Determine the type of the following equation and reduce the PDE to its canonical form
$u_{xx} + 4u_{xy} + 4u_{yy} + u = 0$.
We consider pdes in the form $$a_{11}(x,y)u_{xx}+2a_{12}(x,y)u_{xy}+a_{22}(x,y)u_{yy} +F(x,y,u,u_x,u_y)=0$$
Since in our case $a_{12}^2-a_{11}a_{22}=0$, we have that it is parabolic.
Then I think we find $$\frac{dy}{dx}=\frac{a_{12} \pm \sqrt{a_{12}^2-a_{11}a_{22}}}{a_{11}}=2$$ But then what?
Best Answer
Define $\eta \left( x,y \right)=y-2x,\text{ }$ and choose $\xi \left( x,\text{ }y \right)=x$ such that the Jacobian $J:={{\xi }_{x}}{{\eta }_{y}}-{{\xi }_{y}}{{\eta }_{x}}$ does not vanish. Let $v\left( \xi ,\eta \right)=u\left( x,y \right).$ Substituting the new coordinates $\xi $ and $\eta $ into the given equation, we obtain
$$\left( {{v}_{\xi \xi }}-4{{v}_{\xi \eta }}+4{{v}_{\eta \eta }} \right)+4\left( {{v}_{\xi \eta }}-2{{v}_{\eta \eta }} \right)+4{{v}_{\eta \eta }}+v=0\text{ }.$$ Thus,
$${{v}_{\xi \xi }}+v=0,$$ and this is the desired canonical form.