ABSTRACT
Partial dierential equations
Coordinate systems and separability
In section we have seen that it is possible to separate linear di erential
equations into ordinary di erential equations The setup made this
possible for cartesian coordinates Since theHelmholtz equation is separable
in coordinate systems we now investigate these systems more closely
We consider the line element in cartesian coordinates
ds
dx
dy
dz
X
i
dx
i
Making a transformation x
i
x
i
q
k
to new coordinates q
k
then
dx
i
X
k
x
i
q
k
dq
k
i
This expression depends only on the new coordinates q
k
since the x
i
x
i
q
k
are functions of q
k
If the new coordinates q
k
are such that the expression
X
i
x
i
q
l
x
i
q
m
or e
l
e
m
l m
vanishes then the new coordinate system is called orthogonal In this case
the unit vectors e
l
and e
m
are orthogonal Furthermore if the Jacobian
functional determinant
det jq
i
x
k
q
k
j
does not vanish an inverse transformation exists and
dq
l
X
i
q
l
x
i
dx
i
Since the cartesian coordinates are independent from each other one has
x
i
x
k
ik
Kronecker symbol and one can write
x
i
q
l
ik
q
i
ik
Inserting into one obtains
dq
l
X
i
q
l
x
i
X
k
x
i
q
k
dq
k
X
ik
q
l
x
i
x
i
q
k
dq
k
lk
dq
k
dq
l
Raising to the second power results in
dx
i
X
kl
x
i
q
k
x
i
q
l
dq
k
dq
l
Dening now the metric tensor
g
kl
X
i
x
i
q
k
x
i
q
l
one may write the line element in the form
ds
X
kl
g
kl
dq
k
dq
l
X
i
dx
i
Mathematica is of great help In the standard Add On Packages located at
usrlocalmathematicaAddOnsStandardPackages one nds the package
VectorAnalysis The command
<<CalculusVectorAnalysis loads the package In this package the default coordinate system is Cartesian with coordinate variables Xx Y y Zz The command
Coordinates[Cartesian] yields fXx Y y Zzg and
Coordinates[Spherical] gives fRr T theta Pphig
To set a special coordinate system we type
SetCoordinates[Spherical] so that now spherical coordinates represent the default system Some more
commands are useful
CoordinateRanges[ ] gives the intervals over where each of the coordinate variables of the last
dened default system may range Rr T theta Pphi and
CoordinateRanges[Cylindrical] gives the result for the Cylindrical system etc Mathematica o ers transfor
results in
p
p
which are the cartesian coordinates of the point whose spherical coordinates
are On the other hand
r
p
x
y
ArcTanyx or ArcTanx y z
To obtain the formulae for the transformation between cartesian and spherical
coordinates we give the commands
p
x
y
z
ArcCos
z
p
x
y
z
ArcTanx y
and
CoordinatesToCartesian[{Rr,Ttheta,Pphi},Spherical] is yielding
Rr CosPphi SinT theta Rr SinPphi SinT theta Rr CosT theta
Mathematica also understands how to calculate the functional determinant
The command
MatrixForm[JacobianMatrix[Spherical[r,theta,phi]]] forces Mathematica to give the result in the form of a matrix
B
B
Cosphi Sintheta rCosphi Costheta r Sinphi Sintheta
Sinphi Sintheta rCostheta Sinphi rCosphi Sintheta
Costheta r Sintheta
C
C
A
and calculate the determinant of
Simplify[Det[%]] results in g r
sin
where g is the determinant of the metric tensor Using the
tensor itself is then represented by the matrix
g
kl
r
A
and
dx sin cos dr r cos cos d r sin sin d
dy sin sin dr r cos sin d r sin cos d
dz cos dr r sin d
dr sin cos dx sin sin dy cos dz
d cos cos dx cos sin dy sin dz
d sin dx cos dy
The line element now reads
ds
dx
dy
dz
P
kl
g
kl
dq
l
dq
k
dr
r
d
r
sin
