18.7 Changes of Coordinates

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

18.7.1 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.
• ...

18.7.2 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 $x_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$, $x_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y$, and $x_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$, and $\vec{x}$ is the position vector ${\skew0\vec r}$. Or we might have a problem in four-di­men­sion­al space-time, in which case $x_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$, $x_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y$, $x_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$, and $x_4$ $\vphantom0\raisebox{1.5pt}{$=$}$ $t$.

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,

$$\xi_1 = \xi_1(x_1,x_2,\ldots,x_n)\quad \xi_2 = \xi_2(x_1... ...,x_n)\quad \dots\quad \xi_n = \xi_n(x_1,x_2,\ldots,x_n) %$$ (18.10)

and vice-versa.

The change of coordinates is characterized by Jacobian matrices

$$\fbox{$\displaystyle {\cal J} \equiv \frac{\partial \vec... ...^{-1} \equiv \frac{\partial \vec \xi}{\partial \vec x} $} %$$ (18.11)

These matrices are inverses of each other, as the above notation indicates. The determinant $\vert{\cal J}\vert$ is the Jacobian $J$ of the transformation from $\vec{x}$ to $\vec\xi$. It is used in converting volume integrals. In particular

$${\rm d}x_1 {\rm d}x_2 \ldots {\rm d}x_n = J {\rm d}\xi_1 {\rm d}\xi_2 \ldots {\rm d}\xi_n$$

The complete Jacobian matrices describes how a small change in $\vec\xi$ relates to the corresponding change in $\vec{x}$, and vice versa:

$$\fbox{$\displaystyle {\rm d}\vec x = {\cal J} {\rm d}\ve... ... \qquad {\rm d}\vec\xi = {\cal J}^{-1} {\rm d}\vec x $} %$$ (18.12)

Note that while $\cal{J}$ is called the transformation matrix from $\vec{x}$ to $\vec\xi$, it really allows you to compute ${\rm d}\vec{x}$ from ${\rm d}\vec\xi$; the opposite of what you would expect.

In index notation, the components of the Jacobian matrices are

$$\fbox{$\displaystyle \Big({\cal J}\Big)_{ik} = \left(\... ...ght)_{ki} \equiv \frac{\partial \xi_k}{\partial x_i} $} %$$ (18.13)

For any values of $i$ and $k$ between 1 and $n$. Note that the first index is the one of the top vector, and the second index the one of the bottom vector.

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 $\vec{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x_1,x_2,\ldots)$ is to be replaced by a cleverly chosen different set $\vec{\xi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\xi_1,\xi_2,\ldots)$ 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:

$$\fbox{$\displaystyle A' = {\cal J}^{-1} A {\cal J}^{-\rm... ...partial x_i} a_{ij} \frac{\partial\xi_l}{\partial x_j} $} %$$ (18.14)

Here $-\rm {T}$ means the transpose of ${\cal{J}}^{-1}$. The transpose matrix has the columns of ${\cal{J}}^{-1}$ as its rows.

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

$$\fbox{$\displaystyle d' = d - \sum_{k=1}^n \left( ... ...ial x_j} \right) \frac{\partial u}{\partial \xi_k} $} %$$ (18.15)

Using equations (18.14) and (18.15) 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:

$$\fbox{$\displaystyle \frac{\partial u}{\partial \vec x} ... ...{\partial \xi_k} \frac{\partial \xi_k}{\partial x_i} $} %$$ (18.16)

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 as outlined above exactly, 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. That is a lot of work. Also, 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.

Now differentiating the $x_1,x_2,\ldots$ with respect to the $\xi_1,\xi_2,\ldots$ will give you matrix $\cal{J}$ instead of ${\cal{J}}^{-1}$. But you can invert the matrix relatively easily using the method of minors. While that is a bit of work, you also save a lot of work because you no longer have to convert $x_1,x_2,\ldots$ in the results to $\xi_1,\xi_2,\ldots$ and clean up the mess.

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.4} gives the derivation of the various formulae above.

18.7.3 Rotation of coordinates

The purpose of this section is to simplify second order partial differential equations by rotating the coordinate system to a different orientation. This allows you to simplify the matrix $A$ of the partial differential equation considerably. In particular, in this way you can bring the new matrix $A'$ into the form of a diagonal matrix:

$$\fbox{$\displaystyle A' = \left( \begin{array}{cccc}... ...dots & \ddots \end{array} \right) \equiv \Lambda $} %$$ (18.17)

So the partial differential equation simplifies in the new coordinates to:

$$\fbox{$\displaystyle \lambda_1 u_{\xi_1\xi_1} + \lambda_... ...\xi_2\xi_2} + \ldots + \lambda_n u_{\xi_n\xi_n} = d' $} %$$ (18.18)

There are no longer any mixed derivatives. And the remaining coefficients of the PDE are the eigenvalues of the original matrix $A$.

One limitation to the procedure in this section should be stated right away. It concerns the case that the matrix $A$ is not constant, but varies from point to point. For such a partial differential equation, you can select a point, any point you like, and bring the equation in the above diagonal form at that one selected point. At other points there will then still be mixed derivatives in the transformed equation.

To figure out how to convert a partial differential equation to the above diagonal form, first a brief review of linear algebra is needed. First recall that in three dimensions, you can define basis vectors ${\hat\imath}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(1,0,0)$, ${\hat\jmath}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(0,1,0)$, and ${\hat k}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(0,0,1)$. And you can write any other three dimensional vector in terms of these three basis vectors, like for example $3{\hat\imath}-2{\hat\jmath}+4{\hat k}$. Similarly in $n$ dimensions you can define $n$ basis vectors ${\hat\imath}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(1,0,0,\ldots)$, ${\hat\imath}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(0,1,0,\ldots)$, ...${\hat\imath}_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\ldots,0,0,1)$.

