5. Similarity Solutions for PDE’s
For linear partial differential equations there are various techniques for reducing the pde to
an ode (or at least a pde in a smaller number of independent variables). These include
various integral transforms and eigenfunction expansions. Such techniques are much less
prevalent in dealing with nonlinear pde’s. However, there is an approach which identifies
equations for which the solution depends on certain groupings of the independent variables
rather than depending on each of the independent variables separately. We will illustrate
this technique first for a linear pde. Consider the heat equation
/ t uÝx, tÞ ? D / xx uÝx, tÞ = 0
Ý5.1Þ
and introduce the dilation transformation
z = P a x,
vÝz, sÞ = P c uÝP ?a z, P ?b sÞ
s = P b t,
Ý5.2Þ
Then / t uÝx, tÞ = P ?c / s vÝz, sÞ s v ÝtÞ = P b?c / s vÝz, sÞ, etc and so (5.1) becomes
P b?c / s vÝz, sÞ ? D P 2a?c / zz vÝz, sÞ = 0.
Now if b ? c = 2a ? c, i.e., b = 2a then the equation (5.1) is invariant under this
transformation; i.e., if uÝx, tÞ solves the heat equation in the variables x,t then for z, s, vÝz, sÞ
given by (5.2), vÝz, sÞ solves the heat equation in the variables z,s. Note that
v s ?c/b = ÝP c uÞÝP b tÞ
and
z
s
a/b
=
?c/b
= u t ?c/b
Pax = x .
a/b
t a/b
ÝP b tÞ
i.e., both groupings of variables are invariant under the transformation (5.2) for all choices
of a,b,c. This suggests that we look for a solution for (5.1) that is of the form
u = t c/b yÝYÞ
i.e.,
for
Y=
x = x
t a/b
t
since a/b =
1
2
.
Ý5.3Þ
/ t u = c t c/2a?1 yÝYÞ + t c/2a y v ÝYÞ / t Y
2a
= c t c/2a?1 yÝYÞ + t c/2a y v ÝYÞ ? 12
2a
= t c/2a?1
x
t3
c yÝYÞ ? Y y v ÝYÞ
2
2a
and
1
/ x u = t c/2a y v ÝYÞ / x Y = t c/2a?1/2 y v ÝYÞ
/ xx u = t c/2a?1/2 y ” ÝYÞ / x Y = t c/2a?1 y ” ÝYÞ.
Then (5.1) becomes
Y
t c/2a?1 D y ”ÝYÞ ? c yÝYÞ + y v ÝYÞ
2a
2
= 0;
Ý5.4Þ
i.e., the pde has been reduced to an ode. If (5.1) is satisfied for x > 0, t > 0 and if, in
addition, u(x,t) satisfies
uÝx, 0Þ = 0, x > 0,
then since
uÝx, tÞ ¸ 0, x ¸ K,
/ x uÝ0, tÞ = Q, t > 0,
yÝYÞ = t ?c/2a uÝx, tÞ and Y = x , it follows that
t
Y ¸ K if x ¸ K, or t ¸ 0
and
Y = 0 if x = 0
hence the first two side conditions reduce to yÝKÞ = 0. Now / x uÝ0, tÞ = t c/2a?1/2 y v Ý0Þ and this
can equal the constant Q if and only if c = a. Then the initial boundary value problem for
u(x,t) reduces to the following problem for yÝYÞ
Y
D y ”ÝYÞ ? 1 yÝYÞ + y v ÝYÞ = 0, y v Ý0Þ = Q, yÝYÞ ¸ 0 as Y ¸ K
2
2
Ý5.5Þ
On the other hand, if the boundary condition at x = 0 is uÝ0, tÞ = u 0 , then uÝ0, tÞ = t c/2a yÝ0Þ
can equal the constant u 0 if and only if c = 0. In this case the initial boundary value problem
for u(x,t) reduces to
D y ”ÝYÞ +
Y v
y ÝYÞ = 0, y Ý0Þ = u 0 ,
2
yÝYÞ ¸ 0 as Y ¸ K
Ý5.6Þ.