d
The Helmholtz equation
ux y z k
ux y z
and the threedimensional Laplace equation
ux y z u
xx
u
yy
u
zz
are separable in coordinate systems As long as a boundary coincides with
the coordinate lines of these coordinate systems it is easy to solve the
boundary value problem For other partial di erential equations and other
coordinate systems the question of separability is answered by the Staeckel
determinant and
The coordinate systems are in Mathematica notation
Cartesian x y z Cylindrical r z
Spherical r ParabolicCylindrical u v z
Paraboloid u v EllipticCylindrical u v z a
ProlateSpheroidal a OblateSpheroidal a
Conical a b ConfocalEllipsoidal a b c
ParabolicCylindrical uvz
The parameters a b c refer on the centre of the system and other geometric
properties axes eccentricity Other useful coordinate systems are Bipolar
u v z a Bispherical u v a and various toroidal systems
Vector analysis is also supported byMathematica In cartesian coordinates
a vector
K will be dened by
K K
x
e
x
K
y
e
y
K
z
e
z
and in spherical coordinates one has
K K
r
e
r
K
e
K
e
whereK
x
K
y
etc andK
r
are the vector components in the actual coordinate
cartesian into other coordinates necessitates a transformation not only of the
coordinates and the components but also of the unit vectors e
i
e
l
according
to the scheme
cartesian new system new system cartesian
dx
i
X
k
x
i
q
k
dq
k
dq
l
X
i
q
l
x
i
dx
i
e
i
X
k
x
i
q
k
p
g
kk
e
k
e
l
X
i
q
l
x
i
e
i
K
i
X
l
q
l
x
i
K
l
K
l
X
i
x
i
q
l
p
g
ll
K
i
Due to the di erentials will also be transformed We now give an
example for the transformation of a vector between cartesian and spherical
coordinates
e
x
sin cos
e
r
cos cos
e
sin
e
e
y
sin sin
e
r
cos sin
e
cos
e
e
z
cos
e
r
sin
e
e
r
sin cos e
x
sin sin e
y
cose
z
e
cos cos e
x
cos sin e
y
sine
z
e
sin e
x
cos e
y
K
x
K
r
sin cos K
cos cos K
sin
K
y
K
r
sin sin K
cos sin K
cos
K
z
K
r
cosK
sin
K
r
K
x
sin cos K
y
sin sin K
z
cos
K
K
x
cos cos K
y
cos sin K
z
sin
K
K
x
sin K
y
cos
These calculations have been made using the two matrices
M
li
x
i
q
l
B
B
sin cos sin sin cos
r cos cos r cos sin r sin
r sin sin r sin cos
C
C
A
N
il
q
l
x
i
B
sin cos cos cos sin
sin sin cos sin cos
C
A
K
i
X
l
N
il
K
l
and according to
K
l
X
i
M
li
g
ll
K
i
Now we can elaborate vector algebra and vector analysis The scalar dot
product of two vectors v
and v
is computed in the default coordinates using
CalculusVectorAnalysis
v
v
DotProduct[v1,v2] where v
v
x
v
y
v
z
denes a vector If the scalar product is to be cal
culated in another coordinate system the command DotProduct[v1,v2] is useful
Let v
axe
x
e
y
e
z
v
e
x
e
y
or
v
a x is obtained by DotProduct[v1,v2] The vector cross product is dened by
v
v
CrossProduct[v1,v2] so that one obtains the new vector f axg
Mathematica calculates the scalar triple product using
ScalarTripleProduct[v1,v2,v3] Next we dene the del nabla operator by
r e
x
x
e
y
y
e
z
z
rU gradU
or more generally
r
X
l
e
l
p
g
ll
q
l
rU gradU
X
l
e
l
p
g
ll
U
q
l
Mathematica uses Grad[f,coordsys] Thus Grad[7*xˆ3+yˆ2-zˆ4,Cartesian[x,y,z]] delivers the vector f x
yz
g
The curlv is elaborated by
Curl[v,Spherical[r,theta,phi]] Let v={rˆ2*Sin[theta],r*Cos[phi]*Sin[theta],r*Sin[phi]}; then yields the vector
fCottheta Sinphi Sinphir Costheta Cosphi Sinthetag
Using the palette with Greek and other special symbols we may rewrite
Csc r
Cos Sin r
Sin Sin
r
Sin
r
Cos r Cos Sin
r
