Picard Iteration

For the initial value problem y' = ay, y(0) = 1, Picard iteration constructs an integral sequence of successive approximations. It builds an ODE solution by repeatedly integrating the previous approximation. The process generates the Taylor series for the exact solution.

Worked Example

Consider the problem y' = ay, with initial condition y₀(t) = 1.

The Picard iteration formula is y_n+1(t) = 1 + ∫₀^t a y_n(s) ds.

n=0: y₀(t) = 1
n=1: y₁(t) = 1 + ∫₀^t a(1) ds = 1 + at
n=2: y₂(t) = 1 + ∫₀^t a(1 + as) ds = 1 + at + (at)²/2

By induction, the n-th iterate is y_n(t) = Σ_j=0ⁿ (at)^j/j!. The exact solution is the limit as n → ∞, which is exp(at).

The ODE residual for the n-th iterate is y_n'(t) - a y_n(t) = -a(at)ⁿ/n!. A finite iterate is not an exact ODE solution even when the initial condition agrees. For example, if a = 1 and t = 1, then the second iterate (n=2) gives y₂(1) = 1 + 1 + 1/2 = 2.5, which is an approximation to the exact solution e ≈ 2.7182818. The absolute error is approximately 0.2182818. The ODE residual is -1(1)²/2 = -0.5.

Experiment

Use the Picard Iteration module to visualize successive integral iterates.

Set Rate a to 1 and Time t to 1. Observe that for Iteration n = 2, the Approximation is 2.5 and ODE residual is -0.5.
Change Iteration n to 6. Observe that the Approximation improves to ~7.355 and absolute error drops significantly.

Limitations

While Picard iteration demonstrates the existence and uniqueness of solutions theoretically, it is rarely used as a practical numerical method due to the difficulty of symbolically integrating arbitrary functions repeatedly. The visual domain here is bounded to show the properties of polynomial approximation, not a general differential equation solver.

References

See more about differential equations and Picard iteration at the following source:

J. Lebl, Notes on Diffy Qs

Picard Iteration