Solve following system of nonlinear equations (Newton method diverges)

newton raphsonnonlinear systemsystems of equations

I'm interested into a method that would solve a system of 4 equations in the following form:

$$
\frac{1}{y} + \frac{6}{z} = w,\\
\frac{3}{y} + \frac{10}{z} = x, \\
\frac{3}{w} + \frac{2.5}{x} = y, \\
\frac{1}{w} + \frac{17.5}{x} = z \\
$$

Note that the solution is (2, 5, 2, 4). However, I am doing a research project that will require me to calculate the general form of this system of equations with parameters as constants in the numerators (so the current numbers are just placeholders).

I tried substitution, transformations, and a Newton-Rapshon algorithm in R (which unfortunately diverged).

Does anyone know of a way to compute a system of equations in this form? Thanks!

Best Answer

Your $(2, 5, 2, 4)$ is not a solution: it doesn't satisfy the second equation.

Substituting $w$ and $x$ from the first two equations into the last two and simplifying gives you $$ \eqalign{-{\frac {y \left( 120\,{y}^{2}-34\,yz-17\,{z}^{2} \right) }{ \left( 2\,z+12\,y \right) \left( 3\,z+10\,y \right) }}&=0\cr {\frac {z \left( 110\,{y}^{2}-15\,yz-6\,{z}^{2} \right) }{ \left( 2\,z +12\,y \right) \left( 3\,z+10\,y \right) }}&=0\cr}$$ Of course $y$ and $z$ can't be $0$, so that leaves you with $$ \eqalign{ 120\,{y}^{2}-34\,yz-17\,{z}^{2} &= 0\cr 110\,{y}^{2}-15\,yz-6\,{z}^{2} &= 0\cr}$$ Each of these is the equation of two lines through the origin in the $yz$ plane, and the slopes are all different, so the intersection is only $(0,0)$, which is not a solution of the original system. Thus your system has no solutions.

Related Solutions

[Math] Constrained Newton-Raphson method

Put:

$$X = \log\left(\frac{\alpha}{\alpha + \beta}\right) = \log\left(\frac{1}{1 + \frac{\beta}{\alpha}}\right)$$

Then working in terms of X will ensure that the logarithm is always well defined. You then need to define another independent variable Y such that a linarization of Y in terms of small changes in $\alpha$ and $\beta$ does not become almost linearly dependent on the way the change in X depends on small changes in $\alpha$ and $\beta$. Since X depends on the ratio of $\alpha$ and $\beta$, you can choose Y to be a function of the product of $\alpha$ and $\beta$, so:

$$Y = \alpha\beta$$

might work well, at least you'll have elminated two potential problems with Newton-Raphson.

[Math] How to symbolically solve a system of linear equations

One way is to use the free Maxima computer algebra system.

You can enter your matrix like this:

A: matrix(
 [1,1,1], 
 [(3/4)*a+(1/4)*b,(1/2)*a+(1/2)*b,(1/4)*a+(3/4)*b], 
 [((3/4)*a+(1/4)*b)^2,((1/2)*a+(1/2)*b)^2,((1/4)*a+(3/4)*b)^2]
 );

$$ \begin{pmatrix}1 & 1 & 1\\ \frac{b}{4}+\frac{3a}{4} & \frac{b}{2}+\frac{a}{2} & \frac{3b}{4}+\frac{a}{4}\\ {{\left( \frac{b}{4}+\frac{3a}{4}\right) }^{2}} & {{\left( \frac{b}{2}+\frac{a}{2}\right) }^{2}} & {{\left( \frac{3b}{4}+\frac{a}{4}\right) }^{2}}\end{pmatrix} $$
and the result vector like that:

bcol: matrix(
 [b-a], 
 [(b^2-a^2)/2], 
 [(b^3-a^3)/3]
);

$$ \begin{pmatrix}b-a\\ \frac{{{b}^{2}}-{{a}^{2}}}{2}\\ \frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix} $$ This command will solve your linear system by $LU$ decomposition:

ls : linsolve_by_lu(A,bcol);

You extract and simplify the solution vector like this:

xycol: ratsimp(first(ls));

$$ \begin{pmatrix}\frac{2b-2a}{3}\\ -\frac{b-a}{3}\\ \frac{2b-2a}{3}\end{pmatrix} $$ where first picks the first element of the result and ratsimp is one of Maxima's simplification functions.

With this you can verify the result:

ratsimp(A . xycol);

$$ \begin{pmatrix}b-a\\ \frac{{{b}^{2}}-{{a}^{2}}}{2}\\ \frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix} $$ Note that Maxima uses a dot for the matrix multiplication operator.

Best Answer

Related Solutions

[Math] Constrained Newton-Raphson method

[Math] How to symbolically solve a system of linear equations

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