Finally the Laplacian operator can be realized by the command
Laplacian[f] and the divergence of a vector v is given by Div[v] Behind these commands there are some complicated calculations In general
coordinates q
l
one has
r
X
l
e
l
p
g
ll
q
l
rU gradU
X
l
e
l
p
g
ll
U
q
l
In spherical coordinates this reads
r e
r
r
e
r
e
r sin
Applying the operator div on a vector yields
div eU U dive e gradU
div
K div
X
l
e
l
K
l
X
l
K
l
dive
l
e
l
gradK
l
or in spherical coordinates
div
K div K
r
e
r
K
e
K
e
K
r
dive
r
K
dive
K
dive
e
r
gradK
r
e
gradK
e
gradK
It is thus clear that the basic vectors e have also to be transformed But
expressions like
e
r
r
e
r
e
sin cos e
x
sin sin e
y
cose
z
are quite cumbersome It is hence of advantage to use cartesian coordinates
for such calculations We can write
dive
r
e
x
e
y
e
z
x
e
x
y
e
z
e
l
k
lk
dive
r
x
x
r
y
y
r
z
z
r
r
and
dive
cot
r
dive
Then
div
K
K
r
r
r
K
r sin
K
r
K
r
r
K
cot
or more generally
div
K r
K
X
l
p
g
ll
K
l
q
l
p
g
X
l
K
l
q
l
r
g
g
ll
Similar situations appear for curl and One has
curl
K r
K K
curle
K
curle
K
curle
gradK
e
gradK
e
gradK
e
and in spherical coordinates this reads
curl
K
r
r
K
r sin
K
r
K
cot
curl
K
r sin
K
r
K
r
r
K
curl
K
K
r
r
K
r
r
K
and more generally
r
K
p
g
K
q
p
g
K
q
K
p
g
g
p
g
q
K
p
g
g
p
g
q
r
K
p
g
K
q
p
g
K
q
K
p
p
g
K
p
p
g
r
K
p
g
q
p
g
q
K
p
g
g
p
g
q
K
p
g
g
p
g
q
For the Laplacian one nds
U div gradU
p
g
X
l
q
l
p
g
p
g
ll
U
q
l
or in spherical coordinates
U
U
r
r
U
r
r
sin
U
r
U
cos
r
sin
U
Now it is quite simple to separate the Helmholtz equation Uk
U
Using the separation setup
Ur Rr
one gets after division by U
R
r
R
k
ar
R
cot
a bsin
b
where a and b are separation constants These ordinary di erential equations
may be solved using the methods discussed in chapter Again Mathematica
commands are of great help Using and applying
Expand[(Laplacian[U[r,,],Spherical[r, ,]]) *rˆ2/U[r,,]]; results in the yet unseparated equations in the form
r R
r
Rr
Cot
r
R
r
Rr
Csc
The command Expand is used to expand out products or positive integer powers in an expression Please remember to load the package VectorAnal
ysis
The vector Helmholtz equation which can be found in electromagnetism
presents diculties In general coordinate systems it is dened by
three equations
K
x
k
K
x
K
x
x
K
x
y
K
x
z
k
K
x
K
y
k
K
y
K
y
x
K
y
y
K
y
z
k
K
y
K
z
k
K
z
K
z
x
K
z
y
K
z
z
k
K
x
but in other coordinate systems problems appear the equivalent equations
are coupled In spherical coordinates one nds
K
r
k
K
r
K
r
r
r
K
r
r
sin
K
r
r
K
r
r
cot
r
K
r
K
r
r
r
K
K
cot
r
r
sin
K
k
K
r
But a trick may help use the cartesian vector components as functions
of the general spherical coordinates so that
K
l
K
l
Solve the
three equations and then transform the cartesian coordinates into the gen
eral spherical etc coordinates which had been chosen originally to t a
boundary surface
In two dimensions the problem of separability of partial di erential equa
tions looks nicer Due to conformal mapping the Laplace equation becomes
separable in an innite number of twodimensional systems and for axially
symmetric problems threedimensional conformal mapping There are also
some separable twodimensional toroidal systems
x a sinh cosT y a sinh sinT
z a sinT T cosh cos
or quasitoroidal systems like the helical coordinate system or mag
netic eld lines coordinate according to Boozer or Hamada
It might be of interest to state that there are also nonlinear partial di er
