1.4.3.1 Solution rotcoor-a

Question:

Simplify the partial differential equation

$$10 u_{xx} + 6 u_{xy} + 2 u_{yy} = u_x + x + 1$$

by rotating the coordinate system. Classify the equation. Draw the original and rotated coordinate system and identify the angle of rotation.

Answer:

Identify the matrix $A$. Find the eigenvalues and orthonormal eigenvectors. You find

$$\lambda_1=11 \quad{\hat\imath}'=\left(\begin{array}{c}3 1\... ...ath}'=\left(\begin{array}{c}-1 3\end{array}\right)/\sqrt{10}$$

The eigenvalues are of the same sign, so it is elliptic.

You then find the relation between the coordinates to be

$$x = \frac{3}{\sqrt{10}} x' - \frac{1}{\sqrt{10}} y' \qquad y = \frac{1}{\sqrt{10}} x' + \frac{3}{\sqrt{10}} y'$$

$$x' = \frac{3}{\sqrt{10}} x + \frac{1}{\sqrt{10}} y' \qquad y' = - \frac{1}{\sqrt{10}} x' + \frac{3}{\sqrt{10}} y'$$

So using the conversion rule for the first derivative $u_x$, the PDE becomes

$$11 u_{x'x'} + u_{y'y'} = \frac{3}{\sqrt{10}} u_{x'} - \frac{... ...} u_{y'} + \frac{3}{\sqrt{10}} x' - \frac{1}{\sqrt{10}} y' + 1$$