Canonical form of second-order linear PDEs
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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 (136) we get
where
The reader can derive this as partial differentiation practice.
Hyperbolic case#
PDE (136) is hyperbolic if \(b^2 - 4ac > 0\) so the obvious choice is to set \(A = C = 0\) in eq. (137) (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 (136) 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 (136) 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 (137) 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.