1.4 Changes of Coordinates

Changes of coordinates are a primary way to understand, simplify, and sometimes even solve, partial differential equations.

1.4.1 Introduction

It is possible to simplify many partial differential equation problems by using coordinate systems that are special to the problem:

• In unsteady pipe flows, use the lines along which sound waves propagate (characteristic lines) as coordinate lines to simplify the partial differential equation.
• In steady supersonic flows, use the Mach lines along which disturbances propagate (characteristic lines) as coordinate lines to simplify the partial differential equation.
• In problems with anisotropic properties, rotate your coordinate system along the principal or physical directions.
• In problems with spherical symmetry, spherical coordinates are usually easier than Cartesian ones.
• ...

1.4.2 The formulae for coordinate transformations

Assume the purpose is to address a problem in an $n$-dimensional space. The coordinates in this space form a vector

$$\vec x=(x_1,x_2,\ldots,x_n)$$

For example, we may have a problem in three-dimensional Cartesian coordinates $x$, $y$, and $z$. Then $x_1=x$, $x_2=y$, and $x_3=z$, and $\vec{x}$ is the position vector ${\skew0\vec r}$. Or we might have a problem in four-dimensional space-time, in which case $x_1=x$, $x_2=y$, $x_3=z$, and $x_4=t$.

The idea is now to switch to some new set of independent coordinates

$$\vec \xi = \xi_1,\xi_2,\ldots,\xi_n$$

that simplify the problem. Of course, these new coordinates will have to be some sort of functions of the old ones,

$$\xi_1 = \xi_1(x_1,x_2,\ldots,x_n)\quad \xi_2 = \xi_2(x_1... ...x_n)\quad \dots\quad \xi_n = \xi_n(x_1,x_2,\ldots,x_n) %$$ (1.5)

and vice-versa.

The change of coordinates is characterized by Jacobian matrices that have coefficients

$$\fbox{$\displaystyle \left(\frac{\partial \vec \xi}{\par... ...ht)_{ik} \equiv \frac{\partial x_i}{\partial \xi_k} $} %$$ (1.6)

Note that the first index is the one of the top vector, and the second the one of the bottom vector. The two matrices are inverses of each other:

$$\frac{\partial \vec \xi}{\partial \vec x} \frac{\partial \vec x}{\partial \vec \xi} = I$$

The determinant of either matrix is the Jacobian for the mapping from the top vector to the bottom vector. It appears in converting the volume integrals in one coordinate system into volume integrals in the other coordinate system.

The purpose is now to simplify second order quasi-linear partial differential equations using coordinate transforms. As noted in the previous section, second order quasi-linear equations are of the form

$$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \frac{\partial^2 u}{\partial x_i \partial x_j} = d$$

The set of independent coordinates $\vec{x}=(x_1,x_2,\ldots)$ is to be replaced by a cleverly chosen different set $\vec{\xi}=(\xi_1,\xi_2,\ldots)$ to simplify the equation.

Of course, before you can do anything clever like that, you have to first know what happens to the partial differential equation when the coordinates are changed. It turns out that the form of the equation remains the same in the new coordinates:

$$\sum_{k=1}^n \sum_{l=1}^n a'_{ij} \frac{\partial^2 u}{\partial \xi_k \partial \xi_l} = d'$$

The coefficients $a_{kl}$ do again form a symmetric matrix, call it $A'$. However, the matrix $A'$ is different from the matrix $A$ in the original coordinates. Also, the right hand side $d'$ is different from $d$.

The expression for the new matrix $A'$ can be written in either matrix notation or index notation. It is

$$\fbox{$\displaystyle A' = \frac{\partial \vec \xi}{\part... ...ial x_i} a_{ij} \frac{\partial \xi_l}{\partial x_j} $} %$$ (1.7)

The expression for the new right hand side $d'$ is best written in index notation:

$$\fbox{$\displaystyle d' = d - \sum_{k=1}^n \left( ... ...al x_j} \right) \frac{\partial u}{\partial \xi_k} $} %$$ (1.8)

Using equations (1.7) and (1.8) above, you can figure out what the new matrix and right hand side are. However, that may not yet be enough to fully transform the problem to the new coordinates. Recall that the coefficients $a_{ij}$ and $d$ might involve first order derivatives with respect to $x_1,x_2,\ldots$. These derivatives must be converted to derivatives with respect to $\xi_1,\xi_2,\ldots$. To do that, use:

$$\fbox{$\displaystyle \frac{\partial u}{\partial \vec x} ... ...\partial \xi_k} \frac{\partial \xi_k}{\partial x_i} $} %$$ (1.9)

in vector and index notation respectively.

