of it as having two different tools for the same job. Sometimes a screwdriver works better than a wrench. Knowing how to switch between representations gives you flexibility.

A First-Order Example

Let’s start simple. Consider this first-order ODE:

$$\frac{dy}{dx} + 5y + 1 = 0$$

The conversion process: integrate once with respect to \( x \).

$$\int \frac{dy}{dx} \, dx + 5 \int y \, dx + \int 1 \, dx = c$$

This simplifies to:

$$y + 5 \int y \, dx + x = c$$

Rearrange into standard integral equation form:

$$y(x) = (c – x) – 5 \int y(t) \, dt$$

Notice I switched the integration variable from \( x \) to \( t \). This is standard practice. It prevents confusion between the free variable \( x \) and the dummy variable of integration.

Removing the Arbitrary Constant

That constant \( c \) is annoying. We remove it using a boundary condition.

Suppose \( y(0) = 1 \). Plug \( x = 0 \) into our equation:

$$y(0) = (c – 0) – 5 \int_0^0 y(t) \, dt$$

$$1 = c – 0$$

$$c = 1$$

Wait, that’s only true if we’re integrating from 0. Let me be more careful. With limits from 0 to \( x \):

$$y(x) = (c – x) – 5 \int_0^x y(t) \, dt$$

At \( x = 0 \): \( y(0) = c – 0 – 5(0) = c \). So \( c = 1 \).

Substituting back:

$$y(x) = (1 – x) – 5 \int_0^x y(t) \, dt$$

Or equivalently:

$$y(x) = (1 – x) + 5 \int_0^x (1 – y(t)) \, dt$$

That’s it. We’ve converted a first-order differential equation into an integral equation. The boundary condition \( y(0) = 1 \) is now baked into the equation itself.

Second-Order Equations: Two Types

Second-order equations require two boundary conditions. How those conditions are specified determines what type of integral equation you get.

There are two types:

Initial Value Problem (IVP)

Definition: Both conditions are specified at the same point. You’re given \( y(a) \) and \( y'(a) \) for some fixed value \( a \).

Example:

$$\frac{d^2 y}{dx^2} + ky = tx$$

with \( y(0) = 2 \) and \( y'(0) = 5 \).

See how both conditions are at \( x = 0 \)? That’s the “initial” point. Both the function and its derivative are pinned down at the start.

The kicker: Initial value problems convert to Volterra integral equations. The upper limit of integration is the variable \( x \).

Boundary Value Problem (BVP)

Definition: Conditions are specified at different points. You’re given \( y(a) \) and \( y(b) \) for two different values \( a \) and \( b \).

Example:

$$\frac{d^2 y}{dx^2} + \ell y = mx$$

with \( y(a) = A \) and \( y(b) = B \).

The function is pinned at two separate points. You don’t know the derivative at either endpoint.

The kicker: Boundary value problems convert to Fredholm integral equations. The limits of integration are both constants.

Here’s the comparison:

Converting an IVP to a Volterra Integral Equation

Let’s work through a complete example.

Problem: Convert to an integral equation:

$$y” + y = 0$$

with \( y(0) = 0 \) and \( y'(0) = 0 \).

Step 1: Rewrite in normal form.

Put the highest derivative on one side:

$$y”(x) = -y(x)$$

Step 2: Integrate once from 0 to \( x \).

$$\int_0^x y”(x) \, dx = -\int_0^x y(x) \, dx$$

The left side is straightforward:

$$\left[ y'(x) \right]_0^x = -\int_0^x y(x) \, dx$$

$$y'(x) – y'(0) = -\int_0^x y(x) \, dx$$

Step 3: Apply the first boundary condition.

We know \( y'(0) = 0 \), so:

$$y'(x) – 0 = -\int_0^x y(x) \, dx$$

$$y'(x) = -\int_0^x y(x) \, dx$$

Step 4: Integrate again from 0 to \( x \).

$$\int_0^x y'(x) \, dx = -\int_0^x \left( \int_0^x y(x) \, dx \right) dx$$

Left side:

$$y(x) – y(0)$$

Right side needs the repeated integral formula. For a double integral:

$$\int_0^x \int_0^s f(t) \, dt \, ds = \int_0^x (x – t) f(t) \, dt$$

Applying this:

$$y(x) – y(0) = -\int_0^x (x – t) y(t) \, dt$$

Step 5: Apply the second boundary condition.

We know \( y(0) = 0 \):

$$y(x) – 0 = -\int_0^x (x – t) y(t) \, dt$$

Final answer:

$$y(x) = -\int_0^x (x – t) y(t) \, dt$$

This is a Volterra integral equation of the second kind. The kernel is \( K(x, t) = -(x – t) \), and the upper limit is the variable \( x \).

Converting a BVP to a Fredholm Integral Equation

This is trickier. The unknown derivative at the boundary introduces an extra step.

