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Mathematics LibreTexts

3.1.1: Normal Form in Two Variables

[ "article:topic", "showtoc:no" ]
  • Page ID
    2348
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    Consider the differential equation

    \begin{equation}
    \label{eqnplane}\tag{3.1.1.1}
    a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}+\mbox{terms of lower order}=0
    \end{equation}

    in \(\Omega\subset\mathbb{R}^2\). The associated characteristic differential equation is

    \begin{equation}
    \label{planechar}\tag{3.1.1.2}
    a\chi_x^2+2b\chi_x\chi_y+c\chi_y^2=0.
    \end{equation}

    We show that an appropriate coordinate transform will simplify equation (\ref{eqnplane}) sometimes in such a way that we can solve the transformed equation explicitly.

    Let \(z=\phi(x,y)\) be a solution of (\ref{planechar}). Consider the level sets \(\{(x,y):\ \phi(x,y)=const.\}\) and assume  \(\phi_y\not=0\) at a point \((x_0,y_0)\) of the level set. Then there is a function \(y(x)\) defined in a neighborhood of \(x_0\) such that \(\phi(x,y(x))=const.\) It follows

    $$y'(x)=-\dfrac{\phi_x}{\phi_y},$$

    which implies, see the characteristic equation (\ref{planechar}),

    \begin{equation}
    \label{quadratic}\tag{3.1.1.3}
    ay'^2-2by'+c=0.
    \end{equation}

    Then, provided  \(a\not=0\), we can calculate \(\mu:=y'\) from the (known) coefficients \(a\), \(b\) and \(c\):

    \begin{equation}
    \label{mu}\tag{3.1.1.4}
    \mu_{1,2}=\dfrac{1}{a}\left(b\pm\sqrt{b^2-ac}\right).
    \end{equation}

    These solutions are real if and only of \(ac-b^2\le0\).

    Equation (\ref{eqnplane}) is hyperbolic if \(ac-b^2<0\), parabolic if  \(ac-b^2=0\) and ellipticif  \(ac-b^2>0\). This follows from an easy discussion of the eigenvalues of the matrix

    $$\left(\begin{array}{cc}
    a&b\\
    b&c
    \end{array}\right),$$

    see an exercise.

    Normal form of a hyperbolic equation

    Let \(\phi\) and \(\psi\) are solutions of the characteristic equation (\ref{planechar}) such that

    \begin{eqnarray*}
    y_1'\equiv\mu_1&=&-\dfrac{\phi_x}{\phi_y}\\
    y_2'\equiv\mu_2&=&-\dfrac{\psi_x}{\psi_y},
    \end{eqnarray*}

    where \(\mu_1\) and \(\mu_2\) are given by (\ref{mu}). Thus \(\phi\) and \(\psi\) are solutions of the linear homogeneous equations of first order

    \begin{eqnarray}
    \label{phi}\tag{3.1.1.5}
    \phi_x+\mu_1(x,y)\phi_y&=&0\\
    \label{psi}\tag{3.1.1.6}
    \psi_x+\mu_2(x,y)\psi_y&=&0.
    \end{eqnarray}

    Assume \(\phi(x,y)\), \(\psi(x,y)\) are solutions such that \(\nabla\phi\not=0\) and \(\nabla\psi\not=0\), see an exercise for the existence of such solutions.

    Consider two families of level sets defined by \(\phi(x,y)=\alpha\) and \(\psi(x,y)=\beta\), see Figure 3.1.1.1.


    Figure 3.1.1.1: Level sets

    These level sets are characteristic curves of the partial differential equations (\ref{phi}) and (\ref{psi}), respectively, see an exercise of the previous chapter.

    Lemma. (i) Curves from different families can not touch each other

    (ii) \(\phi_x\psi_y-\phi_y\psi_x\not=0\).

