Let $Lu = A\frac{\partial^2 u}{\partial x^2} + 2B\frac{\partial^2 u}{\partial x\partial y} + C\frac{\partial^2 u}{\partial y^2}$ be a second-order linear differential operator, and let $(\xi, \eta) = (\xi(x,y), \eta(x,y))$ be a smooth change of coordinates with nonvanishing Jacobian.

Step 1: Apply the chain rule to compute second derivatives. We have: $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}$

Computing $\frac{\partial^2 u}{\partial x^2}$ involves terms like $\frac{\partial^2 u}{\partial \xi^2}\left(\frac{\partial \xi}{\partial x}\right)^2$, mixed terms $\frac{\partial^2 u}{\partial \xi\partial \eta}\frac{\partial \xi}{\partial x}\frac{\partial \eta}{\partial x}$, and first-order derivative terms.

Step 2: The transformed operator has the form: $\tilde{L}u = \tilde{A}\frac{\partial^2 u}{\partial \xi^2} + 2\tilde{B}\frac{\partial^2 u}{\partial \xi\partial \eta} + \tilde{C}\frac{\partial^2 u}{\partial \eta^2} + \text{lower order}$

where the coefficients $\tilde{A}, \tilde{B}, \tilde{C}$ are given by: $\begin{pmatrix} \tilde{A} \\ \tilde{B} \\ \tilde{C} \end{pmatrix} = \begin{pmatrix} \xi_x^2 & 2\xi_x\xi_y & \xi_y^2 \\ \xi_x\eta_x & \xi_x\eta_y + \xi_y\eta_x & \xi_y\eta_y \\ \eta_x^2 & 2\eta_x\eta_y & \eta_y^2 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \end{pmatrix}$

Step 3: The discriminant transforms as: $\tilde{B}^2 - \tilde{A}\tilde{C} = (B^2 - AC)\det\begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix}^2$

Step 4: Since the Jacobian determinant is nonzero by assumption, we have: $\det\begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix}^2 > 0$

Therefore, $\tilde{B}^2 - \tilde{A}\tilde{C}$ and $B^2 - AC$ have the same sign.