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Here’s a clean and structured Markdown version of your handwritten lecture notes on Perturbation Methods for Algebraic Equations , preserving all key points, mathematical expressions, and logical flow.
Part I: Perturbation Methods for Algebraic Equations
I. Series Method
Example 1:
x 2 + ε x − 1 = 0 ⇒ x = − 1 2 ε ± 1 + 1 4 ε 2 = x ( ε ) x^2 + \varepsilon x - 1 = 0
\Rightarrow x = -\frac{1}{2}\varepsilon \pm \sqrt{1 + \frac{1}{4}\varepsilon^2} = x(\varepsilon) x 2 + ε x − 1 = 0 ⇒ x = − 2 1 ε ± 1 + 4 1 ε 2 = x ( ε ) Exact solution:
x = { 1 − 1 2 ε + 1 8 ε 2 − 1 128 ε 4 + O ( ε 6 ) − 1 − 1 2 ε − 1 8 ε 2 + 1 128 ε 4 + O ( ε 6 ) x =
\begin{cases}
1 - \frac{1}{2}\varepsilon + \frac{1}{8}\varepsilon^2 - \frac{1}{128}\varepsilon^4 + O(\varepsilon^6) \\
-1 - \frac{1}{2}\varepsilon - \frac{1}{8}\varepsilon^2 + \frac{1}{128}\varepsilon^4 + O(\varepsilon^6)
\end{cases} x = { 1 − 2 1 ε + 8 1 ε 2 − 128 1 ε 4 + O ( ε 6 ) − 1 − 2 1 ε − 8 1 ε 2 + 128 1 ε 4 + O ( ε 6 ) (1) Expansion Method
Assume expansion of x ( ε ) x(\varepsilon) x ( ε ) x = 1 x = 1 x = 1
x ( ε ) = 1 + ε x 1 + ε 2 x 2 + ⋯ x(\varepsilon) = 1 + \varepsilon x_1 + \varepsilon^2 x_2 + \cdots x ( ε ) = 1 + ε x 1 + ε 2 x 2 + ⋯ Substitute back into original equation:
x 2 = 1 + 2 ε x 1 + ( ε 2 x 1 2 + 2 x 2 ) ε 2 + ⋯ ε x = ε + ε 2 x 1 + ε 3 x 2 + ⋯ x^2 = 1 + 2\varepsilon x_1 + (\varepsilon^2 x_1^2 + 2x_2)\varepsilon^2 + \cdots \\
\varepsilon x = \varepsilon + \varepsilon^2 x_1 + \varepsilon^3 x_2 + \cdots x 2 = 1 + 2 ε x 1 + ( ε 2 x 1 2 + 2 x 2 ) ε 2 + ⋯ ε x = ε + ε 2 x 1 + ε 3 x 2 + ⋯ So the equation becomes:
x 2 + ε x − 1 = 0 ⇒ (Collect terms by powers of ε ) x^2 + \varepsilon x - 1 = 0
\Rightarrow
\text{(Collect terms by powers of } \varepsilon) x 2 + ε x − 1 = 0 ⇒ (Collect terms by powers of ε ) At ε = 0 \varepsilon = 0 ε = 0
O ( ε 0 ) : 1 − 1 = 0 (OK) O(\varepsilon^0): \quad 1 - 1 = 0 \quad \text{(OK)} O ( ε 0 ) : 1 − 1 = 0 (OK) O ( ε 1 ) O(\varepsilon^1) O ( ε 1 )
2 x 1 − 1 = 0 ⇒ x 1 = 1 2 2x_1 - 1 = 0 \Rightarrow x_1 = \frac{1}{2} 2 x 1 − 1 = 0 ⇒ x 1 = 2 1 O ( ε 2 ) O(\varepsilon^2) O ( ε 2 )
