# Canonical form of second-order linear PDEs

## Contents

# Canonical form of second-order linear PDEs#

Mathematics for Scientists and Engineers 2

Here we consider a general second-order PDE of the function \(u(x, y)\):

Recall from a previous notebook that the above problem is:

**elliptic**if \(b^2 - 4ac > 0\)**parabolic**if \(b^2 - 4ac = 0\)**hyperbolic**if \(b^2 - 4ac < 0\)

Any elliptic, parabolic or hyperbolic PDE can be reduced to the following **canonical forms** with a suitable coordinate transformation \(\xi = \xi(x, y), \qquad \eta = \eta(x,y)\)

Canonical form for hyperbolic PDEs: \(u_{\xi \eta} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta}) \) or \( u_{\xi \xi} - u_{\eta \eta} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta})\)

Canonical form for parabolic PDEs: \(u_{\eta \eta} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta}) \) or \( u_{\xi \xi} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta})\)

Canonical form for elliptic PDEs: \(u_{\xi \xi} + u_{\eta \eta} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta})\)

We find the coordinate transformation

Plugging this back into (134) we get

where

The reader can derive this as partial differentiation practice.

## Hyperbolic case#

PDE (134) is hyperbolic if \(b^2 - 4ac > 0\) so the obvious choice is to set \(A = C = 0\) in eq. (135) (note that we could have also chosen for example \(A = 1, C = -1\)). We get a system of ODEs

Dividing the first equation by \((\xi_y)^2\) and the second by \((\eta_y)^2\) we get

These are two identical quadratic equations with roots

Where \(\lambda_1 = \xi_x / \xi_y\) and \(\lambda_2 = \eta_x / \eta_y\) and they need to be different for the transformation to make sense. Because \(b^2 - 4ac > 0\) we know that they will be two distinct real numbers.

But what do the quantities \(\xi_x / \xi_y \) and \( \eta_x / \eta_y\) actually represent? They are the slopes of the **characteristics** \(\xi(x, y) = \text{const.}\) and \(\eta(x, y) = \text{const.}\) Notice that if we hadn’t divided the equations by \(\xi_y\) and \(\eta_y\) we would have

whose characteristic curves satisfy the ODEs

The solutions of these ODEs are

where \(c_1, c_2\) are integration constants, so we choose \(\xi\) and \(\eta\) to equal them

Finally, going back to the canonical form

we integrate w.r.t. \(\eta\) and \(\xi\) to get the solution

where \(\phi\) and \(\psi\) are arbitrary functions.

### Example: d’Alembert’s solution#

The d’Alembert’s solution encountered in lectures is an example of the method of characteristics. Here we will show this. Let us transform the 1-D wave equation

to canonical form. Comparing with (134) we see that \(a=1, b=0, c=-v^2\). This leads to

and the characteristics are given by

The solution \(u\) is given by

or in terms of \(x\) and \(y\):

which is the d’Alembert’s solution of the wave equation.

## Parabolic case#

PDE (134) will be parabolic if \(b^2 - 4ac = 0\). We therefore require \(B = 0\) and either \(A = 0\) or \(C = 0\). Let us choose \(A = 0\) and \(C \neq 0\), so dividing (135) by \(C\) we get the canonical form

Note: If we chose \(C=0\) and \(A \neq 0\) we would get \(u_{\xi \xi} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta})\).

Since \(A = 0\):

Therefore the equation

has two equal roots

but we still need \(\xi\) and \(\eta\) to be independent for the transformation to make sense. So we let \(\xi\) be a solution of

i.e.

and we can choose

so that \(\xi\) and \(\eta\) are independent. Then going back to the canonical form and integrating it twice, we get the solution

We could have chosen \(\xi\) and \(\eta\) the other way around, of course.

### Example: \(u_{xx} + 2u_{xy} + u_{yy} = 0\)#

Kreyszig problem set 12.4, question 11.

This is a parabolic PDE because \(2^2 - 4 = 0\). Therefore we have a single root

Then \(\xi = y - x\) and we can choose \(\eta = x\). So the solution is

or in original coordinates

where \(\phi\) and \(\psi\) are arbitrary functions.

### Example: \(u_{xx} - 4u_{xy} + 4u_{yy} = \cos (2x+y)\)#

The PDE is parabolic and we have a single root

And we choose \(\eta = y + 2x\) and \(\xi = x\). The canonical form is

Integrating twice w.r.t. \(\xi\)

Which is

Or in original coordinates

where \(\phi\) and \(\psi\) are arbitrary functions.

## Elliptic case#

We will not use method of characteristics to solve elliptic equations because the PDE gets only marginally reduced, i.e. the canonical form is the Poisson’s equation.