We can integrate this once to obtain
y v ÝYÞ = C 1 e ?Y
and
Y
2 /4D
yÝYÞ = C 1 X e ?V
0
= C 3 erf
where erfÝxÞ :=
,
2 /4D
dV + C 2
Y
4D
+ C2
2 X x e ?R 2 dR.
^ 0
2
Then the boundary conditions lead to
yÝYÞ = u 0 ? u 0 erf
Y
4D
,
and
x
4Dt
uÝx, tÞ = u 0 ? u 0 erf
x
4Dt
= u 0 erf c
.
where erfcÝsÞ = 1 ? erfÝsÞ. The solution of the equation (5.5) is given by
yÝYÞ = C 1 Y + C 2 2 ^D e ?Y
and since
2 /4D
+ ^ Y erf
Y
4D
Y
4D
y v ÝYÞ = C 1 + C 2 ^ erf
the auxiliary conditions are satisfied for
Q
yÝYÞ = QY ? ^
= Q Y erf c
2 ^D e ?Y
Y
4D
2 /4D
+ ^ Y erf
Y
4D
?Y 2 /4D
? 2Q D
.
^ e
Of course these problems could have been solved by integral transform methods as well,
and in fact the integral transform is able to deal with much more general boundary
conditions than the ones treated here. On the other hand, consider the example
/ t uÝx, tÞ ? / x ßu / x uÝx, tÞà= 0,
uÝx, 0Þ = NÝxÞ,
uÝx, tÞ ¸ 0 as | x| ¸ K
and
x 5 R, t > 0,
XR uÝx, tÞ dx = 1, for t > 0.
Ý5.7Þ
Ý5.8Þ
The transformation (5.2) reduces the pde to
P b?c / s v ? P 2a / zz
1
2
P ?2c v 2
= 0,
and the equation is invariant under the transformation if b ? c = 2Ýa ? cÞ, i.e., c = 2a ? b.
Then if we let
3
uÝx, tÞ = t Ý2a?bÞ/b yÝYÞ,
Y=
x ,
t a/b
the condition (5.8) becomes
x
t a/b
XR uÝx, tÞ dx = t 2Ýa/bÞ?1 XR y
t a/b dY = t 3Ýa/bÞ?1 X yÝYÞ dY = 1
R
This condition can hold for all t > 0 if 3Ýa/bÞ = 1; i.e., 3a = b and then
Y= x ,
3 t
uÝx, tÞ = t ?1/3 yÝYÞ,
so that
/ t u = ? 13 t ?4/3 yÝYÞ + t ?1/3 y v ÝYÞ / t Y = ? 13 t ?4/3 ßyÝYÞ + Y y v ÝYÞà
/ xx
1
2
2
= d2
dY
u2
1 ?2/3 2
t y
2
Ý/ x YÞ 2 =
1 ?2/3
t Ý2y y”
2
+ 2y v Þ t ?2/3
and
? 13 t ?4/3 ßyÝYÞ + Y y v ÝYÞà ?
1 ?4/3
t Ý2y y”
2
+ 2y v Þ = 0;
i.e.,
y y” + y v +
or
Then
1
3
ßyÝYÞ + Y y v ÝYÞà = 0
3ÝyÝYÞ y v ÝYÞÞ v + ÝY yÝYÞÞ v = 0.
3yÝYÞ y v ÝYÞ + Y yÝYÞ = C 0 ,
and since we want the solution to be an even function of Y, we impose the condition
y v Ý0Þ = 0, which leads to C 0 = 0. Then 3y v ÝYÞ + Y = 0, or y ÝYÞ = ? 16 Y 2 + C 1 . If we write
this solution in the form
yÝYÞ =
1
6
ÝA 2 ? Y 2 Þ
then in order to have yÝYÞ ¸ 0 as Y 2 ¸ K, we must have
yÝYÞ =
1
6
0
ÝA 2 ? Y 2 Þ
if Y 2 < A 2
if
Y2 > A2
Then this is a weak solution since the derivative is discontinuous at Y 2 = A 2 . The condition
(5.8) requires that A = 3 92 and then
4
uÝx, tÞ = t
?1/3
1 ?1/3
t
6
yÝYÞ =
A2 ?
x
3 t
2
0
if
if
| x| < At 1/3
| x| < At 1/3
The solution is plotted versus x for several times
Note that unlike solutions to linear parabolic equations, solutions to nonlinear equations
may have finite speed of propagation.
