Solving an Elliptic PDE Using the Characteristic Equation: A Beginner's Guide

[maverick280857]

Hello

In our math course, we encountered the following elliptic PDE:

<br /> y^{2}u_{xx} + u_{yy} = 0<br />

In order to solve it, we converted it to the characteristic equation,

<br /> y^{2}\left(\frac{dy}{dx}\right)^{2} + 1 = 0<br />

Next, we wrote:

\frac{dy}{dx} = \frac{i}{y}

My question is: the characteristic equation has no solution in \mathbb{R} but we went ahead and mechanically wrote the expression for dy/dx. Does this mean that we should regard x and y as complex variables? If so, how does one reconcile with the fact that some solution to the PDE as u(x,y) = c is a surface in (x,y,u) space? Perhaps this is a trivial question, but I'm just starting to learn PDEs. Does this also mean that we should not ascribe a physical significance to x and y as coordinates in \mathbb{R}^2?

Thanks.

 

[HallsofIvy]

Yes, the fact that it is an elliptic equation tells you that the characteristic equation does not have real roots.

 

[maverick280857]

Oops yes, of course...I didn't see that.

Also, in such a case, do y and x lose their "physical significance" of being real variables in real space?

 

[arildno]

Mmm..no, it means that the characteristic curve itself doesn't lie in real space.
That's quite a different thing from saying that x and y cannot be regarded as real variables.

 

[maverick280857]

Mmm..no, it means that the characteristic curve itself doesn't lie in real space.
That's quite a different thing from saying that x and y cannot be regarded as real variables.

Could you please elaborate? And where can I read more about such issues?

 

[arildno]

Well, my memory on characteristics has gone hazy, so it would be helpful if you posted the precise procedure utilized in the particular example.

However, as a general trait, the method of characteristics is a trick whereby we get a family of curves along everyone of which the u-signal propagates in a simple manner (say, by being conserved).

If therefore that family of curves lie in the complex plane, it means that there aren't a set of real curves y(x) along which u propagates. For example, y cannot be solved entirely as a function of x when we constrict ourselves to the real plane.

Please post a few details about the specific procedure.

 

[maverick280857]

Ok, so the elliptic equation is

au_{xx} + 2bu_{xy} + cu_{yy} = 0

and its characteristic equation is

a\left(\frac{dy}{dx}\right)^{2} -2b\frac{dy}{dx} + c = 0

Here, b^{2}-ac &lt;0 so it has complex roots, and the characteristic curves are:

\zeta(x,y) = c_{1}
\eta(x,y) = c_{2}