Nonlinear Systems

Figure 7: A pair of equations.

• A pair of equations:
  

  $x^2 + y^2 = 4$

  $e^x + y = 1$

• Graphically, the solution to this system is represented by the intersections of the circle $x^2 + y^2 = 4$ with the curve $y = 1 -e^x$ (see Fig. 7)

• Newton's method can be applied to systems as well as to a single nonlinear equation. We begin with the forms

  $f(x, y)=0$,

  $g(x, y)= 0$

• Let
  
  $$x = r, y=s$$
  
  
  be a root.

• Expand both functions as a Taylor series about the point $(x_i,y_i)$ in terms of
  
  $$(r - x_i), (s - y_i)$$
  
  
  where $(x_i,y_i)$ is a point near the root:
• Taylor series expansion of functions;

$f(r,s)=0=f(x_i,y_i)+f_x(x_i,y_i)(r-x_i)+f_y(x_i,y_i)(s-y_i)+\ldots$

$g(r,s)=0=g(x_i,y_i)+g_x(x_i,y_i)(r-x_i)+g_y(x_i,y_i)(s-y_i)+\ldots$

$\bullet$ Truncating both series gives

$0=f(x_i,y_i)+f_x(x_i,y_i)(r-x_i)+f_y(x_i,y_i)(s-y_i)$

$0=g(x_i,y_i)+g_x(x_i,y_i)(r-x_i)+g_y(x_i,y_i)(s-y_i)$

$\bullet$ which we can rewrite as

$f_x(x_i,y_i)\Delta x_i+f_y(x_i,y_i)\Delta y_i=-f(x_i,y_i)$

$g_x(x_i,y_i)\Delta x_i+g_y(x_i,y_i)\Delta y_i=-g(x_i,y_i)$

• where $\Delta x_i$ and $\Delta y_i$ are used as increments to $x_i$ and $y_i$;
  
  $$x_{i+1}=x_i+\Delta x_i$$
  
  
  
  
  $$y_{i+1}=y_i+\Delta y_i$$
  
  
  are improved estimates of the $(x,y)$ values.
• We repeat this until both $f(x, y)$ and $g(x,y)$ are close to zero.

Example:

$f(x, y) = 4 - x^2-y^2=0$

$g(x, y) = 1 -e^x - y =0$

The partial derivatives are

$$f_x=-2x,f_y=-2y,$$

$$g_x=-e^x,g_y=-1$$

$\bullet$ Beginning with $x_0 = 1, y_0 = - 1.7$, we solve

$-2\Delta x_0+3.4\Delta y_0=-0.1100$

$-2.7183 \Delta x_0 - 1.0 \Delta y_0 = 0.0183$

$\bullet$ This gives

$\Delta x_0=0.0043$,

$\Delta y_0=-0.0298$

$\bullet$ from which

$x_1 = 1.0043$,

$y_1=-1.7298$.

• These agree with the true value within 2 in the fourth decimal place. Repeating the process once more:

  $x_2 = 1.004169$,

  $y_2 = -1.729637$.

  Then,

  f(1.004169,-1.729637)=-0.0000001,

  g(1.004169,-1.729637)=-0.00000001,