Burger’s Equation
We will consider the effect of the transformation (5.2) on the Burger’s equation
/ t uÝx, tÞ + uÝx, tÞ / x uÝx, tÞ = 0.
Under the transformation (5.2), this becomes
P b?c / s vÝz, sÞ + P a?2c vÝz, sÞ / z vÝz, sÞ = 0,
and the equation is invariant if c = a ? b. Then we look for a solution of the form (5.3)
u = t a/b?1 yÝYÞ,
Y=
x ,
t a/b
or, letting a/b ? 1 = m, so that m is the free parameter,
u = t m yÝYÞ,
Y=
x .
t m+1
5
x
/ t u = m t m?1 yÝYÞ + t m y v ÝYÞ ?Ým + 1Þ m+2
t
Then
= t m?1 ßm yÝYÞ ? Ým + 1Þ Y y v ÝYÞà,
/ x u = t m y v ÝYÞ Ýt ?m?1 Þ,
/ t u + u / x u = t m?1 ßm yÝYÞ ? Ým + 1Þ Y y v ÝYÞ + yÝYÞ y v ÝYÞà = 0.
and
For m = 0 this reduces to
yÝYÞ y v ÝYÞ ? Y y v ÝYÞ = 0, or
yÝYÞ = Y
uÝx, tÞ = x .
t
A more interesting example is provided by the KdV equation
/ t u + 6u / x u + / xxx u = 0
which is transformed by (5.2) into
P b?c / s v + 6P a?2c v / z v + P 3a?c / zzz v = 0.
For invariance, we must have
b ? c = a ? 2c = 3a ? c,
or
c = a ? b,
i.e.,
a= b
3
3a = b;
and
c = ? 23 b,
b =free parameter.
Looking for a solution of the form
u = Ý3tÞ ?2/3 yÝYÞ
Y=
x ,
Ý3tÞ 1/3
leads to
/ t u + 6u / x u + / xxx u = ?Ý3tÞ ?5/3 ß2yÝYÞ + Y y v ÝYÞ ? 6yÝYÞ y v ÝYÞ ? y vvv ÝYÞà = 0
i.e.,
y vvv ÝYÞ + ß6yÝYÞ + Yà y v ÝYÞ ? 2yÝYÞ = 0.
6
This ordinary differential equation can be shown to have the so called Painlevé property,
meaning that it does not have a movable singular point. A movable singular point is a
point where the solution becomes singular at a point whose location depends on the
arbitrary constants of integration. The equation y v ÝYÞ = yÝYÞ 2 , has the solution
1
C?Y
which has a singular point whose location depends on the arbitrary constant of integration,
C. Then this equation does not have the Painleve property. There is a conjecture, as yet not
proved, that pde’s that reduce under transformations of the form (5.2) to ode’s having the
Painleve property are equations that admit soliton solutions and are solvable by the inverse
scattering transform.
yÝYÞ =
6. Darcy’s Law and the Richard’s Equation
In deriving a mathematical model for flow in a porous medium, we begin with a
conservation statement expressed in terms of the state variables
SÝx, tÞ = moisture content (percentage of available pore space at position x
that is filled with fluid at time t)
QÝx, tÞ = fluid flow rate at x, t
Conservation of fluid requires that in each ball, B, in the porous medium, we have
XB / t SÝx, tÞ dx + X/B Q 6 n dS = 0.
Then, using the divergence theorem to convert the boundary integral to a volume integral,
and using the fact that B is arbitrary, we arrive at the conservation equation
/ t SÝx, tÞ + div Q = 0 at each Ýx, tÞ.
