Consider the second-order PDE:

​$$a u_{xx} + b u_{xy} + a u_{yy} = 0 \quad \text{in } \mathbb{R}^2,$$

for $$a,b\in \mathbb{R}$$ . Which of the following is true? ​

Correct option is C

​$$\text{Sol. The given PDE is:} \\[10pt]a u_{xx} + b u_{xy} + a u_{yy} = 0. \\[10pt]\text{To classify the PDE, we compute the discriminant:} \\[10pt]S^2 - 4RT, \\[10pt]\text{where:} \\[10pt]S = b \, (\text{coefficient of } u_{xy}), \quad R = a \, (\text{coefficient of } u_{xx}), \quad T = a \, (\text{coefficient of } u_{yy}). \\[10pt]\text{Step 1: Compute } S^2 - 4RT \\[10pt]S^2 - 4RT = b^2 - 4a^2. \\[10pt]\text{Step 2: Classification of the PDE} \\[10pt]\text{The PDE is classified as follows:} \\[10pt]1. \text{Parabolic: } S^2 - 4RT = 0, \text{i.e.,} \\[10pt]b^2 - 4a^2 = 0 \quad \Rightarrow \quad |b| = 2|a|. \\[10pt]2. \text{Hyperbolic: } S^2 - 4RT > 0, \text{i.e.,} \\[10pt]b^2 - 4a^2 > 0 \quad \Rightarrow \quad |b| > 2|a|. \\[10pt]3. \text{Elliptic: } S^2 - 4RT < 0, \text{i.e.,} \\[10pt]b^2 - 4a^2 < 0 \quad \Rightarrow \quad |b| < 2|a|. \\[10pt]\text{Hence, Option C is correct.}$$​​