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The two dimensional Newton-Raphson method is to find a zero root $(x_0,y_0)$ which satisfy
\begin{array}{*{20}{c}} {f\left( {{x_0},{y_0}} \right) = 0} \\ {g\left( {{x_0},{y_0}} \right) = 0} \end{array} by iteration. However, now I have only one equation $f(x,y)=0$ and I want to find its zero root $(x_0,y_0)$, what improvement should I do to the the Newton-Raphson method?

If I add another equation $g(x,y)=0$, then the determinant of Jacobian matrix will be zero and the inverting of the Jacobian matrix wouldn't exist.

Mhenni Benghorbal
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Pengpeng
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1 Answers1

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$$ {\rm f}\left(\vec{r}\right) + \delta\vec{r}\cdot\nabla{\rm f}\left(\vec{r}\right) \approx 0 $$ $$ {\rm f} + \left(\delta x\quad\delta y\right){{\rm f}_{x} \choose {\rm f}_{y}} \approx 0 $$ $$ {\rm f}\left({\rm f}_{x}\quad{\rm f}_{y}\right) + \left(\delta x\quad\delta y\right) \left({\rm f}_{x}^{2} + {\rm f}_{y}^{2}\right) \approx 0 $$ $$ \left(\delta x\quad\delta y\right) \approx -\, {{\rm f}\left({\rm f}_{x}\quad{\rm f}_{y}\right) \over \left({\rm f}_{x}^{2} + {\rm f}_{y}^{2}\right)} $$ $$ \left\lbrace% \begin{array}{rcl} \delta x & \approx & -\, {{\rm f}\left(\vec{r}\right) \over \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial x}\right\rbrack^{2} + \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial y}\right\rbrack^{2} }\, {\partial{\rm f}\left(\vec{r}\right)\over\partial x} \\[3mm] \delta y & \approx & -\, {{\rm f}\left(\vec{r}\right)\, \over \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial x}\right\rbrack^{2} + \left\lbrack{\partial{\rm f}\left(\vec{r}\right)\over\partial y}\right\rbrack^{2} }\, {\partial{\rm f}\left(\vec{r}\right)\over\partial y} \end{array}\right. $$

Equivalent to $$ \delta\vec{r} \approx -\,{\nabla{\rm f}\left(\vec{r}\right) \over \left\vert\nabla{\rm f}\left(\vec{r}\right)\right\vert^{2}} \,{\rm f}\left(\vec{r}\right) $$

Felix Marin
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