To complete the model, we need a constituitive relation between the fluid flux, Q, and the
other state variables. An empirical law known as Darcy’s law provides this relationship. Just
as Maxwell initiated the mathematical treatment of electromagnetism, a French engineer
named H. Darcy initiated in 1856 the mathematical theory of flow in porous media. He
conducted a series of experiments on columns of various soils that led to the formulation of
what is now called Darcy’s law. Imagine a vertical soil column which is totally saturated with
water. The flow of water through the column is assumed to be driven by gravity and external
pressure and we can define the hydraulic head, H(x), at a point, x, in the column to be the
sum of the gravity head (proportional to the elevation of the water above some datum level)
and the pressure head (due to atmospheric or other applied pressure fields)
7
If we denote the hydraulic head by H = HÝxÞ, and the flow rate through the saturated
column by Q, then Darcy’s law asserts that
Q = ?K
H Top ? H Bottom
Ax
or
Q = ?K 4H
where K denotes a soil dependent parameter called the hydraulic conductivity. Note the
similarity with Fourier’s law of heat conduction, Fick’s law of diffusion and Ohm’s law
relating current flow to potential gradient.
For saturated flow, the moisture content is constant and the flow equation obtained by
combining Darcy’s law with the conservation equation therefore becomes
divÝ?K 4HÞ = 0.
In order to describe partially saturated flow, we note first that in the saturated situation the
fluid flow is driven by the hydraulic head. To extend Darcy’s law to the case of partially
saturated flow, it is necessary to postulate the existence of some other driving potential. For
this purpose we can assert that fluid flow in a partially saturated medium is driven by a
capillary suction force that propels fluid from regions of low suction toward regions where
the suction is high. Empirically, we expect the highest suction at the interface between a
wet and a dry region, at a so called ”wetting front”. We define the total head as the sum of
the suction head, f, the hydraulic head, H. Having done this, we immediately decide to
neglect the gravity and pressure heads in order to concentrate on the role of the capillary
suction head. In a physical setting, it is legitimate to neglect the gravity head if the flow is in
a horizontal direction (since in this case the entire flow is at the same elevation above the
datum) and we assume further that the external pressure is the same at each point in the
porous medium. Then Darcy’s law for partially saturated flow is assumes the form
Q = ?K 4f
where K is a function of suction, f, or of water content S..
8
Typical K vs Psi or K vs Theta curves
Then, combining this version of Darcy’s law with the conservation equation leads to the so
called Richard’s equation. If we choose to view moisture content as a function of suction,
S = SÝfÞ, we obtain
/ t SÝfÝx, tÞÞ ? 4ÝKÝfÞ 4fÞ = 0,
or
CÝfÞ / t fÝx, tÞ ? 4ÝKÝfÞ 4fÞ = 0,
where
CÝfÞ = S v ÝfÞ.
Ý6.1Þ
On the other hand, if we view the suction as a function of moisture content, f = fÝSÞ, then
/ t SÝx, tÞ ? 4ÝKÝfÝSÞÞ 4fÞ = / t SÝx, tÞ ? 4ÝKÝfÝSÞÞ f v ÝSÞ 4SÞ = 0
or
/ t SÝx, tÞ ? 4ÝDÝSÞ 4SÞ = 0
where
DÝSÞ = KÝfÝSÞÞ f v ÝSÞ.
Ý6.2Þ
Empirical measurements for S and f suggest that the relationship between these two
quantities has the following form
The equations Ý6.1Þ and Ý6.2Þ are referred to respectively as the pressure formulation and
moisture content formulation of Richard’s equation. Empirical evidence suggests that the
9
diffusivity coefficient, D = DÝSÞ, in the moisture content formulation behaves approximately
like S m for some m > 1. We will use similarity transformations to show that if this is the
case, then solutions to Richard’s equation behave much differently than solutions to the
heat equation which is superficially similar to Richard’s equation.
/ t SÝx, tÞ ? / xx ÝS 2 Þ = / t SÝx, tÞ ? / x Ý2S / x SÞ = 0.
