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MA 201, Mathematics III, July-November 2018,
Part II: Partial Differential Equations
Laplace’s equation as steady-state heat conduction
equation
Lecture 15
Lecture 15 MA 201, PDE (2018) 1 / 37
Steady state heat conduction: Laplace’s equation
Laplace’s equation in two or three dimensions
usually arises in two types of physical problems:
1. As steady state heat conduction.
2. As equation of continuity for incompressible potential flow.
However, here we will emphasize only on the first type.
Steady state solution here means
1. the solution for large time.
2. the solution does not depend anymore on time.
Lecture 15 MA 201, PDE (2018) 2 / 37
Laplace’s equation
Laplace’s equation in two dimensions and three dimensions in Cartesian
coordinates, are , respectively, given by
uxx + uyy = 0, (1)
uxx + uyy + uzz = 0. (2)
The above equations can be obtained from the two-dimensional and three-dimensional
transient heat conduction equations when u does not depend on t.
Hence Laplace’s equation models
steady heat flow in a region where the temperature is fixed on the boundary.
Lecture 15 MA 201, PDE (2018) 3 / 37
Maximum Principle
Theorem: Let u(x, y) satisfy Laplace’s equation in D, an open, bounded, connected
region in the plane; and let u be continuous on the closed domain D ∪ ∂D consisting
of D and its boundary. If u is not a constant function, then the maximum and
minimum values of u are attained on the boundary of D and nowhere inside D.
. �
This is called maximum principle theorem for Laplace’s equation.
Lecture 15 MA 201, PDE (2018) 4 / 37
Steady state heat conduction in two dimensions
We consider steady state heat conduction
in a two-dimensional rectangular region.
To be specific,
consider the equilibrium temperature inside a rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b.
Here
the temperature is a prescribed function of position on the boundary.
In general the Dirichlet BVP will be like
uxx + uyy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b
u(0, y) = g1(y), u(a, y) = g2(y), 0 ≤ y ≤ b
u(x, 0) = f1(x), u(x, b) = f2(x), 0 ≤ x ≤ a.
Lecture 15 MA 201, PDE (2018) 5 / 37
Steady state heat conduction in two dimensions
where
f1(x), f2(x), g1(y), g2(y) are given functions.
Though the equation is linear and homogenous,
the BCs are not homogenous.
Hence
the BVP is needed to be split into four BVPs with each containing one
non-homogenous BC.
Take
u = u1 + u2 + u3 + u4, 0 ≤ x ≤ a, 0 ≤ y ≤ b.
Lecture 15 MA 201, PDE (2018) 6 / 37
Steady state heat conduction in two dimensions
BVP I and BVP II:
u1,xx + u1,yy = 0; u2,xx + u2,yy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b;
u1(0, y) = 0, 0 ≤ y ≤ b; u2(0, y) = 0, 0 ≤ y ≤ b;
u1(a, y) = 0, 0 ≤ y ≤ b; u2(a, y) = 0, 0 ≤ y ≤ b;
u1(x, 0) = f1(x), 0 ≤ x ≤ a; u2(x, 0) = 0, 0 ≤ x ≤ a;
u1(x, b) = 0, 0 ≤ x ≤ a; u2(x, b) = f2(x), 0 ≤ x ≤ a;
BVP III and BVP IV:
u3,xx + u3,yy = 0; u4,xx + u4,yy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b;
u3(0, y) = g1(y), 0 ≤ y ≤ b; u4(0, y) = 0, 0 ≤ y ≤ b;
u3(a, y) = 0, 0 ≤ y ≤ b; u4(a, y) = g2(y), 0 ≤ y ≤ b;
u3(x, 0) = 0, 0 ≤ x ≤ a; u4(x, 0) = 0, 0 ≤ x ≤ a;
u3(x, b) = 0, 0 ≤ x ≤ a; u4(x, b) = 0, 0 ≤ x ≤ a.
We will consider only one of them.......take u1 = u for convenience
Lecture 15 MA 201, PDE (2018) 7 / 37
Steady state heat conduction in two dimensions
Consider the steady state heat conduction in a rectangular region
0 ≤ x ≤ a, 0 ≤ y ≤ b
where three boundaries along x = 0, x = a, y = b are kept at 00C
while
the temperature along the boundary y = 0 is f(x).
To find the temperature at any point (x, y).