x 1 2 + 2 x 2 + x 1 = 0 ⇒ ( 1 2 ) 2 + 2 x 2 + 1 2 = 0 ⇒ x 2 = − 1 8 x_1^2 + 2x_2 + x_1 = 0 \Rightarrow \left(\frac{1}{2}\right)^2 + 2x_2 + \frac{1}{2} = 0 \Rightarrow x_2 = -\frac{1}{8} x 1 2 + 2 x 2 + x 1 = 0 ⇒ ( 2 1 ) 2 + 2 x 2 + 2 1 = 0 ⇒ x 2 = − 8 1 Thus,
x ( ε ) = 1 + 1 2 ε − 1 8 ε 2 + ⋯ x(\varepsilon) = 1 + \frac{1}{2}\varepsilon - \frac{1}{8}\varepsilon^2 + \cdots x ( ε ) = 1 + 2 1 ε − 8 1 ε 2 + ⋯
Note: This matches the series expansion up to ε 2 \varepsilon^2 ε 2
(2) Iterative Method (Fixed Point)
Given f ( x ) = 0 ⇔ x = g ( x ) f(x) = 0 \Leftrightarrow x = g(x) f ( x ) = 0 ⇔ x = g ( x )
Iterative scheme:
x n + 1 = g ( x n ) x_{n+1} = g(x_n) x n + 1 = g ( x n ) Recall: Converges when x ∈ [ a , b ] x \in [a,b] x ∈ [ a , b ] ∣ g ′ ( x ) ∣ ≤ L < 1 |g'(x)| \leq L < 1 ∣ g ′ ( x ) ∣ ≤ L < 1
For our example:
x = 1 − ε x = g ( x ) ⇒ ∣ g ′ ( x ) ∣ = ∣ ε 2 ⋅ 1 1 − ε x ∣ < 1 x = \sqrt{1 - \varepsilon x} = g(x)
\Rightarrow
|g'(x)| = \left| \frac{\varepsilon}{2} \cdot \frac{1}{\sqrt{1 - \varepsilon x}} \right| < 1 x = 1 − ε x = g ( x ) ⇒ ∣ g ′ ( x ) ∣ = 2 ε ⋅ 1 − ε x 1 < 1 This implies convergence if x > 1 2 − 1 4 ε x > \frac{1}{2} - \frac{1}{4}\varepsilon x > 2 1 − 4 1 ε
Choose initial guess x 0 = 1 x_0 = 1 x 0 = 1
x 1 = 1 − ε x 0 = 1 − ε ≈ 1 − 1 2 ε − 1 8 ε 2 x_1 = \sqrt{1 - \varepsilon x_0} = \sqrt{1 - \varepsilon} \approx 1 - \frac{1}{2}\varepsilon - \frac{1}{8}\varepsilon^2 x 1 = 1 − ε x 0 = 1 − ε ≈ 1 − 2 1 ε − 8 1 ε 2 x 2 = 1 − 1 2 ε + 1 8 ε 2 + O ( ε 3 ) x_2 = 1 - \frac{1}{2}\varepsilon + \frac{1}{8}\varepsilon^2 + O(\varepsilon^3) x 2 = 1 − 2 1 ε + 8 1 ε 2 + O ( ε 3 ) x 3 = 1 − 1 2 ε + 1 8 ε 2 − 1 128 ε 4 + ⋯ x_3 = 1 - \frac{1}{2}\varepsilon + \frac{1}{8}\varepsilon^2 - \frac{1}{128}\varepsilon^4 + \cdots x 3 = 1 − 2 1 ε + 8 1 ε 2 − 128 1 ε 4 + ⋯
Observation: Each iteration adds one more accurate term in the perturbation series.
Compare with exact solution:
1 − ε = 1 − 1 2 ε − 1 8 ε 2 − 1 16 ε 3 + ⋯ \sqrt{1 - \varepsilon} = 1 - \frac{1}{2}\varepsilon - \frac{1}{8}\varepsilon^2 - \frac{1}{16}\varepsilon^3 + \cdots 1 − ε = 1 − 2 1 ε − 8 1 ε 2 − 16 1 ε 3 + ⋯ So:
x 1 x_1 x 1 ε \varepsilon ε ε 2 \varepsilon^2 ε 2 x 2 x_2 x 2 ε 2 \varepsilon^2 ε 2 ε 3 \varepsilon^3 ε 3 …
✅ Key Insight : For each iteration, we gain one more accurate term in the asymptotic expansion.
