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.