Consider
Ý6.3Þ
Guided by previous examples, we make the ansatz
z = xb .
t
SÝx, tÞ = t a yÝzÞ
Then
Ý6.4Þ
t a?1 ÝayÝzÞ ? bzy v ÝzÞÞ ? 2t 2Ýa?bÞ y v ÝzÞ 2 + yÝzÞ y”ÝzÞ
=0
and equation Ý6.3Þ reduces to an ode in z if a ? 1 = 2Ýa ? bÞ, i.e., a = 2b ? 1. In this case
2 y v ÝzÞ 2 + yÝzÞ y”ÝzÞ
= Ý2b ? 1ÞyÝzÞ ? bzy v ÝzÞ,
and, in particular, if b = 13 , we have 2b ? 1 = ?b and Ý2b ? 1ÞyÝzÞ ? bzy v ÝzÞ =
Then the ode becomes the exact equation,
2ÝyÝzÞ y v ÝzÞÞ v +
1
3
Ý z yÝzÞÞ v .
Ý z yÝzÞÞ v = 0.
If we add the conditions, y v Ý0Þ = 0, and yÝzÞ ¸ 0 as z 2 ¸ K,
then
yÝzÞ =
1
3
to the differential equation,
A 2 ? z 2 if z 2 < A 2
12
0
if z 2 > A 2
satisfies the equation and these conditions. Then
SÝx, tÞ = 1 y
3 t
x
t
3
is a weak solution for the partial differential equation (6.3) (which is Richard’s equation with
DÝSÞ = 2S). Note particularly, this solution has compact support for each t > 0 and the
”front” where x = A 3 t , travels with finite speed equal to x v ÝtÞ = 13 A t ?2/3 , t > 0.
More generally, consider the equation
/ t SÝx, tÞ ? / xx ÝS m Þ = 0,
m > 1,
Ý6.5Þ
10
SÝx, tÞ = SÝ?x, tÞ,
where
and
SÝx, tÞ ¸ 0 as | x| ¸ K
XR SÝx, tÞ dx = 1 for all t > 0.
We make the ansatz (6.4) which leads to
t a?1 ßayÝzÞ ? bz y v ÝzÞà ? t ma?2b mÝm ? 1Þ y m?2 ÝzÞ y v ÝzÞ 2 + m y m?1 y”ÝzÞ
= 0.
The equation reduces to an ODE in z if a ? 1 = ma ? 2b; i.e., a = 2b ? 1 .
m?1
Additionally,
X SÝx, tÞ dx = t a X yÝzÞ t b dz = t a+b X yÝzÞ dz = 1,
R
R
R
and this holds, independent of t, provided a + b = 0. Then
?bßyÝzÞ + z y v ÝzÞà ? ßmÝm ? 1Þ y m?2 + m y m?1 y”ÝzÞà = 0,
or
v
?bÝz yÝzÞÞ v ? Ým y m?1 ÝzÞ y v ÝzÞÞ = 0.
We integrate and use the fact that SÝx, tÞ = SÝ?x, tÞ, to obtain
m y m?1 ÝzÞ y v ÝzÞ + b z yÝzÞ = 0.
We separate and integrate again, obtaining
m
m?1
y m?1 ÝzÞ = ? 12 b z 2 + C 0 .
Now the previously discovered relations
a = 2b ? 1 = ?b
m?1
imply b =
1
m+1
yÝzÞ =
=
=
and so
? 1 b z2
A2 ? mm
2
2
A ? m?1 z
m + 1 2m
2
m ? 1 B2 ? z2
m + 1 2m
1
m?1
1
m?1
1
m?1
.
11
The condition yÝzÞ ¸ 0 as | z| ¸ K is satisfied if
m ? 1 B2 ? z2
m + 1 2m
yÝzÞ =
1
m?1
.
0
if
z2 < B2
if
z2 > B2
and then
SÝx, tÞ = t a y
x
tb
=
m?1
m+1
1
m+1
t
x2
m+1 2
t
2m
B2 ?
1
m?1
.
