Calculating the Principal Part of a Differential Operator

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Definition :L is quasilinear if it has the form (1) $L(u)=\sum_{|k|=n}a_k(u) \nabla^ku+f(u)$ where $a_k$ and $f$ are formal differential operators of order $\leq n-1$

As a example for a quasilinear Differential operator I got (2) $L(u)=\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}$

Now I am trying to write this differential operator like (1):

So let $k=(k_1,k_2) \in \mathbb{N}^2$ and $n=2$

In this case I get

(3) $L(u)=\sum_{|k|=n}a_k(u) \frac{\partial^{|k|}u}{\partial x^{k_1} \partial y^{k_2}}+f(u)$

If (2) has to equal (3), I get

$a_{(2,0)}(u)=1$

$a_{(1,1)}(u)=0$

$a_{(0,2)}(u)=-1$

and $f(u)=0$

Given this Differential equation on $\Gamma$=x-Axis , I tried to calculate the principal part of the Differential operator:

The Principal part is defined as $L_p(v,u):=\sum_{|k|=n}a_k(u)v^k$

As far as I understood it, $v$ is a vector that needs to be orthogonal to $\Gamma$. Because $\Gamma$ is the x Axis I chose $v=\left(\begin{array}{c} 0 \\ 1 \end{array}\right)$ ,with the result $L_p(v,u)=1 \left(\begin{array}{c} 0 \\ 1 \end{array}\right)^{ \left(\begin{array}{c} 2 \\ 0 \end{array}\right)}-1\left(\begin{array}{c} 0 \\ 1 \end{array}\right)^{ \left(\begin{array}{c} 0 \\ 2 \end{array}\right)}=-1$

My Question is: Are my calculations correct or did I misunderstood something. The value of $f(u)$ being $0$ looks kinda suspicious to me. Does anyone know an example where $f(u) \neq 0$. Is my calculation of $L_p$ right? I did not use the fact that $L_p$ is dependent on $u$ so this confuses me a little bit.