# Canonical form of second-order linear PDEs#

Mathematics for Scientists and Engineers 2

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

(134)#$au_{xx} + bu_{xy} + cu_{yy} = f(x, y, u, u_x, u_y)$

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 $$\xi = \xi(x, y), \qquad \eta = \eta(x,y)$$

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

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

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

We find the coordinate transformation

$\begin{split} u_x = u_\xi \xi_x + u_\eta \eta_x, \qquad u_y = u_\xi \xi_y + u_\eta \eta_y \\ \text{and similarly for } u_{xx}, u_{xy}, u_{yy} \end{split}$

Plugging this back into (134) we get

(135)#$A u_{\xi \xi} + B u_{\xi \eta} + C u_{\eta \eta} = F(\xi, \eta, u, u_\xi, u_\eta)$

where

\begin{split} \begin{aligned} & A = a(\xi_x)^2 + b\xi_x \xi_y + c(\xi_y)^2 \\ & B = 2a \xi_x \eta_x + b(\xi_x \eta_y + \xi_y \eta_x) + 2c \xi_y \eta_y \\ & C = a(\eta_x)^2 + b \eta_x \eta_y + c(\eta_y)^2 \end{aligned} \end{split}

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

$\begin{split} A = a(\xi_x)^2 + b\xi_x \xi_y + c(\xi_y)^2 = 0 \\ C = a(\eta_x)^2 + b \eta_x \eta_y + c(\eta_y)^2 = 0 \end{split}$

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

$\begin{split} a \left( \frac{\xi_x}{\xi_y} \right)^2 + b \left( \frac{\xi_x}{\xi_y} \right) + c = 0 \\ a \left( \frac{\eta_x}{\eta_y} \right)^2 + b \left( \frac{\eta_x}{\eta_y} \right) + c = 0 \end{split}$

These are two identical quadratic equations with roots

$\lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$\xi_x = \lambda_1 \xi_y, \qquad \eta_x = \lambda_2 \eta_y$

whose characteristic curves satisfy the ODEs

$\frac{dy}{dx} = -\lambda_1, \qquad \frac{dy}{dx} = -\lambda_2$

The solutions of these ODEs are

$y + \lambda_1 x = c_1, \qquad y + \lambda_2 x = c_2$

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

$\xi = y + \lambda_1 x, \qquad \eta = y + \lambda_2 x$

Finally, going back to the canonical form

$u_{\xi \eta} = F$

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

$u(\xi, \eta) = \int \int F d \eta d \xi + \phi(\xi) + \psi(\eta)$

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

$u_{tt} - v^2 u_{xx} = 0$

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

$\lambda_{1,2} = \frac{0 \pm \sqrt{0 +4v^2}}{2} = \pm v$

and the characteristics are given by

$\xi = x + vt, \qquad \eta = x - vt$

The solution $$u$$ is given by

$u(\xi, \eta) = \phi(\xi) + \psi(\eta)$

or in terms of $$x$$ and $$y$$:

$u(x, y) = \phi(x + vt) + \psi(x - vt)$

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

$u_{\eta \eta} = \phi(\xi, \eta, u, u_{\xi}, u_{\eta})$

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 $$A = 0$$:

$A = a\left( \frac{\xi_x}{\xi_y} \right)^2 + b\left( \frac{\xi_x}{\xi_y} \right) + c = 0$

Therefore the equation

$a \lambda^2 + b\lambda + c = 0$

has two equal roots

$\lambda = \lambda_1 = \xi_x / \xi_y = \eta_x / \eta_y = \lambda_2$

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

$\frac{dy}{dx} = -\lambda$

i.e.

$\xi = y + \lambda x$

and we can choose

$\eta = x$

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

$u(\xi, \eta) = \int \int F d \eta d \eta + \eta \phi(\xi) + \psi(\xi)$

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

$\lambda = \frac{-b}{2a} = -1$

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

$u(\xi, \eta) = \eta \phi(\xi) + \psi(\xi)$

or in original coordinates

$u(x, y) = x \phi(y - x) + \psi(y - x)$

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

$\lambda = \frac{-b}{2a} = 2$

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

$u_{\xi \xi} = \cos (2x+y) = \cos \eta$

Integrating twice w.r.t. $$\xi$$

$u(\xi, \eta) = \int \int \cos \eta d \xi d \xi + \xi \phi(\eta) + \psi(\eta)$

Which is

$u(\xi, \eta) = \frac{\xi^2}{2} \cos \eta + \xi \phi(\eta) + \psi(\eta)$

Or in original coordinates

$u(x, y) = \frac{x^2}{2} \cos (2x + y) + x \phi(x+2y) + \psi(x+2y)$

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.