Intermediate · 10 minute learning note

Variational iteration method

Use the equation’s residual to improve an initial guess.

u(0.5) = 2/3THE MODEL IN THIS NOTEu(t) = 1 / (1 + t)10.51tuEXACT REFERENCE · u′ + u² = 0, u(0) = 1
The exact reference solution used throughout the learning notes.

The idea, in plain language.

Construct a correction functional containing the current residual and a Lagrange multiplier. Determine the multiplier from stationarity conditions for the chosen operator, then iterate. For a simple first-order initial-value problem, this can reduce to a familiar integral iteration.

un+1(t)=un(t)−∫0t[un′(s)+un(s)2] ds\begin{aligned}u_{n+1}(t)&=u_n(t)\\&\quad-\int_0^t[u_n^{\prime}(s)+u_n(s)^2]\,ds\end{aligned}

Origins & connection to Ganji’s work

Developed by Ji-Huan He, with antecedents in earlier variational correction approaches. Ganji’s work applies and compares VIM; he did not originate it.

Ganji and collaborators investigated VIM alongside other analytical methods for thermal problems and nonlinear evolution equations.

Before you begin: Integration · Initial-value problems · Basic variational reasoning · Residuals.

A method you can follow.

  1. Identify the residualWrite the equation as L(u)+N(u)−f=0 and keep track of the prescribed data.
  2. Derive the multiplierConstruct a correction integral and apply the appropriate stationarity conditions. The multiplier depends on the operator; −1 is correct for this example, not for every equation.
  3. Iterate from an admissible guessStart with u₀=1. When each iterate satisfies uₙ(0)=1, the correction becomes uₙ₊₁(t)=1−∫₀ᵗuₙ(s)² ds.
  4. Test convergenceCheck differences between iterates together with the governing-equation residual and an independent reference.
Worked example

Two corrections to nonlinear decay

u′ + u² = 0, u(0) = 1Exact reference: u(t)=1/(1+t)
  1. Choose u₀(t)=1 and the first-order multiplier λ=−1.
  2. The first correction is u₁(t)=1−∫₀ᵗ1 ds=1−t.
  3. The second is u₂(t)=1−∫₀ᵗ(1−s)² ds=1−t+t²−t³/3.
  4. Each iterate satisfies uₙ(0)=1. The iterate index is not the same as the number of retained HPM series terms.

Check the result

At t=0.25, u₂≈0.8072917 and u_exact=0.8.

For this problem VIM reduces to Picard iteration. Different analytical frameworks can lead to the same underlying iterative construction.

Explore the interactive notebook

Know the limits.

  • A multiplier must be derived for the actual linear operator and boundary structure.
  • Convergence of a correction sequence is conditional and may be local.
  • A rapidly changing polynomial degree does not by itself establish greater accuracy.

Where you will encounter it

Heat-transfer equationsNonlinear wave and evolution modelsInitial-value problems

Read the originals.

This is an original educational explanation, not a claim that all illustrated methods were invented by Professor Ganji. The worked model is deliberately shared across notes so the results and limitations can be compared.

Back to all methods