A nonlinear system of n n n n n n x 1 , x 2 , … , x n x_1, x_2, \ldots, x_n x 1 , x 2 , … , x n
f 1 ( x 1 , x 2 , … , x n ) = 0 f 2 ( x 1 , x 2 , … , x n ) = 0 ⋮ f n ( x 1 , x 2 , … , x n ) = 0 \begin{aligned}
f_1(x_1, x_2, \ldots, x_n) &= 0 \\
f_2(x_1, x_2, \ldots, x_n) &= 0 \\
&\;\vdots \\
f_n(x_1, x_2, \ldots, x_n) &= 0
\end{aligned} f 1 ( x 1 , x 2 , … , x n ) f 2 ( x 1 , x 2 , … , x n ) f n ( x 1 , x 2 , … , x n ) = 0 = 0 ⋮ = 0 where each f i f_i f i x ⃗ = ( x 1 , x 2 , … , x n ) T ∈ R n \vec{x} = (x_1, x_2, \ldots, x_n)^T \in \mathbb{R}^n x = ( x 1 , x 2 , … , x n ) T ∈ R n R \mathbb{R} R
In vector form, F ( x ⃗ ) = 0 ⃗ F(\vec{x}) = \vec{0} F ( x ) = 0
F ( x ⃗ ) = ( f 1 ( x ⃗ ) , f 2 ( x ⃗ ) , … , f n ( x ⃗ ) ) T F(\vec{x}) = \big(f_1(\vec{x}), f_2(\vec{x}), \ldots, f_n(\vec{x})\big)^T F ( x ) = ( f 1 ( x ) , f 2 ( x ) , … , f n ( x ) ) T Here:
F F F R n \mathbb{R}^n R n R n \mathbb{R}^n R n f 1 , f 2 , … , f n f_1, f_2, \ldots, f_n f 1 , f 2 , … , f n F F F
Continuity from Bounded Partials
Theorem. Let f f f D ⊆ R n D \subseteq \mathbb{R}^n D ⊆ R n R \mathbb{R} R x ⃗ 0 ∈ D \vec{x}_0 \in D x 0 ∈ D f f f δ > 0 \delta > 0 δ > 0 K > 0 K > 0 K > 0 ∥ x ⃗ − x ⃗ 0 ∥ < δ \|\vec{x} - \vec{x}_0\| < \delta ∥ x − x 0 ∥ < δ x ⃗ ∈ D \vec{x} \in D x ∈ D
∣ ∂ f ( x ⃗ ) ∂ x j ∣ ≤ K , j = 1 , 2 , … , n \left| \frac{\partial f(\vec{x})}{\partial x_j} \right| \le K, \quad j = 1, 2, \ldots, n ∂ x j ∂ f ( x ) ≤ K , j = 1 , 2 , … , n Then f f f x ⃗ 0 \vec{x}_0 x 0
Example
Place the system
3 x 1 − cos ( x 2 x 3 ) − 1 2 = 0 x 1 2 − 81 ( x 2 + 0.1 ) 2 + sin x 3 + 1.06 = 0 e − x 1 x 2 + 20 x 3 + 10 π − 3 3 = 0 \begin{aligned}
3x_1 - \cos(x_2 x_3) - \tfrac{1}{2} &= 0 \\
x_1^2 - 81(x_2 + 0.1)^2 + \sin x_3 + 1.06 &= 0 \\
e^{-x_1 x_2} + 20x_3 + \tfrac{10\pi - 3}{3} &= 0
\end{aligned} 3 x 1 − cos ( x 2 x 3 ) − 2 1 x 1 2 − 81 ( x 2 + 0.1 ) 2 + sin x 3 + 1.06 e − x 1 x 2 + 20 x 3 + 3 10 π − 3 = 0 = 0 = 0 in the form F ( x ⃗ ) = 0 ⃗ F(\vec{x}) = \vec{0} F ( x ) = 0
The coordinate functions are
f 1 ( x 1 , x 2 , x 3 ) = 3 x 1 − cos ( x 2 x 3 ) − 1 2 f 2 ( x 1 , x 2 , x 3 ) = x 1 2 − 81 ( x 2 + 0.1 ) 2 + sin x 3 + 1.06 f 3 ( x 1 , x 2 , x 3 ) = e − x 1 x 2 + 20 x 3 + 10 π − 3 3 \begin{aligned}
f_1(x_1, x_2, x_3) &= 3x_1 - \cos(x_2 x_3) - \tfrac{1}{2} \\
f_2(x_1, x_2, x_3) &= x_1^2 - 81(x_2 + 0.1)^2 + \sin x_3 + 1.06 \\
f_3(x_1, x_2, x_3) &= e^{-x_1 x_2} + 20x_3 + \tfrac{10\pi - 3}{3}
\end{aligned} f 1 ( x 1 , x 2 , x 3 ) f 2 ( x 1 , x 2 , x 3 ) f 3 ( x 1 , x 2 , x 3 ) = 3 x 1 − cos ( x 2 x 3 ) − 2 1 = x 1 2 − 81 ( x 2 + 0.1 ) 2 + sin x 3 + 1.06 = e − x 1 x 2 + 20 x 3 + 3 10 π − 3 so F ( x ⃗ ) = ( f 1 ( x ⃗ ) , f 2 ( x ⃗ ) , f 3 ( x ⃗ ) ) T F(\vec{x}) = \big(f_1(\vec{x}), f_2(\vec{x}), f_3(\vec{x})\big)^T F ( x ) = ( f 1 ( x ) , f 2 ( x ) , f 3 ( x ) ) T
Methods
Numerical methods for solving nonlinear systems generalize the single-variable case.
Fixed-Point Iteration
Rewrites F ( x ⃗ ) = 0 ⃗ F(\vec{x}) = \vec{0} F ( x ) = 0 x ⃗ = G ( x ⃗ ) \vec{x} = G(\vec{x}) x = G ( x ) x ⃗ ( k ) = G ( x ⃗ ( k − 1 ) ) \vec{x}^{(k)} = G(\vec{x}^{(k-1)}) x ( k ) = G ( x ( k − 1 ) ) fixed point .
Newton’s Method
Uses the Jacobian matrix of F F F F ( x ⃗ ) = 0 ⃗ F(\vec{x}) = \vec{0} F ( x ) = 0 Newton’s method .