ential equations that are separable So the equation
u
xx
u
tt
u
xt
ku
n
x
u
xx
is separable by the setup
m
nm m
one obtains
mm GG
m
G
km
n
m
G
n
On the other hand
u
t
u
n
u
x
x
nu
n
u
x
u
n
u
xx
can be separated by ux t T tXx into
T
T
n
X
X
XX
X
where is the separation const Another possibility to separate is
given by u F t x
p
Gt Another separable nonlinear partial di erential
equation of rst order is given by the Jacobi equation of mechanics
S
x
S
y
S
z
fx gy hz
which can be separated by S
x S
y S
z
Problems
Find the transformation formulae between cartesian and cylindrical co
ordinates using and
CoordinatesFromCartesian[{x,y,z},Cylindrical] which should give
p
x
y
ArcTanx y z
and
should yield
Rr CosTtheta Rr SinTthetaPphi
What happens if you replace Rr rTtheta Pphi
r Cos r Sin
Derive the formulas for the transformation between cartesian and elliptic
cylinder coordinates First we look up the elliptic cylindrical system by
CoordinatesFromCartesian[{x,y,z}, EllipticCylindrical] The result is astonishing ReArcCoshx iy ImArcCoshx iy z
but becomes clear by the command
CoordinatesToCartesian[{u,v,z},EllipticCylindrical] giving Cosv Coshu Sinv Sinhu z
Thus the coordinate surfaces are given by elliptic cylinders xa cosh
ya sinh
const by hyperbolic cylinders xa cos
ya sin
const and planes z const The default value
of a
Now we would like to know the interval over which the coordinates u v z
may vary The command
CoordinateRanges[EllipticCylindrical] gives the answer u v z
Let us now consider conical coordinates Typing
f Llambda Mmu
Nnu
g
which gives V v Uu Pphi but
f ReArcCoth
p
x
y
i z ImArcCoth
p
x
y
i z
ArcTanx yg
Investigate cylindrical coordinates r z We rst try
which yields Rr Ttheta Zz
Now let us solve the Laplace equation in these coordinates We dene
a separation setup
and give the command
r Rr
Rr
Zz
r
We now read o the three ordinary equations
R
R
R
rR
b
a
r
or R
R
r
a
r
b
R
Z
Z
b or Z
bZ
r
a
r
or
a
which can be solved according to the methods given earlier
Investigate now the twodimensional Laplace equation If one inserts
the ansatz
one gets the result
x X
x
X x
x
X
x
X x
Y
y
Y y
Solve the two ordinary equations just obtained for Xx and Y y
The result should be
ffX x C Cosha Logx i C Sinha Logxgg
and the solution of the twodimensional Laplacian reads
Demonstrate that the harmonic polynomial
xˆ4-6*xˆ2*yˆ2+yˆ4 is a solution of the twodimensional Laplace equation
The operation Div on a vector may formally be regarded as the DotProduct of the two vectors to calculate this divergence
Methods to reduce partial to ordinary dierential
equations
In the last section we have seen that many but not all partial di erential
equations can be reduced to ordinary di erential equations by the separation
method There exist however other methods to reach this goal Some of them
are very useful for engineering problems The Laplace transform theory has
become a basic part of electronic engineering study Like in all other
integral transformations of ordinary di erential equations
p
zw
z p
zw
z p
zwz
a setup
wz
t
Z
t
Kz tytdt
is made t is a new complex or real variable and Kz t is the kernel of the
transformation Various such kernels have been used
LaplaceTransformation
Z
expzt
i
i
Z
i
expzt
Fourier transformation
Z
expizt
Z
expizt
Hankel transformation
Z
tJ
n
zt
Z
zJ
n
zt
Mellin transformation
Z
t
z
i
i
Z
i
z
t
Euler transformation
Z
z t
