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The question asks us to classify the following partial differential equations as parabolic, hyperbolic, or elliptic. If the differential equation depends on some of the independent variables, find the regions of parabolicity, hyperbolicity, and ellipticity. In each case, $u$ is a function of the two variables explicitly mentioned in the equation.

The equation is $$u_{xx} +2yu_{xy} +x(2−x)u_{yy} = u$$

I understand to the point where when the eigen values of the coefficient matrix are the same sign the equation is elliptic, opposite signs means hyperbolic, and when one is zero means parabolic. This one particular equations is very difficult to determining it's classification due to the algebra involved.

I've found the coefficient matrix to be $A=\begin{pmatrix} 1 & y\\ y & x(2-x) \end{pmatrix}$ and understand that this will have eigen values of $\displaystyle\lambda=\frac{1+2x-x^2\pm\sqrt{x^4-4x^3+6x^2-4x+4y^2+1}}{2}$ or simplified as $\displaystyle\lambda=\frac{2-(x-1)^2\pm\sqrt{(x-1)^4+4y^2}}{2}$. The first eigen value was simple enough to find as we simply try to minimize the equation when we are adding the radical to find that the smallest value for any $x$ and $y$ is 1, so all that's left is to find the signs of the other eigen value, which is giving me a lot of difficulty.

My first step was to set the radical equal to zero and write an equation for $y$ and $x$ which will classify the equation as parabolic.

$\displaystyle0=\frac{2-(x-1)^2-\sqrt{(x-1)^4+4y^2}}{2}$

$\displaystyle0=2-(x-1)^2-\sqrt{(x-1)^4+4y^2}$

$2-(x-1)^2=\sqrt{(x-1)^4+4y^2}$

$4-4(x-1)^2+(x-1)^4=(x-1)^4+4y^2$

$4-4(x-1)^2=4y^2$

$1-(x-1)^2=y^2$

$y=|1-(x-1)^2|$

Assuming my algebra is correct here, then this is when the equation is parabolic. Would this also mean that when $y>|1-(x-1)^2|$ it is elliptic and when $y<|1-(x-1)^2|$ it is hyperbolic?

Any help would be greatly appreciated.

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1 Answer 1

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The equation $$a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}=\Phi(x,y,u,u_x,u_y),$$ where $|a|+|b|+|c|\neq 0$, belongs (at a point or in a region) to the

hyperbolic type if $\;b^2-ac>0$

parabolic type if $\;b^2-ac=0$

elliplic type if $\;b^2-ac<0$

In our case $$a=1,\quad b=y,\quad c=x(2-x),$$ $$b^2-ac=x^2-2x+y^2=(x-1)^2+y^2-1$$

Answer:

The equation $$u_{xx} +2yu_{xy} +x(2−x)u_{yy} = u$$ belongs to the

hyperbolic type if $(x-1)^2+y^2>1$,

parabolic type if $\;(x-1)^2+y^2=1$,

elliplic type if $\;(x-1)^2+y^2<1$.

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  • $\begingroup$ Thank you! In the end, I got to a solution around that. The professor suggested that I should point out the geometrical properties of the characterization. That it's a circle of radius 1 centered at (1,0) where the PDE is hyperbolic outside the circle, parabolic on the circle, and elliptic within the circle. $\endgroup$
    – Peetrius
    Jul 16, 2018 at 13:34

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