Lecture 15 MA 201, PDE (2018) 8 / 37
Steady state heat conduction in two dimensions
BVP will consist of the following:
The governing equation is two-dimensional Laplace’s equation:
uxx + uyy = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b. (3)
The boundary conditions are:
u(0, y) = 0, 0 ≤ y ≤ b, (4a)
u(a, y) = 0, 0 ≤ y ≤ b, (4b)
u(x, 0) = f(x), 0 ≤ x ≤ a, (4c)
u(x, b) = 0, 0 ≤ x ≤ a. (4d)
It being a pure BVP and the solution being a function of x and y,
obviously we will not have any initial conditions.
Hence
This problem is called a steady-state problem.
Lecture 15 MA 201, PDE (2018) 9 / 37
Steady state heat conduction in two dimensions
Assume a solution of the form:
u(x, y) = X(x)Y (y). (5)
Using (5) in (3)
X ′′
X+
Y ′′
Y= 0
On separating the variables x and y,
X ′′
X= −
Y ′′
Y= k(say).
Giving us
X ′′ − kX = 0, (6)
Y ′′ + kY = 0. (7)
The zero and positive values of k will not give rise to solutions conforming to the
boundary conditions.
Lecture 15 MA 201, PDE (2018) 10 / 37
Steady state heat conduction in two dimensions
We consider only the negative values of k, say − λ2, to write the equations (6)
and (7) as
X ′′ + λ2X = 0, (8)
Y ′′ − λ2Y = 0, (9)
so that the solution u(x, y) can be written as
u(x, y) = (A cos λx+B sinλx)(C coshλy +D sinhλy). (10)
Using boundary condition (4a)
A = 0.
Using boundary condition (4b),
λn =nπ
a, n = 1, 2, 3, . . .
⇒
un(x, y) = sinnπx
a
(
An coshnπy
a+Bn sinh
nπy
a
)
.
Lecture 15 MA 201, PDE (2018) 11 / 37
Steady state heat conduction in two dimensions
Using boundary condition (4d)
Bn = −cosh nπb
a
sinh nπba
An
so that the solution u(x, y) can be written as
u(x, y) =
∞∑
n=1
un(x, y) =
∞∑
n=1
An sinnπx
a
(
coshnπy
a−
cosh nπba
sinh nπba
sinhnπy
a
)
=∞∑
n=1
An sinnπx
a
sinh nπ(b−y)a
sinh nπba
(11)
Remaining boundary condition (4c) can be used to evaluate the coefficients An:
f(x) =
∞∑
n=1
An sinnπx
a.
An is obtained as
An =2
a
∫ a
0f(x) sin
nπx
adx. (12)
The solution to the BVP described by equations (3)-(4) is given by
(11) with An given by (12).
Lecture 15 MA 201, PDE (2018) 12 / 37
Steady state heat conduction in two dimensions
Similarly we can find the other solutions u2, u3 and u4 and
write the total solution as u = u1 + u2 + u3 + u4.
This problem with Dirichlet conditions
along all boundaries is called a Dirichlet problem for a rectangle.
The problem with Neumann conditions
along all boundaries is called a Neumann problem for a rectangle.
This new problem can be solved by writing the boundary conditions as
ux(0, y) = 0, (13a)
ux(a, y) = 0, (13b)
uy(x, 0) = f(x), (13c)
uy(x, b) = 0. (13d)
TRY to solve it yourself.
Lecture 15 MA 201, PDE (2018) 13 / 37
Solution in different types of domains
Till now the problems that we have taken up are for bounded domains in
Cartesian coordinates.
There are other types of domain which occur frequently in many physical problems.
We need to rewrite our governing equations and the related conditions.
Some domains of importance are
• circular domain
• spherical domain
• cylindrical domain.
Lecture 15 MA 201, PDE (2018) 14 / 37
Solution in different types of domains
For two dimensional problem with (x, y)
∇2u(x, y) =∂2u
∂x2+
∂2u
∂y2.
For three dimensional problem with (x, y, z)
∇2u(x, y, z) =∂2u
∂x2+
∂2u
∂y2+
∂2u
∂z2.
For two dimensional problem with (r, θ), that is, in polar coordinates
∇2u(r, θ) =∂2u
∂r2+
1
r
∂u
∂r+
1
r2∂2u
∂θ2.
