Physical significance of repetitive roots of a PDE

Question

Consider, the following Linear PDEs with homogenous coefficients, $$(D^4 + D'^4 -2D^2D'^2)z =0; \qquad......{\rm eqn}.1$$
$$(D^3D'^2 + D^2D'^3)z =0; \qquad......{\rm eqn}.2$$
Here z is a function of $z= f(x,y)$ and D & D' respectively represent partial differential operators, i.e $$\frac{∂}{∂}, \quad\frac{∂}{∂y}$$ Upon solving both of them i ended up with repetitive roots and was confused regarding their physical significance. Can someone help me out to understand the significance of repetitive roots in case of Homogenous Linear PDE?

Answer 1

Double roots typically mean that the homogeneous solutions consist not only of exponential solutions but also of exponentials multiplied by linear terms.

To simplify the discussion, I will treat ODE's (not sure why you jumped directly to PDE's).

lets start with an example:

$$(\partial_{x}-1)^{2}f=0$$

the double root is $\{1,1\}$ so that we know that $e^{x}$ is a homogeneous solution (indeed it can be seen by inspection). However, note that $xe^{x}$ also solves this:

$$(\partial_{x}-1)^{2}xe^{x}=(\partial_{x}-1)\left((\partial_{x}-1)xe^{x}\right)=(\partial_{x}-1)\left(e^{x}\right)=0$$

This is a general phenomenon.

There are several ways to understand this, but I like to think of this as a result of a degeneracy. Consider $$(\partial_{x}-a)(\partial_{x}-b)f=0$$

the solution is a linear combination $Ae^{ax}+Be^{bx}$. But what happens when $b$ approaches $a$? call $b=a+\epsilon$ then the general solution becomes

$$e^{ax}[A+Be^{\epsilon x}]$$ We see that by choosing $A=-B=\frac{1}{\epsilon}$, then as $\epsilon$ approaches zero we find an additional solution $xe^{ax}$. This is the remanent of the “non degenerate” solution we had when $a\neq b$.

As an exercise, ask yourself what happens with a triple root.