0
if
if
x2 < B2
x2 > B2
m+1
m+1
t2
t2
This is a weak solution for the equation (6.5) which is generally referred to as the porous
medium equation. The notable features of the solution are the lack of smoothness and the
compact support (also interpreted as finite speed of propagation). Both of these properties
are in contrast to what we find in solving the heat equation. Note that in the first example,
equation (6.3) we chose b = 1/3 in order to make the ode exactly integrable while in the
example (6.5) the choice b = ?a was made in order to have the area under the graph of
SÝx, tÞ equal one for all positive t. In this case, the ode indicentally turned out to be
integrable when b was chosen in this way. Since (6.3) is a special case of (6.5), it appears
that the condition on the area under the graph automatically forces the ode obtained by the
similarity transformation to become exactly integrable. Then it seems to be a fortuitous
accident that the conditions that must be enforced to get the similarity transformation to go
through are also conditions that make the resulting solution physically relevant.
It is mildly surprising that the similarity method we used to construct a solution for the
1-dimensional porous flow equation also succeeds in n dimensions. The n dimensional
porous flow equation is
/ t uÝx, tÞ ? div grad Ýu m Þ = 0
We assume
uÝx, tÞ = t a yÝzÞ
x 5 R n , t > 0.
z = xb ,
t
x 5 R n , t > 0. Then
/ t Ýt a yÝzÞÞ = t a?1 ßayÝzÞ ? b4 z yÝzÞ 6 zà
4 2x Ýu m Þ = 4 2z Ýt am yÝzÞ m Þ | 4 x z| 2 = t ma?2b 4 2z Ý yÝzÞ m Þ
In order to simplify the spatial term, let us now suppose that yÝzÞ = wÝrÞ where r = z 6 z ,
i.e., the solution is radial. Then we are left with just the radial part of the Laplacian,
4 2z Ý yÝzÞ m Þ = ÝwÝrÞ m Þ” + n ?r 1 ÝwÝrÞ m Þ v
12
and the equation reduces to
t a?1 ßawÝrÞ ? b r w v ÝrÞà ? t ma?2b ÝwÝrÞ m Þ” + n ?r 1 ÝwÝrÞ m Þ v
=0
The equation becomes independent of t if a ? 1 = am ? 2b; i.e., a = 2b?1
. In addition, the
m?1
1
and
area under the graph of u equals one for all t > 0 if a + nb = 0. Then b =
Ým ? 1Þn + 2
if we multiply the equation by r n?1 , we get
b ßn r n?1 wÝrÞ + r n w v ÝrÞà + ßr n?1 ÝwÝrÞ m Þ” + Ýn ? 1Þr n?2 ÝwÝrÞ m Þ v à = 0.
This can be rewritten as
v
Ýr n?1 ÝwÝrÞ m Þ v Þ + bÝr n wÝrÞÞ v = 0.
Then
r n?1 ÝwÝrÞ m Þ v + br n wÝrÞ = C 0 ,
and the conditions wÝrÞ, w v ÝrÞ ¸ 0 as r ¸ K
and
imply that C 0 = 0. Then
ÝwÝrÞ m Þ v = m wÝrÞ m?1 w v ÝrÞ = ?b r wÝrÞ
m wÝrÞ m?2 w v ÝrÞ =
so
wÝrÞ m?1 = C 1 ?
m
wÝrÞ m?1
m?1
m?1
2m
v
= ?b r
b r2
Finally, the conditions at infinity lead to the solution
wÝrÞ =
C1 ?
m?1
2m
b r2
1
m?1
0
if
if
2m
Ým ? 1Þb
2m
r2 > C1
Ým ? 1Þb
r2 < C1
and
uÝx, tÞ = t ?nb
C1 ?
0
m?1
2m
|x| 2
b b
t
1
m?1
if
if
|x| 2
2m
< C1
Ým ? 1Þb
tb
|x| 2
2m
> C1
Ým ? 1Þb
tb
where
b=
1
.
Ým ? 1Þn + 2
This is called Barenblatt’s solution for the PM equation. Like the 1-dimensional example, it
has compact support for each t > 0 and the set where u > 0 grows at a finite speed for all
t > 0.
13