Enter Variational Data
Formula Used
General second variation form
δ²J[η] = ∫ab [A(x)η′(x)² + B(x)η(x)η′(x) + C(x)η(x)²] dx
where A(x) = Fy′y′, B(x) = 2Fyy′, and C(x) = Fyy.
The calculator samples the interval uniformly, approximates η′ with finite differences, and integrates numerically using the trapezoidal rule.
Legendre check: a necessary local minimum condition is A(x) ≥ 0 along the interval. The page reports whether that condition appears strongly satisfied, weakly satisfied, or violated on the sampled grid.
Boundary admissibility: admissible perturbations should satisfy η(a) = 0 and η(b) = 0. The tool checks both numerically using a tolerance scaled to the perturbation magnitude.
How to Use This Calculator
- Enter the interval bounds and the number of grid points.
- Provide the coefficient functions A(x), B(x), and C(x).
- Enter an admissible perturbation η(x) that vanishes at both endpoints.
- Click the compute button to evaluate the sampled second variation.
- Review the total value, component contributions, Legendre status, and plotted traces.
- Export the complete numeric dataset to CSV or save the result section as a PDF.
Example Data Table
Illustrative sample for a = 0, b = 1, A(x) = 1 + x², B(x) = 0.2x, C(x) = 2 + x, and η(x) = x(1 − x).
| x | A(x) | B(x) | C(x) | η(x) | η′(x) | Integrand |
|---|---|---|---|---|---|---|
| 0.00 | 1.0000 | 0.0000 | 2.0000 | 0.0000 | 1.0000 | 1.0000 |
| 0.25 | 1.0625 | 0.0500 | 2.2500 | 0.1875 | 0.5000 | 0.3115 |
| 0.50 | 1.2500 | 0.1000 | 2.5000 | 0.2500 | 0.0000 | 0.1563 |
| 0.75 | 1.5625 | 0.1500 | 2.7500 | 0.1875 | -0.5000 | 0.4521 |
| 1.00 | 2.0000 | 0.2000 | 3.0000 | 0.0000 | -1.0000 | 2.0000 |
Frequently Asked Questions
1. What does this calculator actually measure?
It estimates the second variation of a variational functional for one chosen perturbation. The result helps assess whether the reference path behaves like a local minimum, maximum, or neutral candidate along that perturbation direction.
2. Why must the perturbation vanish at the endpoints?
In fixed endpoint problems, admissible perturbations satisfy η(a) = 0 and η(b) = 0. Without those boundary conditions, the second variation test may not reflect the true constrained variational problem.
3. What is the Legendre condition here?
The Legendre condition requires A(x) = Fy′y′ to remain nonnegative along the interval for a minimum candidate. The calculator checks that numerically on the selected grid and reports the observed status.
4. Does a positive result prove a strict minimum?
Not by itself. A positive sampled value supports local minimality for the tested perturbation, but a full proof may also require broader perturbation families, Jacobi analysis, and attention to conjugate points.
5. What does a negative second variation imply?
A negative value suggests the reference curve is not a strict local minimum against that perturbation. It may indicate instability, a saddle type behaviour, or the need to recheck the coefficient model.
6. Which function syntax is supported?
Use x, numbers, parentheses, commas, and standard functions like sin, cos, tan, exp, sqrt, log, abs, min, and max. Use explicit multiplication, such as 2*x instead of 2x.
7. How accurate is the numerical result?
Accuracy depends on smooth expressions, appropriate grid density, and stable perturbations. Increasing the number of points generally improves the trapezoidal estimate, though very sharp functions may require additional refinement.
8. Why are the component terms shown separately?
Separating the Aη′², Bηη′, and Cη² contributions helps reveal which part of the model drives positivity or negativity. That is especially useful when diagnosing boundary effects or checking coercivity patterns.