Nonlinear Systems

Figure 5: 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. 5)
• Newton's method can be applied to systems as well as to a single nonlinear equation. We begin with the forms
  
  $$f(x, y)=O, g(x, y)= O$$
  
  

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

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

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

  
  
  Truncating both series gives
  
  

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

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

  
  
  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 $y_i$ so that $x_{i+1}=x_i+\Delta x_i$ and $y_{i+1}=y_i+\Delta y_i$ 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
  
  $$f_x=-2x,f_y=-2y,g_x=-e^x,g_y=-1$$
  
  
  Beginning with $x_0 = 1, y_0 = - 1.7$, we solve
  
  $$-2\Delta x_0+3.4\Delta y_0=-0.1100$$
  
  
  
  
  $$-2.7183 \Delta x_0 - 1.0 \Delta y_0 = 0.0183$$
  
  

• This gives $\Delta x_0=0.0043,\Delta y_0=-0.0298$, 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 produces $x_2 = 1.004169, y_2 = -1.729637$. The function values at this second iteration are approximately -0.000000l and -0.00000001.