2.2.1: A Linearization Method
- Page ID
- 2168
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We can transform the inhomogeneous Equation (2.2.1) into a homogeneous linear equation for an unknown function of three variables by the following trick.
We are looking for a function \(\psi(x,y,u)\) such that the solution \(u=u(x,y)\) of Equation (2.2.1) is defined implicitly by \(\psi(x,y,u)=const.\) Assume there is such a function \(\psi\) and let \(u\) be a solution of (2.2.1), then
$$
\psi_x+\psi_uu_x=0,\ \ \psi_y+\psi_uu_y=0.
$$
Assume \(\psi_u\not=0\), then
$$
u_x=-\frac{\psi_x}{\psi_u},\ \ u_y=-\frac{\psi_y}{\psi_u}.
$$
From (2.2.1) we obtain
\begin{equation}
\label{homthree}\tag{2.2.1.1}
a_1(x,y,z)\psi_x+a_2(x,y,z)\psi_y+a_3(x,y,z)\psi_z=0,
\end{equation}
where \(z:=u\).
We consider the associated system of characteristic equations
\begin{eqnarray*}
x'(t)&=&a_1(x,y,z)\\
y'(t)&=&a_2(x,y,z)\\
z'(t)&=&a_3(x,y,z).
\end{eqnarray*}
One arrives at this system by the same arguments as in the two-dimensional case above.
Proposition 2.2. (i) Assume \(w\in C^1\), \(w=w(x,y,z)\), is an integral, i. e., it is constant along each fixed solution of (\ref{homthree}), then \(\psi=w(x,y,z)\) is a solution of (\ref{homthree}).
(ii) The function \(z=u(x,y)\), implicitly defined through \(\psi(x,u,z)=const.\), is a solution of (2.2.1), provided that \(\psi_z\not=0\).
(iii) Let \(z=u(x,y)\) be a solution of (2.2.1) and let \((x(t),y(t))\) be a solution
of
$$
x'(t)=a_1(x,y,u(x,y)),\ \ y'(t)=a_2(x,y,u(x,y)),
$$
then \(z(t):=u(x(t),y(t))\) satisfies the third of the above characteristic equations.
Proof. Exercise.
Contributors and Attributions
Integrated by Justin Marshall.