What is the point of canonical form in PDEs?

I have been looking at canonical forms of second order PDEs:

My question is: why is that useful? It doesn’t seem to make the PDEs any easier to solve.

A hyperbolic equation for example becomes(Copied from above link)
wξν+L[w]=Gw_{\xi\nu}+L[w]=G, where L is some first order operator and G some function.
That doesn’t seem much easier to solve than a hyperbolic equation to me.



1 Answer


There is probably not a unique or complete answer to this, but a useful insight could be that in the Hyperbolic case this form highlights much more clearly the characteristic curves of the equations, i.e., the curves along which the data remains costant. This is very useful in practice.

If you write the equation as wξν+L[w]=Gw_{\xi\nu}+L[w]=G with the change of variable ξ=ϕ(x,y)\xi=\phi(x,y) and ν=ψ(x,y)\nu=\psi(x,y), you have indeed that the equations ϕ(x,t)=k1\phi(x,t)=k_1 and ψ(x,t)=k2\psi(x,t)=k_2 are the characteristic functions, which is, ξ\xi and ν\nu are constant along these functions.

For example, in the wave equation uxx+c2uyy=0u_{xx} + c^2u_{yy} = 0, which is the base case of hyperbolic equations, setting ξ=x+cy\xi=x+cy and ν=x−cy\nu=x-cy the equation simply becomes uξν=0u_{\xi\nu}=0 which has solution u=F(x+ct)+G(x−ct)u=F(x+ct)+G(x-ct), and x+cyx+cy and x−cyx-cy are the characteristics for the equation.

In the Elliptic case, the solutions of the canonical form includes a complex term, which represents the fact that there isn’t any explicit characteristic curve for this equations.

This shows why elliptic equations are used to model stationary systems, while Hyperbolic ones model a Transport of the initial data throught the domain.