The right column presents the inverse transformation Now we will demon
strate the application of the Laplace transformation on partial di erential
equations As an example we choose the heat conduction equation for heat
conductivity
T
T
T x x
Here T x t represents the temperature at point x at time t Multiplication
of by the Laplace kernel expzt and integration results in
Z
expzt
T
x
dt
x
Z
expztT x tdt
Z
expzt
T
t
dt
Due to one has the Laplace transform wz x of T x t
wz x
Z
expztT x tdt
Partial integration of the rhs term of
R
udv uv
R
vdu yields
Z
expzt
T
t
dt
expztT x t
Z
zT x t expztdt
the Laplace transform of Tt due to exp and exp Since
z and t are independent variables we may rewrite as
Z
expzt
T
t
dt T x z
Z
T x t expztdt
Using the denitions we thus have
Z
expzt
T
t
dt zwx z x
Now we consider
Tx
The next step is
Z
expzt
T
x
dt
d
dx
Z
expztT x tdt
d
dx
wx z
and thus
Z
expzt
T
x
dt
d
wx z
dx
Collecting the terms and inserting we obtain from
d
wx z
zwz x x
solving one needs to apply the inverse transformationwz x to obtain
the solution T x t of
It is remarkable that the initial condition x became the inhomogeneous
term of the ordinary di erential equation For the solution of
reads
wx z C
z expx
p
z C
z expx
p
z
C
C
are the integration constants depending on z They allow to satisfy the
boundary conditions of the partial di erential equation They might
be
T t t T t
or
w z
Z
expztT tdt
Z
expzt tdt !wz
so that C
z C
z !wz To obtain the correct full solution of the
partial di erential equation one has to apply the inverse transformation on
wx z with respect to z
Mathematica may help with these calculations It has a special package
that is to be loaded into the memory of your computer by the command
In some newer versions of Mathematica the package is autoloading and this
command no longer needed
To understand howMathematica handles the problem we start step by step
with which now reads
z LaplaceTransformT x t t z T x
what is equivalent with The next step is
LaplaceTransform[D[T[x,t],{x,2}],t,z] results in
LaplaceTransformT
x t t z
what is equivalent with Mathematica sometimes uses the notation
w[z][x] for wz x to point to the di erence between the variable and the parameter z
The integral transformations work only for linear ordinary or partial di er
ential equations Nonlinear equations may be handled by similarity transfor
itself on temperature Boltzmann suggested the pseudo Laplacian operator
u
t
x
fux t
u
x
u
t
to replace Now a similarity variable
x t x
t
has been dened Insertion of into results in
u
x
du
d
x
u
x
t
u
t
u
x
t
u
xx
x
t
u
x
t
u
and assuming the transition ux t u
x
t
u
fuu
fuu
In this expression all terms depend on with the exception of x
t To
continue one assumes
x
t or x
t
which yields and The physical interpretation and the usefulness of
even for the linear equation const will be discussed later on
problems and in section
The disadvantage of similarity transformations is hidden in the fact that
similarity solutions may only satisfy very restricted boundary conditions
Birkhoff has developed several methods to nd similarity transforma
tions He uses transformations of two independent
variables occurring in linear or nonlinear partial di erential equations
!x
i
a
i
x
i
i m
!y
j
a
j
y
j
j n
Let us look at an example We consider the nonlinear partial di erential
equation of second order for ux y t
u
p
u
xx
u
yy
pu
p
u
x
u
y
u
t
Using !x a
m
x !y a
n
y
!