Lecture 15 MA 201, PDE (2018) 15 / 37
Solution in different types of domains
For three dimensional problem with (r, θ, z), that is, in cylindrical coordinates
∇2u(r, θ, z) =∂2u
∂r2+
1
r
∂u
∂r+
1
r2∂2u
∂θ2+
∂2u
∂z2.
For three dimensional problem with (r, θ, φ), that is, in spherical coordinates
∇2u(r, θ, φ) =∂2u
∂r2+
2
r
∂u
∂r+
1
r2∂2u
∂θ2+
cot θ
r2∂u
∂θ+
1
r2 sin2 θ
∂2u
∂φ2.
Lecture 15 MA 201, PDE (2018) 16 / 37
Laplace’s equation in polar coordinates
For a problem involving circular disk, polar coordinates are more appropriate than
rectangular coordinates.
Let us formulate the steady-state heat flow problem in polar coordinates r, θ,
where x = r cos θ, y = r sin θ.
A circular plate of radius a can be simply represented by
r ≤ a with 0 ≤ θ ≤ 2π.
The unknown temperature inside the plate is now u = u(r, θ),
The given temperature on the boundary of the plate is
u(a, θ) = f(θ), where f is a known function.
Lecture 15 MA 201, PDE (2018) 17 / 37
Laplace’s equation in polar coordinates
Now we have the following boundary value problem:
urr +1
rur +
1
r2uθθ = 0, 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π, (14)
u(a, θ) = f(θ), 0 ≤ θ ≤ 2π. (15)
There is a periodic boundary condition which is implicit in nature:
u(r, θ) = u(r, θ + 2π). (16)
Using the separation of variables method, assume a solution:
u(r, θ) = R(r)T (θ).
Using this in equation (14),
r2R′′
R+ r
R′
R+
T ′′
T= 0 (17)
Lecture 15 MA 201, PDE (2018) 18 / 37
Laplace’s equation in polar coordinates
Separating the variables
r2R′′
R+ r
R′
R= −
T ′′
T= k (18)
which give rise to the following ODEs:
r2R′′ + rR′ − kR = 0, (19)
T ′′ + kT = 0. (20)
We cannot consider negative values of k since if k is negative,
then the ODE in T (θ) has exponential solutions, and exponential solutions cannot
satisfy periodicity conditions.
Since we are looking for a periodic solution in θ,
we must take k = λ2.
But we should keep in mind that λ = 0, corresponding to k = 0, is also an
eigenvalue with corresponding eigenfunction u0(r, θ) = constant.
Lecture 15 MA 201, PDE (2018) 19 / 37
Laplace’s equation in polar coordinates
Hence the equations reduce to
r2R′′ + rR′ − λ2R = 0, (21)
T ′′ + λ2T = 0. (22)
(22) has the general solution
T (θ) = A cos λθ +B sinλθ. (23)
The Dirichlet periodic boundary condition (16) will give us
cos 2πλ = 1,
i.e., λn = n.
Lecture 15 MA 201, PDE (2018) 20 / 37
Laplace’s equation in polar coordinates
Tn(θ):
Tn(θ) = An cosnθ +Bn sinnθ (24)
Equation (21) takes the form
r2R′′ + rR′ − n2R = 0 (25)
which is of Cauchy-Euler form and the solution is
Rn = Cnr−n +Dnr
n (26)
We set Cn = 0 since we are seeking a bounded solution in 0 ≤ r ≤ a.
and r−n is not bounded when r → 0.
Lecture 15 MA 201, PDE (2018) 21 / 37
Laplace’s equation in polar coordinates
Solution to the BVP:
u(r, θ) =A0
2+
∞∑
n=1
rn(An cosnθ +Bn sinnθ). (27)
Using the given boundary condition (15),
f(θ) =A0
2+
∞∑
n=1
an(An cosnθ +Bn sinnθ), (28)
The coefficients are given by
An =1
πan
∫ 2π
0f(θ) cosnθ dθ, n = 0, 1, 2, 3, . . . (29a)
Bn =1
πan
∫ 2π
0f(θ) sinnθ dθ, n = 1, 2, 3, . . . (29b)
(27) is the solution of Laplace’s equation with the coefficients given by (29a) and
(29b).