That may still not be enough, because the resulting equation will probably still contain $x_1,x_2,\ldots$ themselves. You will also need to express these in terms of $\xi_1,\xi_2,\ldots$ to get the final partial differential equation completely in terms of the new coordinates.

But that is it. You are now done. At least with the partial differential equation. There might also be boundary and/or initial conditions to invert. That can be done in a similar way, but we will skip it here.

One additional point should be made. If you follow the procedure exactly as outlined above, you will have to express $\xi_1,\xi_2,\ldots$ in terms of $x_1,x_2,\ldots$, and differentiate these expressions. You will also need to express $x_1,x_2,\ldots$ in terms of $\xi_1,\xi_2,\ldots$ to get rid of $x_1,x_2,\ldots$ in the equations. But if you are, say, switching from Cartesian to sperical coordinates, the expressions for the spherical coordinates in terms of the Cartesian ones are awkward. You would much rather just deal with the expressions of the Cartesian coordinates in terms of the spherical ones.

Just dealing with $x_1,x_2,\ldots$ in terms of $\xi_1,\xi_2,\ldots$ can be done with a bit of manipulation. To get the Jacobian matrix that you want, evaluate it as the inverse of the other one:

$$\frac{\partial \vec \xi}{\partial \vec x} = \left( \frac{\partial \vec x}{\partial \vec \xi} \right)^{-1}$$

This only requires you to differentiate the original coordinates $\vec{x}$ with respect to the new ones, $\vec\xi$. And the result will already be fully in terms of the new coordinates. To convert $d$ into $d'$, as described above, you will need to evaluate the second order derivatives of $\xi_1,\xi_2,\ldots$ in it. Do that as

$$\frac{\partial^2 \xi_k}{\partial x_i\partial x_j} = \sum... ...\partial x_i}\right) \; \frac{\partial \xi_l}{\partial x_j}$$

Take the two first order derivatives at the end of this expression from the inverse matrix that you already computed.

Derivation {D.3} gives the derivation of the various formulae above.

1.4.3 Rotation of coordinates

The purpose of this section is to simplify second order partial differential equations by rotating the coordinate system over a suitable angle. It should be noted straight away that this procedure tends to be largely limited to constant coefficient linear equations.

If you simply rotate the coordinate system, Cartesian coordinates stay Cartesian. For example, suppose that the original coordinates are Cartesian coordinates $x$, $y$, and $z$, and that the unit vectors along the axes are ${\hat\imath}$, ${\hat\jmath}$, and ${\hat k}$. Then the new coordinates, call them $x'$, $y'$, and $z'$, are also Cartesian, but with different unit vectors ${\hat\imath}'$, ${\hat\jmath}'$, and ${\hat k}'$.

In any number of dimensions, the original coordinates will be indicated by $x_1,x_2,\ldots$ and the rotated coordinates $x'_1,x'_2,\ldots$. (Using $x'$ instead of $\xi$ like in the previous subsection seems more appropriate for coordinates that are merely rotated.) The axial unit vectors of the original coordinate system will be called ${\hat\imath}_1,{\hat\imath}_2,\ldots$, and those of the rotated coordinates ${\hat\imath}'_1,{\hat\imath}'_2,\ldots$.

The original partial differential equation takes the form

$$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \frac{\partial^2 u}{\partial x_i \partial x_j} = d$$

where the coefficients $a_{ij}$ form a symmetric matrix $A$. For the method to work as described here, matrix $A$ must be a constant matrix.

The partial differential equation in the rotated coordinates takes a similar form as in the original system:

$$\sum_{k=1}^n \sum_{l=1}^n a'_{kl} \frac{\partial^2 u}{\partial x'_k \partial x'_l} = d$$

However, the coefficients $a'_{kl}$ will be different.