Problem: Convert to an integral equation:

$$\frac{d^2 y}{dx^2} + \lambda y = 0$$

with \( y(0) = 0 \) and \( y(\ell) = 0 \).

Step 1: Rewrite in normal form.

$$y”(x) = -\lambda y(x)$$

Step 2: Integrate once from 0 to \( x \).

$$\int_0^x y”(x) \, dx = -\lambda \int_0^x y(x) \, dx$$

$$y'(x) – y'(0) = -\lambda \int_0^x y(x) \, dx$$

Step 3: Handle the unknown \( y'(0) \).

Here’s the difference from IVPs. We don’t know \( y'(0) \). So we introduce a constant:

Let \( y'(0) = c \) (unknown for now).

$$y'(x) = c – \lambda \int_0^x y(x) \, dx$$

Step 4: Integrate again from 0 to \( x \).

$$\int_0^x y'(x) \, dx = c \int_0^x dx – \lambda \int_0^x \left( \int_0^x y(x) \, dx \right) dx$$

$$y(x) – y(0) = cx – \lambda \int_0^x (x – t) y(t) \, dt$$

Using \( y(0) = 0 \):

$$y(x) = cx – \lambda \int_0^x (x – t) y(t) \, dt$$

Step 5: Use the second boundary condition to find \( c \).

Set \( x = \ell \) and use \( y(\ell) = 0 \):

$$0 = c\ell – \lambda \int_0^\ell (\ell – t) y(t) \, dt$$

Solve for \( c \):

$$c = \frac{\lambda}{\ell} \int_0^\ell (\ell – t) y(t) \, dt$$

Step 6: Substitute back and simplify.

$$y(x) = \frac{\lambda x}{\ell} \int_0^\ell (\ell – t) y(t) \, dt – \lambda \int_0^x (x – t) y(t) \, dt$$

After some algebra (splitting the first integral into two parts), this simplifies to:

$$y(x) = \lambda \int_0^\ell K(x, t) y(t) \, dt$$

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

$$K(x, t) = \begin{cases} \dfrac{t(\ell – x)}{\ell} & \text{if } 0 < t < x \\[1em] \dfrac{x(\ell – t)}{\ell} & \text{if } x < t < \ell \end{cases}$$

This is a Fredholm integral equation of the second kind. Both limits are constants (0 and \( \ell \)). The kernel is piecewise-defined, which is typical for BVP conversions.

Pro Tip: This kernel is symmetric: \( K(x, t) = K(t, x) \). That’s not a coincidence. It’s a consequence of the self-adjoint nature of the original differential operator.

The Complete Strategy

Here’s the step-by-step method for converting any second-order ODE into an integral equation:

1. Write the ODE and boundary conditions. Know whether you have an IVP or BVP.
2. Rewrite in normal form. Isolate the highest derivative: \( y” = f(x, y, y’) \).
3. Integrate once from 0 to \( x \). Apply the fundamental theorem of calculus to the left side.
4. Apply known boundary conditions. For IVPs, substitute \( y'(0) \). For BVPs, introduce \( y'(0) = c \).
5. Integrate again from 0 to \( x \). Use the repeated integral formula to convert double integrals.
6. Apply remaining boundary conditions. For BVPs, use the second condition to solve for \( c \).
7. Simplify. Change integration variables to \( t \), combine terms, identify the kernel.

Red Flag: If you end up with arbitrary constants that can’t be eliminated, you’ve probably missed a boundary condition or made an integration error. Go back and check.

The Repeated Integral Formula

This formula is crucial. You’ll use it every time.

For integrating twice:

$$\int_0^x \int_0^s f(t) \, dt \, ds = \int_0^x (x – t) f(t) \, dt$$

For integrating \( n \) times:

$$\int_0^x \int_0^{s_{n-1}} \cdots \int_0^{s_1} f(t) \, dt \, ds_1 \cdots ds_{n-1} = \frac{1}{(n-1)!} \int_0^x (x – t)^{n-1} f(t) \, dt$$

This transforms nested integrals (hard to work with) into single integrals (much easier). Memorize it.

Quick Reference

Common Mistakes to Avoid

• Forgetting to change variables. Use \( t \) inside the integral, \( x \) outside. Mixing them up causes confusion.
• Wrong limits. For IVPs from initial point \( a \), integrate from \( a \) to \( x \). Don’t default to 0 if your initial condition is elsewhere.
• Dropping the kernel. The \( (x – t) \) factor in repeated integrals is essential. It’s not just bookkeeping; it encodes how the system responds to inputs at different times.
• Ignoring piecewise kernels. BVP kernels are almost always piecewise. Don’t try to force them into a single formula.

Frequently Asked Questions

Next: Converting Integral Equations Back to Differential Equations