Lecture 15 MA 201, PDE (2018) 22 / 37
Laplace’s equation in polar coordinates
What we have just solved is called
Interior Dirichlet problem for a circle since we have used Dirichlet condition fin the
region r ≤ a.
If we change only the condition to Neumann
we have what is called Interior Neumann problem for a circle.
Now if we change the region to r > a with Dirichlet condition
we have what is called Exterior Dirichlet problem for a circle.
If in above condition is replaced by Neumann
we have what is called Exterior Neumann problem for a circle.
Lecture 15 MA 201, PDE (2018) 23 / 37
Diffusion in a disk
Consider a circular, planar disk of radius a for which
• initial temperature is a function of the radial distance r alone
• boundary is held at zero degrees.
Intuition tells that
the temperature u in the disk depends only on time and the distance r from the centre.
To be precise, the initial temperature u will be the same for some r = rn irrespective
of what value of θ is assigned.
That is, if the initial temperature u is, say, u1 for some r = r1, then it is imperative
that the temperature is same for that specific r.
Consider now that the initial temperature u is, say, u2 for some r = r2, then it is
imperative that the temperature is same for that specific r.
which, in turn means that
the heat flow will take place either from r = r1 to r = r2 or from r = r2 to r = r1
depending on which has higher temperature.
Lecture 15 MA 201, PDE (2018) 24 / 37
Diffusion in a disk
It then allows us to consider u simply as
u = u(r, t) though the given equation initially seemed to contain r, θ, t as independent
variables.
This assumption looks perfectly alright
because there is nothing in the initial condition or boundary condition to cause heat to
diffuse in an angular direction
Heat will flow only along rays emanating from the origin.
Now it is obvious that
the diffusion equation looks simpler than what it was originally.
Lecture 15 MA 201, PDE (2018) 25 / 37
Diffusion in a disk
Diffusion equation is now as simple as follows:
ut = α(urr +1
rur), 0 ≤ r ≤ a, t > 0. (30)
Boundary condition
u(a, t) = 0, t > 0, (31)
Initial condition
u(r, 0) = f(r), 0 ≤ r < a, (32)
f is a given initial radial temperature distribution.
Lecture 15 MA 201, PDE (2018) 26 / 37
Diffusion in a disk
Assume a solution in the form:
u(r, t) = R(r)T (t).
From given equation
R′′
R+
1
r
R′
R=
T ′
αT= k.
k = −λ2 gives rise to the following pair of ODEs:
r2R′′ + rR′ + λ2r2R = 0, (33)
T ′ + αλ2T = 0. (34)
Equation (34) can be easily solved to write as
T (t) = Ce−αλ2t. (35)
Can you recognize equation (33)??
Lecture 15 MA 201, PDE (2018) 27 / 37
Diffusion in a disk
Equation (33) is Bessel’s equation of order 0.
Its solution can be written as
R(r) = AJ0(λr) +BY0(λr), (36)
where
J0 and Y0 are, respectively, Bessel’s function of first kind and second kind of order zero.
We are looking for a bounded solution as r → 0,
we must take B = 0 as Y0(λr) → −∞ as r → 0.
Lecture 15 MA 201, PDE (2018) 28 / 37
Diffusion in a disk
Solution u(r, t) can be written as
u(r, t) = AJ0(λr)e−αλ2t. (37)
Applying boundary condition (31),
0 = AJ0(λa) implying
J0(λa) = 0.
Hence λna = νn,
where νn are the zeros of J0.
Hence the eigenvalues are given by
λn =νn
a. (38)
Lecture 15 MA 201, PDE (2018) 29 / 37
Diffusion in a disk
⇒
un(r, t) = AnJ0
(νn
ar)
e−α
ν2n
a2t.
The solution u(r, t) is
u(r, t) =
∞∑
n=1
un(r, t)
=∞∑
n=1
AnJ0
(νn
ar)
e−α
ν2n
a2t. (39)
Now using the initial condition (32):
f(r) =∞∑
n=1
AnJ0
(νn
ar)
(40)
Note the difference between the orthogonal properties of sine/cosine functions
and Bessel functions.
Lecture 15 MA 201, PDE (2018) 30 / 37
Diffusion in a disk
Bessel functions {Jµ(λr)} form an orthogonal set with respect to the weight function
r.
