Subsections

### 1.5 Two-Dimensional Coordinate Transforms

More powerful simplifications by changing coordinates are possible in 2D.

Assume that in terms of coordinates and , we have a partial differential equation:

Then, if we transform to new coordinates, call them and , we will get a new partial differential equation of the form:

The idea is again to choose the new coordinates and so that the new partial differential equation is as simple as possible.

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}:

 (1.12)

#### 1.5.1 Characteristic Coordinates

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:

That can be considered to be a partial differential equation for . A nonlinear first order equation, to be sure. Similarly for to vanish,

Note that and must satisfy the exact same equation, but they must be different solutions. Otherwise they are not valid independent coordinates.

To solve the equation for ( goes the same way), divide by :

and note that, from your calculus or thermo,

So the lines of constant should satisfy the ordinary differential equation

We can achieve this by taking to be the integration constant in the solution of this ordinary differential equation! Integration constants are, like the word says, constant for solutions.

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:

1. Since integration constants are not unique, the characteristic coordinates are not. But the lines of constant and are unique, and are called characteristic lines or characteristics.
2. Elliptic equations do not have characteristics, because the square root in the ordinary differential equation would be imaginary. The coordinates and must be real; you do not want to deal with partial differential equations in complex coordinates.
3. Parabolic equations have only one family of characteristic lines. That is because the square root is zero, so taking the other root does not make a difference.

Example

Question: 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

 (1.14)

Note that could be divided out, so there is no need to figure out what it is.

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.

Example

Question: 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.

Example

Question: 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.

Example

Question: 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.

#### 1.5.2 Parabolic equations in two dimensions

In the parabolic case there is only one equation for the characteristics because the discriminant is zero:

So you can only find one characteristic coordinates, call it .

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.

Example

Question: 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

#### 1.5.3 Elliptic equations in two dimensions

Characteristic lines are solutions to the ordinary differential equation

Elliptic equations have no real characteristics, because the square root is imaginary. However, elliptic equations can still be simplified, assuming that the above ordinary differential equation can be solved analytically.

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)

where and are real and . In other words, take and .

Using and as the new coordinates, it turns out that the partial differential equation takes the two-dimensional canonical form:

 (1.18)

You may note that this is quite similar to what you can get from rotating the coordinate system, as in the previous section. However, the above procedure works even if the coefficients , , and of the original partial differential equation are not constants.

There are significant limitations on this procedure, however, {D.5}

Example

Question: 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

1.5.3 Review Questions
1

Convert the equation

to two-dimensional canonical form.

Using rotation and stretching of the coordinates you would get

Do you get the same equation? Should you? Comment.