Next, a simple linear transformation of coordinates (which leaves the origin unchanged) takes the form, by definition,

$$\fbox{$\displaystyle \vec x = P \vec \xi \qquad \vec \xi = P^{-1} \vec x $} %$$ (18.19)

Here $P$ is a matrix that is called the transformation matrix from $\vec{x}$ to $\vec\xi$. (Although it really computes $\vec{x}$ from $\vec\xi$.)

Matrix $P$ consists of the basis vectors of the new coordinate system, viewed from the old coordinate system. So for the special case that the transformation is a simple rotation of the coordinate system, matrix $P$ consists of the rotated basis vectors ${\hat\imath}_1,{\hat\imath}_2,\ldots,{\hat\imath}_n$, call them ${\hat\imath}'_1,{\hat\imath}'_2,\ldots,{\hat\imath}'_n$, viewed from the original coordinate system. (Conversely, $P^{-1}$ consists of the original basis vectors ${\hat\imath}_1,{\hat\imath}_2,\ldots,{\hat\imath}_n$ when viewed from the new coordinate system.) The important thing to remember is that for the special case of coordinate rotation, the inverse of $P$ is just its transpose:

$$\fbox{$\displaystyle \mbox{Coordinate system rotation:} \quad P^{-1}=P^{\rm T} $} %$$ (18.20)

Note further that for linear transformations, the Jacobian matrices are just

$${\cal J} = \frac{\partial \vec x}{\partial \vec \xi} = P ... ... J}^{-1} = \frac{\partial \vec \xi}{\partial \vec x} = P^{-1}$$

So the expression of the previous subsection for the new matrix $A'$ becomes in terms of $P$:

$$A' = {\cal J}^{-1} A {\cal J}^{-\rm T} = P^{-1} A P^{-\rm T}$$

Now it is known from linear algebra that this becomes the diagonal matrix $\Lambda$ given at the start of the subsection if you take $P^{-\rm T}$ as the matrix $E$ of eigenvectors of $A'$. (If $A$ varies from point to point, that means more specifically the eigenvectors of the selected point.) But for a rotation of coordinates, $P^{-\rm T}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $P^{\rm TT}$ is just $P$. So the needed coordinate transform is

$$\fbox{$\displaystyle \vec x = E \vec \xi \qquad \vec \xi = E^{\rm T} \vec x $} %$$ (18.21)

where the eigenvectors of $A$ are the columns of matrix $E$. The eigenvalues in diagonal matrix $A'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Lambda$ will be in the same order as the corresponding eigenvectors in $E$.

Warning: you must normalize the eigenvectors (divide them by their length) because the basis vectors ${\hat\imath}_1,{\hat\imath}_2,\ldots,{\hat\imath}_n$ are all of length one. And first you must make sure every one is orthogonal to all the others. Fortunately, that is normally automatic. However, if you have a double eigenvalue, any two corresponding eigenvectors are not necessarily orthogonal; you must explicitly make them so. Similarly, for a triple eigenvalue, you will need to create three corresponding orthogonal eigenvectors. And then divide each by its length. (To check whether vectors are orthogonal, check that their dot product is zero.)

Some books, like [3], 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. You would have to find the inverse of $P^{-1}$ using minors. Using orthonormal rather than just orthogonal eigenvectors is recommended.

You might wonder, if $A$ varies from point to point, why can we not simply set

$${\cal J} \equiv \frac{\partial \vec x}{\partial \vec \xi} = E$$

at every point, where matrix $E$ consists of the eigenvectors of $A$ at that point. That would make $A'$ diagonal at every point, instead of just a selected point. Unfortunately however, this does not work, because it is equivalent to $n^2$ scalar differential equations for the $n$ scalar components of $\vec{x}$. If the number of equations is larger than the number of unknowns, there is normally no solution.

Do recall that you will also have to transform the right hand side $d$ to the new coordinates. However, the second formula in (18.21) 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 (18.15) are zero.

You will need the first formula in (18.21), 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 (18.16) 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 (18.21) in terms of its components.

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 $\vert A-\lambda I\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0:

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

Hence $\lambda_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $\lambda_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, $\lambda_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4.

The eigenvectors are solutions of $(A-\lambda I)\vec v$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 that are normalized to length one. For $\lambda_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 $\lambda_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 $\lambda_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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:

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

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:

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

18.7.3 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

18.7.4 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 18.6.2.

From the above, it already starts to become clearer why the classification of second order partial differential equations is in terms of the eigenvalues $A$. If two different second order partial differential equations have the same eigenvalues of their matrix $A$, then you can simply rotate the coordinate system to make their matrices $A$ equal. And the highest order derivatives make the biggest difference for the physical behavior of the system. For short-scale effects, which include singularities, the highest order derivatives dominate. For them, the right hand side $d$ is relatively unimportant. And since the highest order terms are now equal for the two partial differential equations, they must behave very similarly. So they should be classified as being in the same group.

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 $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $v e^{a\bar\xi+b\bar\eta+c\bar\theta}$. 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 $\partial\vec{x}'/\partial\vec{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $({\hat\imath}'_1,{\hat\imath}'_2,\ldots)^T$. 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.

18.7.4 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