18.7.3.1 Solution rotcoor-a

Question:

Simplify the partial differential equation

\begin{displaymath}
10 u_{xx} + 6 u_{xy} + 2 u_{yy} = u_x + x + 1
\end{displaymath}

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

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

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

You then find the relation between the coordinates to be

\begin{displaymath}
x = \frac{3}{\sqrt{10}} x' - \frac{1}{\sqrt{10}} y' \qquad y = \frac{1}{\sqrt{10}} x' + \frac{3}{\sqrt{10}} y'
\end{displaymath}


\begin{displaymath}
x' = \frac{3}{\sqrt{10}} x + \frac{1}{\sqrt{10}} y' \qquad y' = - \frac{1}{\sqrt{10}} x' + \frac{3}{\sqrt{10}} y'
\end{displaymath}

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

\begin{displaymath}
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
\end{displaymath}