More powerful simplifications by changing coordinates are possible in 2D.
Assume that in terms of coordinates and , we have a partial
differential equation:
For example, for a hyperbolic equation, you may like coordinates
and such that and are zero. To find out for what
coordinates and that is the case, expressions for the new
coefficients , , and in terms of the new coordinates
are needed. These can be found by writing out the general
transformation formulae from section 1.4.2 for the special
case of two dimensions. You get, {D.4}:
Characteristic coordinates are coordinates so that the the
and derivatives are eliminated. That
leaves only the derivative, greatly simplifying the
partial differential equation. It reduces to the two-dimensional
canonical form:
(1.13) |
The first thing is to find out how this may be achieved. In terms of
the coefficients of the transformed equation as discussed above,
and must vanish. The condition requires, according to
the given formulae:
To solve the equation for ( goes the same way), divide by
:
By taking the other sign for the square root, you can get a second independent coordinate .
Bottom line, to get characteristic coordinates, solve the plus and minus sign ordinary differential equations above, and equate the integration constants to and .
A couple of notes:
ExampleQuestion: Use characteristic coordinates to reduce the wave equation in multi-dimensional canonical form
to its equivalent two-dimensional canonical form. Then solve it.Solution:
First find the characteristics by solving the ordinary differential equation given above:
Note that the final is the wave propagation speed, not the coefficient in the generic second order equation.The solution is simple:
where and are the integration constants (as well as the characteristic coordinates). So the lines are one set of characteristic lines, and the lines are the other set.Now find the coefficient . The coefficient was zero, and the second order derivatives of and in the formula for are also zero, so is zero too. So the wave equation in characteristic coordinates is
Note that could be divided out, so there is no need to figure out what it is.
(1.14) The wave equation can now easily be solved. Integration with respect to gives
where the integration constant can be any arbitrary function of . Integrating with respect to gives the final solution:
Here is an antiderivative of , so it is arbitrary just like . The additional integration constant is an arbitrary function of .However, you would surely want the solution in terms of the physical coordinates and , rather than the mathematical characteristic coordinates. So substitute for and using the obtained equations for the characteristics. That gives the final solution:
(1.15) That is the general solution of the wave equation. In order to solve a particular problem, you will still need to figure out what and are using whatever the initial and boundary conditions are. One special case, in which the -range is doubly infinite, will be solved in detail later.
ExampleQuestion: Find and sketch the characteristics of the equation
Solution:
Figure out the coefficients in the the characteristic equation by looking at the partial differential equation:
Note that there are only characteristics for negative . For positive the equation is elliptic. And for zero there will only be one direction for the characteristics, horizontal.Use separation of variables to solve. In other words, take the factors to one side and the -factors to the other side:
Squaring both sides to get rid of the square root gives
These are parabolae.
ExampleQuestion: Reduce the equation
to two-dimensional canonical form.Solution:
Two-dimensional canonical form means characteristic form. Find the ordinary differential equation for the characteristics:
Solve it using separation of variables:
The integration constants are the new coordinates:
Work out the partial differential equation in these coordinates using the formulae given at the start of this section:
so
The partial differential equation becomes
Get rid of and completely using the equations for the characteristics:
The resulting partial differential equation is
It does not look easily solvable.
ExampleQuestion: Find the characteristic coordinates of the equation
Solution:
Find the ordinary differential equation for the characteristics:
Solve it:
The characteristic coordinates are the integration constants:
It does not like the partial differential is going to be very simple.
In the parabolic case there is only one equation for the
characteristics because the discriminant is zero:
You will need to take the other coordinate something else, say . You want to take something simple, but it should be independent of the other coordinate.
The partial differential equations then simplifies to the
two-dimensional canonical form
(1.16) |
You may be surprised by that. In choosing , all we did was make the coefficient zero. We did not explicitly make zero. But is zero automatically. The reason is that the physical properties of partial differential equations do not change just because you use different coordinates. A parabolic equation should stay parabolic; there are fundamental differences between the physical behaviors of parabolic, elliptic, and hyperbolic equations. And the equation above would not be parabolic if was nonzero.
ExampleQuestion: Reduce the equation
to two-dimensional canonical form.Solution:
Write the equation for the characteristics
The square root is zero, so the equation is parabolic.Solve the equation and call the integration constant :
So take the new coordinates as
The final partial differential equation then becomes
Characteristic lines are solutions to the ordinary differential
equation
Take either sign of the square root. Solve the equation and call the
integration constant, say, . Then write this integration
constant in the form
(1.17) |
Using and as the new coordinates, it turns out that the
partial differential equation takes the two-dimensional canonical form:
(1.18) |
There are significant limitations on this procedure, however, {D.5}
ExampleQuestion: Reduce the equation
to two-dimensional canonical form.Solution:
Write the equation for the characteristics
This is complex, so the equation is elliptic.Solve using separation of variables
The new coordinates can therefore be chosen as
In terms of these coordinates, the equation becomes
The new partial differential equation becomes
Convert the equation
Using rotation and stretching of the coordinates you would get