For the right rotation of the coordinate system, the coefficients $a'_{kl}$ will be much simpler than the original coefficients $a_{ij}$. In particular, all coefficients $a'_{kl}$ with $k$ not equal to $l$ will then be zero.

The question is now how to find this right rotation of coordinates. In other words, the question is how to choose the new unit vectors ${\hat\imath}'_1,{\hat\imath}'_2,\ldots$ so that all coefficients $a'_{kl}$ with $k$ not equal to $l$ are zero.

The answer is simple. The vectors ${\hat\imath}'_1,{\hat\imath}'_2,\ldots$ must be the eigenvectors of matrix $A$. Of course, make sure that you normalize the eigenvectors of matrix $A$ to length 1 before you take them to be ${\hat\imath}'_1,{\hat\imath}'_2,\ldots$. And if there is ambiguity in their direction, (which happens if two or more eigenvalues are equal), make sure that the eigenvectors you choose are mutually orthogonal. (Since $A$ is a symmetric matrix, these requirements can always be met.)

There is really no need to compute the new matrix $A'$ from equation (1.7) of the previous subsection. If you do everything right, $a_{kl}$ will be zero if $k$ and $l$ are not equal. And a coefficient of the form $a_{kk}$ will be $\lambda_k$, the eigenvalue number $k$ of matrix $A$. (That is regardless of whatever numbering system you selected. But number eigenvalues and corresponding eigenvectors the same way.)

In short, the new matrix of coeficients will be a diagonal one, of the form

$$\fbox{$\displaystyle A' = \left( \begin{array}{cccc}... ...th}'_2\ldots$ are orthonormal eigenvectors of $A$.} $} %$$ (1.10)

So the partial differential equation simplifies to

$$\lambda_1 u_{x'_1x'_1} + \lambda_2 u_{x'_2x'_2} + \ldots + \lambda_n u_{x'_nx'_n} = d$$

There are no longer any mixed derivatives.

The conversion formulae between the coordinates are:

$$\fbox{$\displaystyle \vec x = \frac{\partial \vec x}{\pa... ... \vec x} = ({\hat\imath}'_1,{\hat\imath}'_2,\ldots)^T $} %$$ (1.11)

Here the superscript $T$ stands for transpose. (That means that you take ${\hat\imath}'_1,{\hat\imath}'_2,\ldots$ as the rows of the matrix instead of as the columns.)

The second formula in (1.11) implies that the right hand side $d$ is the same in the transformed equation as in the original one: $x'_1,x'_2,\ldots$ are linear in $x_1,x_2,\ldots$, so their second order derivatives in (1.8) are zero.

You will need the first formula in (1.11), in terms of its components, to get rid of any coordinates $x_1,x_2,\ldots$ in the right hand side $d$ in favor of $x'_1,x'_2,\ldots$. Also, if $d$ contains derivatives with respect to the unknowns $x_1,x_2,\ldots$, you will need to convert those using (1.9) of the previous subsection. To get the derivatives $x'_1,x'_2,\ldots$ with respect to $x_1,x_2,\ldots$ while doing so, write out the second formula in (1.11) in terms of its components.

Some books, like [1], do not bother to normalize the eigenvectors to length one. In that case the coordinate transformation is not just a rotation, but also a stretching of the coordinate system. The matrix $A'$ is still diagonal, but the values on the main diagonal are no longer the eigenvalues of $A$. Also, it becomes messier to find the old coordinates in terms of the new ones. Using orthonormal rather than just orthogonal eigenvectors is recommended.

Example

Question: Classify the equation

$$3 u_{xx} - 2 u_{xy} + 2 u_{yy} - 2 u_{yz} + 3 u_{zz} + 12 u_y - 8 u_z = 0$$

and put it in canonical form.