Example 2: Singular Perturbation
Equation:
ε 3 x 2 + x − 1 = 0 \varepsilon^3 x^2 + x - 1 = 0 ε 3 x 2 + x − 1 = 0 True solution:
x = − 1 2 ε + 1 + 4 ε 2 ε = { 1 − ε + 2 ε 2 − 5 ε 3 + ⋯ − 1 ε − 1 + ε − 2 ε 2 + 5 ε 3 + ⋯ x = -\frac{1}{2\varepsilon} + \frac{\sqrt{1 + 4\varepsilon}}{2\varepsilon}
=
\begin{cases}
1 - \varepsilon + 2\varepsilon^2 - 5\varepsilon^3 + \cdots \\
-\frac{1}{\varepsilon} - 1 + \varepsilon - 2\varepsilon^2 + 5\varepsilon^3 + \cdots
\end{cases} x = − 2 ε 1 + 2 ε 1 + 4 ε = { 1 − ε + 2 ε 2 − 5 ε 3 + ⋯ − ε 1 − 1 + ε − 2 ε 2 + 5 ε 3 + ⋯ At ε = 0 \varepsilon = 0 ε = 0 x = 1 x = 1 x = 1
But note: The second branch has O ( 1 / ε ) \mathcal{O}(1/\varepsilon) O ( 1/ ε ) singular perturbation problem.
(1) Iterative Method Attempt
From:
ε x 2 + x − 1 = 0 ⇒ x 2 = 1 − x ε ⇒ x = 1 ε x − 1 ε = g ( x ) \varepsilon x^2 + x - 1 = 0 \Rightarrow x^2 = \frac{1 - x}{\varepsilon}
\Rightarrow x = \frac{1}{\varepsilon x} - \frac{1}{\varepsilon} = g(x) ε x 2 + x − 1 = 0 ⇒ x 2 = ε 1 − x ⇒ x = ε x 1 − ε 1 = g ( x ) Wait — better form:
x = 1 ε x − 1 ε = g ( x ) x = \frac{1}{\varepsilon x} - \frac{1}{\varepsilon} = g(x) x = ε x 1 − ε 1 = g ( x ) Then:
∣ g ′ ( x ) ∣ = ∣ 1 ε ⋅ ( − 1 x 2 ) ∣ = 1 ε x 2 ⇒ ∣ g ′ ( x ) ∣ < 1 ⟺ ∣ x ∣ > 1 ε |g'(x)| = \left| \frac{1}{\varepsilon} \cdot \left( -\frac{1}{x^2} \right) \right| = \frac{1}{\varepsilon x^2}
\Rightarrow |g'(x)| < 1 \iff |x| > \frac{1}{\sqrt{\varepsilon}} ∣ g ′ ( x ) ∣ = ε 1 ⋅ ( − x 2 1 ) = ε x 2 1 ⇒ ∣ g ′ ( x ) ∣ < 1 ⟺ ∣ x ∣ > ε 1 Summary
Method Applicability Notes Expansion Method Regular perturbations Assumes solution is analytic in ε \varepsilon ε Iterative Method Fixed-point convergence Requires g ′ ( x ) ≤ L < 1 g'(x) \leq L < 1 g ′ ( x ) ≤ L < 1 Singular Case When ε → 0 \varepsilon \to 0 ε → 0 Standard methods fail; need multiple scales
🔍 Key Takeaway :
Regular perturbation works well when x ( ε ) x(\varepsilon) x ( ε ) ε → 0 \varepsilon \to 0 ε → 0
Singular perturbations involve boundary layers or non-uniform behavior — require advanced techniques.
Iterative methods are useful numerically and can recover perturbation series term-by-term.
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