3.1 Classification and characteristics

The general quasi-linear first order equation in two dimensions takes the form

 (3.1)

where , , and may depend on , , and . All scalar first order equation are classified as hyperbolic equations.

The characteristics are defined by

 (3.2)

They will form a single family of lines in the -plane. (In contrast, the characteristics of scalar second order hyperbolic differential equations form two intersecting families of lines.)

In general, the variation of a function of two variables along a line is given by the total differential of calculus,

but along the characteristic lines that becomes

and comparing with the partial differential equation, it is seen that the right hand side is . So the variation of the solution along the characteristic lines is given by
 (3.3)

Note that frequently, you may have to solve the ordinary differential equations in a different form or order. For example, if depends on , you will not be able to solve to find as a function of since in is still an unknown function of . But maybe, say, is not a function of , in which case you could solve ; then you could plug that solution for as a function of into to get an equation for that no longer involves . The bottom line is that it is really best to write the characteristic equations as

 (3.4)

and pick from those proportionalities the ratio that is easiest to solve first.

In all the unsolved problems in the book, there is at least one solvable ratio. But if there is none, you may be forced to try to change variables, e.g. to polar, or eliminate one variable by, say, differentiating a ratio, hopefully producing a second order ordinary differential equation with one variable eliminated.

Example

Question: (5.30) Solve

Solution:

This example wants to solve the partial differential equation

For this equation, a ratio like is not immediately solvable for , since besides would be an unknown function of . The only solvable ratio is in fact that between and :

where is the integration constant. These characteristic lines are hyperbola; they are sketched in figure 3.1.

Now that is a known function of , specifically assuming it is positive, the ordinary differential equation for can be solved

to give

Taking exponentials and noting that the square root equals , this simplifies to

For example, if it is given that at the point shown as a fat dot in figure 3.2, then it follows from the above general expressions that , so the characteristic line is the line shown in grey, and that , so you would get for on the grey line.

Example

Question: (5.6) Solve the nasty example

Solution:

In this case, none of the ratios

is a solvable ordinary differential equation; each involves three variables. You might try to take a derivative of an equation, like

which is indeed an ordinary differential equation for not involving the unknown . But it is an awkward second order nonlinear equation.

The trick is to guess that the combination can be found as a function of :

which produces

This can then be plugged into

to get a separable equation giving as a function of , with another integration constant . However, that becomes a mess, involving either an arctan or logarithm, depending on the value of .