Solution:

$$3 u_{xx} - 2 u_{xy} + 2 u_{yy} - 2 u_{yz} + 3 u_{zz} + 12 u_y - 8 u_z = 0$$

Identify the matrix:

$$A = \left( \begin{array}{rrr} 3 & -1 & 0 \\ -1& 2 & -1 \\ 0 & -1& 3 \end{array} \right)$$

To find the new coordinates (transformation matrix), find the eigenvalues and eigenvectors of $A$:

The eigenvalues are the roots of $\vert A-\lambda I\vert=0$:

$$\vert A-\lambda I\vert = \left\vert \begin{array}{ccc}... ...ight\vert = (3-\lambda)^2(2-\lambda)-(3-\lambda)-(3-\lambda)$$

Hence $\lambda_1=1$, $\lambda_2=3$, $\lambda_3=4$.

The eigenvectors are solutions of $(A-\lambda I)\vec v=0$ that are normalized to length one. For $\lambda_1=1$, writing matrix $A-\lambda_1I$ and applying Gaussian elimination on it produces

$$\left( \begin{array}{rrr} 2& -1& 0 \\ -1& 1& -1 ... ...& -1& 0 \\ 0& 1& -2 \\ 0& 0& 0 \end{array} \right)$$

which gives the normalized eigenvector

$$\vec v_1 = \left( \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right) /\sqrt{6}$$

For $\lambda_2=3$,

$$\left( \begin{array}{rrr} 0& -1& 0 \\ -1& -1& -1 \\... ... -1& -1 \\ 0& -1& 0 \\ 0& 0& 0 \end{array} \right)$$

which gives the normalized eigenvector

$$\vec v_2 = \left( \begin{array}{r} 1 \\ 0 \\ -1 \end{array} \right) /\sqrt{2}$$

For $\lambda_3=4$,

$$\left( \begin{array}{rrr} -1& -1& 0 \\ -1& -2& -1 \... ... -1& 0 \\ 0& -1& -1 \\ 0& 0& 0 \end{array} \right)$$

which gives the normalized eigenvector

$$\vec v_3 = \left( \begin{array}{r} 1 \\ -1 \\ 1 \end{array} \right) /\sqrt{3}$$

The new equation is:

$$u_{\xi\xi} + 3 u_{\eta\eta} + 4 u_{\theta\theta} + 12 u_y - 8 u_z =0$$

However, that still contains the old coordinates in the first order terms. Use the transformation formulae and total differentials to convert the first order derivatives:

$$\left( \begin{array}{r} x y z \end{array} \r... ...ac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \end{array} \right)$$

and its inverse

$$\left( \begin{array}{r} \xi \eta \theta \end{a... ...ft( \begin{array}{r} x y z \end{array} \right)$$

The partial derivatives of $(\xi\eta\theta)$ with respect to $(x,y,z)$ can be read off from the final matrix. So

$$u_y = u_\xi \frac{2}{\sqrt{6}}-u_\theta \frac{1}{\sqrt{3}}$$

$$u_z =u_\xi \frac{1}{\sqrt{6}}-u_\eta \frac{1}{\sqrt{2}} + u_\theta \frac{1}{\sqrt{3}}$$

Hence in the rotated coordinate system, the partial differential equation is:

$$u_{\xi\xi} + 3 u_{\eta\eta} + 4 u_{\theta\theta} + \frac... ...rac{8}{\sqrt{2}} u_\eta - \frac{20}{\sqrt{3}} u_\theta = 0$$

1.4.3 Review Questions

1

Simplify the partial differential equation

$$10 u_{xx} + 6 u_{xy} + 2 u_{yy} = u_x + x + 1$$

by rotating the coordinate system. Classify the equation. Draw the original and rotated coordinate system and identify the angle of rotation.

Solution rotcoor-a

1.4.4 Explanation of the classification

The previous subsection showed how partial differential equations can be simplified by rotating the coordinate system. Using this procedure it is possible to understand why second order partial differential equations are classified as described in section 1.3.2.

Rotation of the coordinate system reduces a partial differential equation of the form

$$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \frac{\partial^2 u}{\partial x_i \partial x_j} = d$$

to

$$\lambda_1 u_{x'_1x'_1} + \lambda_2 u_{x'_2x'_2} + \ldots + \lambda_n u_{x'_nx'_n} = d$$

where $\lambda_1,\lambda_2,\ldots$ are the eigenvalues of the matrix $A$ that has coefficients $a_{ij}$.

