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Common Pitfalls | Mathematics - Wyatt's Notes

Problem. Show that (0,0)(0, 0) is asymptotically stable for the system x=xy3x' = -x - y^3, y=x3yy' = x^3 - y.

Solution

Solution. Critical point: xy3=0-x - y^3 = 0 and x3y=0x^3 - y = 0 gives x=0,y=0x = 0, y = 0.

Choose the Lyapunov function V(x,y)=x22+y22V(x, y) = \frac{x^2}{2} + \frac{y^2}{2}. V(0,0)=0V(0,0) = 0 and V>0V > 0 for (x,y)(0,0)(x, y) \neq (0, 0).

V˙=Vxx+Vyy=x(xy3)+y(x3y)\dot{V} = \frac{\partial V}{\partial x}x' + \frac{\partial V}{\partial y}y' = x(-x - y^3) + y(x^3 - y)

=x2xy3+x3yy2=x2y2+xy(x2y2)= -x^2 - xy^3 + x^3 y - y^2 = -x^2 - y^2 + xy(x^2 - y^2)

This is not negative definite. Let us try V(x,y)=x44+y44V(x, y) = \frac{x^4}{4} + \frac{y^4}{4}:

V˙=x3(xy3)+y3(x3y)=x4x3y3+x3y3y4=(x4+y4)\dot{V} = x^3(-x - y^3) + y^3(x^3 - y) = -x^4 - x^3 y^3 + x^3 y^3 - y^4 = -(x^4 + y^4).

Since V˙=(x4+y4)<0\dot{V} = -(x^4 + y^4) \lt 0 for (x,y)(0,0)(x, y) \neq (0, 0)The origin is asymptotically Stable by Lyapunov’s theorem. \blacksquare

9.10 Worked Example: Van der Pol Oscillator

Section titled “9.10 Worked Example: Van der Pol Oscillator”

Problem. Show that the Van der Pol equation x+μ(x21)x+x=0x'' + \mu(x^2 - 1)x' + x = 0 with μ>0\mu > 0 Has a unique limit cycle.

Solution

_Solution.* Write as a system: x=yx' = y, y=xμ(x21)yy' = -x - \mu(x^2 - 1)y.

The only critical point is (0,0)(0, 0). The Jacobian at the origin is J=(011μ)J = \begin{pmatrix} 0 & 1 \\ -1 & \mu \end{pmatrix}.

tr(J)=μ>0\mathrm{tr}(J) = \mu > 0, det(J)=1>0\det(J) = 1 > 0, τ24Δ=μ24\tau^2 - 4\Delta = \mu^2 - 4.

For 0<μ<20 \lt \mu \lt 2: complex eigenvalues with positive real part (unstable spiral). For μ2\mu \geq 2: two positive real eigenvalues (unstable node).

The origin is always unstable. To show existence of a limit cycle, we use a Liénard’s theorem Or construct a trapping region. Define

L(x)=x3/3xL(x) = x^3/3 - x and write the system as x=yx' = y, y=xμL(x)yy' = -x - \mu L'(x) y.

The function L(x)L(x) has zeros at x=±1x = \pm 1. For μ>0\mu > 0The damping is negative for x<1\lvert x \rvert \lt 1 (energy input) and positive for x>1\lvert x \rvert > 1 (energy Dissipation). This creates a unique stable limit cycle that passes through x=±1x = \pm 1.

A detailed …/1-number-and-algebra/3_proof-and-logic requires constructing an annular region and applying the Poincaré-Bendixson Theorem (the inner boundary encloses the unstable origin; the outer boundary is chosen so that Trajectories point inward). \blacksquare

ODE TypeRecommended MethodWatch out for
SeparableSeparation of variablesMissing h(y)=0h(y) = 0 solutions
First-order linearIntegrating factorCorrect sign in exponent
Constant-coeff linearUndetermined coefficientsOverlap with homogeneous solution
Cauchy-Eulerxrx^r ansatzlnx\ln x terms for repeated roots
Systems near critical ptLinearisationNon-hyperbolic points
Nonlinear / no closed formNumerical (RK4)Step size selection
  • Assuming series solutions always converge everywhere. Frobenius series converge only within the radius of convergence determined by the nearest singularity. For y+p(x)y+q(x)y=0y'' + p(x)y' + q(x)y = 0, the series converges at least up to the nearest singularity of p(x)p(x) or q(x)q(x) in the complex plane, not necessarily for all xx.

  • Confusing the order of a pole in the complex plane with the order of a singular point in an ODE. A regular singular point is one where (xx0)p(x)(x-x_0)p(x) and (xx0)2q(x)(x-x_0)^2q(x) are analytic. An irregular singular point has higher-order singularities. These are unrelated to the order of poles in complex analysis.

  • Forgetting to transform boundary conditions when using Laplace transforms. The Laplace transform incorporates initial conditions at t=0t = 0. If the problem has boundary conditions at t=0t = 0 and t=Lt = L, the Laplace transform may not be the right tool — consider using Fourier series or separation of variables instead.

  • Applying the method of undetermined coefficients when the forcing term is not of the right form. The method works only for forcing terms that are polynomials, exponentials, sines, cosines, or products thereof. For forcing terms like tant\tan t or 1/t1/t, use variation of parameters instead.

  • Ignoring the transient solution in systems. The general solution to a linear ODE is the sum of the complementary solution (transient) and particular solution (steady-state). When studying long-term behaviour, focus on the particular solution, but remember that the transient may be significant at early times.

flowchart TD
A[10_Common Pitfalls] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Differential equations are the language of change. Every pitfall in this file reflects a mismatch between the equation’s structure and the method applied to it. The particular integral must complement, not duplicate, the complementary solution, just as a resonant driving force must be handled differently from a non-resonant one. Separable equations can miss equilibrium solutions because dividing by zero eliminates them. The deeper pattern is that ODE methods are not universal recipes: they work because the equation has specific structural properties, and ignoring those properties produces nonsense.