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Consider the differential operator $L$ defined as:

$$L(x,y,D)=c^2\frac{\partial^2}{\partial x^2}u(x,y)-\frac{\partial^2}{\partial x \partial y}u(x,y)+c \frac{\partial^2}{\partial y^2}u(x,y)+e^{-c^2} \frac{\partial}{\partial x}u(x,y)$$

The symmetric matrix of the coefficients of the highest derivatives is

$$A=\left(\begin{array}{rrr} c^2 & -1/2 \\ -1/2 & c \end{array} \right)$$

Now my question is where the $-1/2$ is coming from. It makes sense to me that $-1/2 + (-1/2)=-1$ which would be the missing coefficient of $\partial x\partial y$ but I don't see why exactly it is possible to write the matrix like this.

illuminatitruthseeker
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1 Answers1

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Do the matrix multiplication: $$ \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} c^2 & -1/2\\ -1/2 & c \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= c^2x^2 -\frac{1}{2}xy -\frac{1}{2}xy + cy^2 = c^2x^2 -xy + cy^2 $$ And you get the polynomial that corresponds to the second order terms of your differential equation.

In general, for every second degree homogeneous polynomial in n variables (that is, a quadratic form), $$ p(x_1,\ldots,x_n) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}{x_i}{x_j} $$ we can find a symmetric matrix $S$ such that $p(x_1,\ldots,x_n) = \mathbf x^\mathrm{T} S \mathbf x,$

camilo
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