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Let's say that we have the second-order PDE which has hyperbolic form

\begin{align}u_{xx}+yu_{yy}=0 \label{1}\tag1 \end{align}

We know $a=1, \ b=0,\ c=y$

Thus, the discriminant is: $$d=b^2-ac=-y$$

So the equation is hyperbolic form when $y<0$.

It is clear that

\begin{align}u_{xx}+yu_{yy}&=0 \Leftrightarrow \\ \frac{\partial^2{u}}{\partial{x}^2}&=-y \frac{\partial^2{u}}{\partial{y}^2} \end{align}

The question is how to formulate the \eqref{1} as a first order PDE system: $$\frac{\partial{Q}}{\partial{x}}+A \frac{\partial{Q}}{\partial{y}}=0$$ where $Q$ vector in $\mathbb{R}$ and A is matrix $2 \times 2$ ?

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1 Answer 1

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Suppose you have $u_{xx} + f(y)u_{yy} = 0$. Then \begin{align} u_x &= -f(y) v_y, \\ v_x &= u_y \end{align} is a simple representation of it as a first-order system. In your case $f(y) = y$.

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  • $\begingroup$ I made the following consideration Let c=-y It is clearly that \begin{align}u_{xx}+yu_{yy}&=0\\ \Leftrightarrow \frac{\partial{u}}{\partial^2{x}}- c \frac{\partial{u}}{\partial^2{y}}&=0\\ \Leftrightarrow \bigg(\frac{\partial{u}}{\partial{x}}\bigg)^2- \bigg(\sqrt{c} \frac{\partial{u}}{\partial{y}}\bigg)^2&=0\\ \bigg( \frac{\partial u }{\partial x} - \sqrt{c} \frac{\partial u }{\partial y} \bigg) \bigg( \frac{\partial u }{\partial x} + \sqrt{c} \frac{\partial u }{\partial y} \bigg)&=0 \end{align} So $$u_{x} + \sqrt{c} u_{y} = 0$$ $$u_{x} - \sqrt{c} u_{y} = 0$$. Is it right? $\endgroup$
    – Athanasios Paraskevopoulos
    Jan 3 at 0:22
  • $\begingroup$ I do not want just the solution, I want to understand the way we solve it. $\endgroup$
    – Athanasios Paraskevopoulos
    Jan 3 at 0:24
  • $\begingroup$ @AthanasiosParaskevopoulos The proposed answer is correct. However, your proposition in the above comments does not look right: $u_{xx} \not \equiv (u_x)^2$. $\endgroup$
    – EditPiAf
    Jan 3 at 10:50
  • $\begingroup$ @EditPiAf ok, I got it! Thank you! $\endgroup$
    – Athanasios Paraskevopoulos
    Jan 3 at 12:44

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