So here’s my higher order PDE, $u_{xxxx}-u_{xxyy}+u_{yyyy}=0$. I’m trying classify this as either elliptic or strictly hyperbolic. It is elliptic if there are no real characteristics at $\vec{x}$ otherwise it is hyperbolic. Thanks for the help, hints are welcome
Update: So consider another equation, $u_{xy}+2u_x=0$. I think I got solution for this right.
$Lu=\frac{\partial^2u}{\partial x\partial y}+2\frac{\partial u}{\partial x}\implies b(x,y)=1$ and $d(x,y)=2$.
$\implies L^p(x,y,i\xi,i\eta)=-\xi\eta=(\xi,\eta)\begin{pmatrix}0&-1/2\\-1/2&0\end{pmatrix}\begin{pmatrix}\xi\\\eta\end{pmatrix}$. So since eigenvalues of the matrix are real and equal then it is definite. Therefore, it is elliptic
Update: this might be useful
Here are definitions from book,
Def: Let L be an operator of the form $L(\vec{x},D)u=\sum_{|\alpha|\leq m}a_\alpha(\vec{x})D^\alpha u$ where $D^{\alpha}=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\partial x_2^{\alpha_2} ...\partial x_n^{\alpha_n}}$. Here, $|\alpha|=\alpha_1+\alpha_2+...+\alpha_n$. Characteristic surfaces`are defined by equation, $L^p(\vec{x},\nabla\phi)=0$.
An equation is elliptic at $\vec{x}$ if there are no real charactersitics at $\vec{x}$, that is, $L^p(\vec{x},i\vec{\xi})\neq 0, \forall \vec{\xi}\neq 0$.
An equation is strictly hyperbolic in direction $\vec{n}$ if
$1.\ L^p(\vec{x},i\vec{n})\neq 0$
$2.$ all the roots $\omega$ of the equation, $L^p(\vec{x},i\vec{\xi}+i\omega\vec{n})=0 $, are real and distinct $\forall \xi\in\mathbb{R}^n$ which is not collinear with $n$