    Proof. (i):
    $$
    y_2'-y_1'\equiv\mu_2-\mu_1=-\dfrac{2}{a}\sqrt{b^2-ac}\not=0.
    $$
    (ii):
    $$
    \mu_2-\mu_1=\dfrac{\phi_x}{\phi_y}-\dfrac{\psi_x}{\psi_y}.
    $$

    \(\Box\)

    Proposition 3.1. The mapping \(\xi=\phi(x,y)\), \(\eta=\psi(x,y)\) transforms equation (\ref{eqnplane}) into
    \begin{equation}
    \label{normhyp}\tag{3.1.1.7}
    v_{\xi\eta}=\mbox{lower order terms},
    \end{equation}
    where \(v(\xi,\eta)=u(x(\xi,\eta),y(\xi,\eta))\).}

    Proof. The proof follows from a straightforward calculation.
    \begin{eqnarray*}
    u_x&=&v_\xi\phi_x+v_\eta\psi_x\\
    u_y&=&v_\xi\phi_y+v_\eta\psi_y\\
    u_{xx}&=&v_{\xi\xi}\phi_x^2+2v_{\xi\eta}\phi_x\psi_x+v_{\eta\eta}\psi_x^2+\mbox{lower order terms}\\
    u_{xy}&=&v_{\xi\xi}\phi_x\phi_y+v_{\xi\eta}(\phi_x\psi_y+\phi_y\psi_x)+v_{\eta\eta}\psi_x\psi_y+\mbox{lower order terms}\\
    u_{yy}&=&v_{\xi\xi}\phi_y^2+2v_{\xi\eta}\phi_y\psi_y+v_{\eta\eta}\psi_y^2+\mbox{lower order terms}.
    \end{eqnarray*}
    Thus
    $$
    au_{xx}+2bu_{xy}+cu_{yy}=\alpha v_{\xi\xi}+2\beta v_{\xi\eta}+\gamma v_{\eta\eta}+l.o.t.,
    $$
    where
    \begin{eqnarray*}
    \alpha:&=&a\phi_x^2+2b\phi_x\phi_y+c\phi_y^2\\
    \beta:&=&a\phi_x\psi_x+b(\phi_x\psi_y+\phi_y\psi_x)+c\phi_y\psi_y\\
    \gamma:&=&a\psi_x^2+2b\psi_x\psi_y+c\psi_y^2.
    \end{eqnarray*}
    The coefficients \(\alpha\) and \(\gamma\) are zero since \(\phi\) and \(\psi\) are solutions of the characteristic equation. Since
    $$
    \alpha\gamma-\beta^2=(ac-b^2)(\phi_x\psi_y-\phi_y\psi_x)^2,
    $$
    it follows from the above lemma that the coefficient \(\beta\) is different from zero.

    \(\Box\)

    Example 3.1.1.1:

    Consider the differential equation
    $$
    u_{xx}-u_{yy}=0.
    $$
    The associated characteristic differential equation is
    $$
    \chi_x^2-\chi_y^2=0.
    $$
    Since \(\mu_1=-1\) and \(\mu_2=1\), the functions \(\phi\) and \(\psi\) satisfy differential equations
    \begin{eqnarray*}
    \phi_x+\phi_y&=&0\\
    \psi_x-\psi_y&=&0.
    \end{eqnarray*}
    Solutions with \(\nabla\phi\not=0\) and \(\nabla\psi\not=0\) are
    $$
    \phi=x-y,\ \ \psi=x+y.
    $$
    Then the mapping
    $$
    \xi=x-y,\ \ \eta=x+y
    $$
    leads to the simple equation
    $$
    v_{\xi\eta} (\xi,\eta)=0.
    $$
    Assume \(v\in C^2\) is a solution, then \(v_\xi=f_1(\xi)\) for an arbitrary \(C^1\) function \(f_1(\xi)\). It follows
    $$
    v(\xi,\eta)=\int_0^\xi\ f_1(\alpha)\ d\alpha+g(\eta),
    $$
    where \(g\) is an arbitrary \(C^2\) function. Thus each \(C^2\)-solution of the differential equation can be written as

    \((\star)\)  \(v(\xi,\eta)=f(\xi)+g(\eta)\),

    where \(f,\ g\in C^2\). On the other hand, for arbitrary \(C^2\)-functions \(f\), \(g\) the function \((\star)\) is a solution of the differential equation \(v_{\xi\eta}=0\). Consequently every \(C^2\)-solution of the original equation \(u_{xx}-u_{yy}=0\) is given by
    $$
    u(x,y)=f(x-y)+g(x+y),
    $$
    where \(f,\ g\in C^2\).

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