For finding the coefficient An, we multiply (40) by rJ0
(
νnar)
and then integrate with respect to r from 0 to a to get
An =
∫ a
0 rf(r)J0(
νnar)
dr∫ a
0 r(
J0(
νnar))2
dr. (41)
The following orthogonality property is used:∫ a
0rJ0
(νn
ar)
J0
(νm
ar)
dr = 0, n 6= m.
Lecture 15 MA 201, PDE (2018) 31 / 37
Steady-state heat conduction in a circular cylinder
Consider a right circular cylinder of radius a and height l having
(a) its convex surface and base in the xy-plane at temperature 00C,
(b) the top end z = l is kept at temperature f(r)0C.
To find the steady-state temperature at any point of the cylinder.
The governing equation for this problem will be Laplace’s equation in r, θ, z.
∇2u(r, θ, z) =∂2u
∂r2+
1
r
∂u
∂r+
1
r2∂2u
∂θ2+
∂2u
∂z2.
For simplicity, we will consider Radially Symmetric Solution for the Laplace’s
equation.
Radially symmetric solution means that u(r, θ, z) = u(r, z) that is the solution doesn’t
depend on the polar angle θ.
In other sense, solutions are symmetric under rotation.
Lecture 15 MA 201, PDE (2018) 32 / 37
Steady-state heat conduction in a circular cylinder
But assuming that the cylinder is symmetrical about its axis, Laplace’s equation
takes the form:
urr +1
rur + uzz = 0, 0 < r ≤ a, 0 ≤ z ≤ l. (42)
The boundary conditions are:
(on the side) u(a, z) = 0, 0 ≤ z ≤ l (43a)
(on the bottom) u(r, 0) = 0, 0 < r ≤ a (43b)
(on the top) u(r, l) = f(r), 0 < r ≤ a. (43c)
Assume a solution in the form
u(r, z) = R(r)Z(z)
Applying it to the governing equation (42):
R′′
R+
1
r
R′
R+
Z ′′
Z= 0.
Lecture 15 MA 201, PDE (2018) 33 / 37
Steady-state heat conduction in a circular cylinder
By separating the variables:
R′′
R+
1
r
R′
R= −
Z ′′
Z= k.
Observing that only the negative value of the separation constant will give rise to
nontrivial solutions,
we get the following ODEs by considering k = −λ2:
Z ′′ − λ2Z = 0, (44)
R′′ +1
rR′ + λ2R = 0, (45)
The solutions of the above equations are, respectively, given by
Z(z) = A cosh λz +B sinhλz, (46)
R(r) = CJ0(λr) +DY0(λr), (47)
Lecture 15 MA 201, PDE (2018) 34 / 37
Steady-state heat conduction in a circular cylinder
The solution u(r, z):
u(r, z) = (A cosh λz +B sinhλz)(CJ0(λr) +DY0(λr)) (48)
We are looking for a bounded solution in 0 ≤ r ≤ a,
we must take D = 0 since Y0 → −∞ as r → 0.
Equation (48) can be written as
u(r, z) = J0(λr)(A cosh λz +B sinhλz). (49)
Now applying the boundary condition (43a), we get 0 = AJ0(λa)
implying
J0(λa) = 0.
Lecture 15 MA 201, PDE (2018) 35 / 37
Steady-state heat conduction in a circular cylinder
Hence
λna = νn,
where νn are the zeros of J0.
The eigenvalues are given by
λn =νn
a. (50)
un(r, z) = AnJ0
(νn
ar)
coshνn
az +BnJ0
(νn
ar)
sinhνn
az.
By superimposing all the solutions,
u(r, z) =
∞∑
n=1
(
AnJ0
(νn
ar)
A coshνn
az +BnJ0
(νn
ar)
sinhνn
az)
. (51)
Lecture 15 MA 201, PDE (2018) 36 / 37
Steady-state heat conduction in a circular cylinder
Using the boundary condition (43b),
we get A = 0 thereby reducing the solution to
u(r, z) =∞∑
n=1
BnJ0
(νn
ar)
sinhνn
az. (52)
The coefficient Bn can be obtained by using the boundary condition (43c):
f(r) =∞∑
n=1
BnJ0
(νn
ar)
sinhνn
al (53)
giving us
Bn =
∫ a
0 rf(r)J0(
znar)
dr
sinh νnal∫ a
0 r(
J0(
znar))2
dr. (54)
Lecture 15 MA 201, PDE (2018) 37 / 37