That immediately explains why only the eigenvalues of matrix $A$ are of importance for the classification. Rotating the mathematical coordinate system obviously does not make any difference for the physical nature of the solutions. And in the rotated coordinates, all that is left of matrix $A$ are its eigenvalues.

The next question is why the classification only uses the signs of the eigenvalues, not their magnitudes. The reason is that the magnitude can be scaled away by stretching the coordinates. That is demonstrated in the next example.

Example

Question: The previous example reduced the elliptic partial differential equation

$$3 u_{xx} - 2 u_{xy} + 2 u_{yy} - 2 u_{yz} + 3 u_{zz} + 12 u_y - 8 u_z = 0$$

to the form

$$u_{\xi\xi} + 3 u_{\eta\eta} + 4 u_{\theta\theta} + \frac... ...rac{8}{\sqrt{2}} u_\eta - \frac{20}{\sqrt{3}} u_\theta = 0$$

Reduce this equation further until it becomes as closely equal to the Laplace equation as possible.

Solution:

The first step is to make the coefficients of the second order derivatives equal in magnitude. That can be done by stretching the coordinates. If

$$\xi = \bar \xi \quad \eta = \sqrt 3 \bar \eta \quad \theta = 2 \bar \theta$$

then

$$u_{\bar\xi\bar\xi} + u_{\bar\eta\bar\eta} + u_{\bar\theta\... ...{6}} u_{\bar\eta} - \frac{10}{\sqrt{3}} u_{\bar\theta} = 0$$

Note that all that is left in the second order derivative terms is the sign of the eigenvalues.

You can get rid of the first order derivatives by changing to a new independent variable $v$. To do so, set $u=v e^{a\bar\xi+b\bar\eta+c\bar\theta}$. Plug this into the differential equation above and differentiate out the product. Then choose $a$, $b$, and $c$ so that the first derivatives drop out. You will find that you need:

$$a = -\frac{8}{\sqrt{6}} \quad b = -\frac{4}{\sqrt{6}} \quad c = \frac{5}{\sqrt{3}} \quad$$

Then the remaining equation turns out to be:

$$v_{\bar\xi\bar\xi} + v_{\bar\eta\bar\eta} + v_{\bar\theta\bar\theta} - \frac{65}{3} v = 0$$

It is not exactly the Laplace equation because of the final term. But the final term does not even involve a first order derivative. It makes very little difference for short-scale phenomena. And short scale phenomena (such as singularities) are the most important for the qualitative behavior of the partial differential equation.

As this example shows, the values of the nonzero eigenvalues can be normalized to 1 by stretching coordinates. However, the sign of the eigenvalues cannot be changed. And neither can you change a zero eigenvalue into a nonzero one, or vice-versa, by stretching coordinates.

You might wonder why all this also applies to partial differential equations that have variable coefficients $a_{ij}$ and $d$. Actually, what $d$ is does not make much of a difference. But generally speaking, rotation of the coordinate system only works if the coefficients $a_{ij}$ are constant. If they depend on position, the eigenvectors ${\hat\imath}'_1,{\hat\imath}'_2\ldots$ at every point can still be found. So it might seem logical to try to find the new coordinates $x'_1,x'_2,\ldots$ from solving $\partial\vec{x}'/\partial\vec{x}=({\hat\imath}'_1,{\hat\imath}'_2,\ldots)^T$. But the problem is that that are $n^2$ equations for only $n$ unknown coordinates. If the unit vectors are not constant, these equations normally mutually conflict and cannot be solved.

The best that can normally be done for arbitrary $A$ is to select a single point that you are interested in. Then rotate the coordinate system to diagonalize the partial differential equation at that one point. In that case, $A$ is diagonal near the considered point. And that is enough to classify the equation at that point. For, the most important feature that the classification scheme tries to capture is what happens to short scale phenomena. Short scale phenomona will “see” the locally diagonal equation. So the classification scheme continues to work.

1.4.4 Review Questions

1

Convert the equation

$$11 u_{x'x'} + u_{y'y'} = \frac{3}{\sqrt{10}} u_{x'} - \frac{... ...} u_{y'} + \frac{3}{\sqrt{10}} x' - \frac{1}{\sqrt{10}} y' + 1$$

to be as close as possible to the Laplace equation.

Solution expclass-a