Chapter 7
Heat Equation
Partial differential equation for temperature u(x, t ) in a heat conducting insulated rod
along the x-axis is given by the Heat equation:
ut = k ux x ,
x ∈ R, t > 0
(7.1)
Here k is a constant and represents the conductivity coefficient of the material used to
make the rod. Since we assumed k to be constant, it also means that material properties
are constant and do not depend on the location in the rod. It means we are considering a
homogeneous material.
Note that rod is a three dimensional object and we considered x ∈ R in the above
model. It means that we are assuming that material properties of the rod vary only in the
x-direction and are constant w.r.t. y, z directions; and thereby also expecting temperature
distribution to respect these symmetries.
In a more realistic model we have to consider that conductivity coefficient will vary
from point to point in the rod and thus k has to be a function of the variables (x, y, z) and
thus we have to look at the following model:
u t = k(x, y, z) u x x + uyy + u z z , x ∈ R3 , t > 0
(7.2)
In Section 7.1 we introduce fundamental solution associated to heat equation, and derive formulae for solutions of Cauchy problems for homogeneous and nonhomogeneous
heat equation posed for x ∈ R. We deal the case of heat equation posed on bounded xintervals and corresponding initial-boundary value problems (IBVP) in Section 7.2. We
obtain a formal solution to IBVP using method of separation of variables, and uniqueness
is proved using energy method. Maximum principle for solutions to heat equation will
be discussed in Section 7.3, and as an application we prove that formal solution to IBVP
derived in Section 7.2 is indeed a solution (under some conditions on the data), and also
prove uniqueness of solutions to IBVP.
7.1 Fundamental solution and its applications
Let a > 0. Sicne heat equation is invariant under the change of coordinates z = ax and
x2
z2
τ = a 2 t (see Exercise 7.1), and τ = t , we look for solutions to Laplace equation having such symmetric
Let us look for a solution of heat equation having the
Š
€ x properties.
form u(x, t ) = v p t . Substituting this ansatz into heat equation yields the following
187
188
Chapter 7. Heat Equation
differential equation satisfied by v
d 2v z d v
+
= 0.
d z2 2 d z
(7.3)
Such transformations which convert a PDE into an ODE are called similarity transformations. The equation (7.3) can be solved explicitly , and its general solution is given
by
2
∫z
s
d s + C2 ,
v(z) = C1
exp −
(7.4)
4
0
where C1 , C2 are arbitrary constants. Since the constant term C2 plays no role in determining a fundamental solution to heat equation, we take C2 = 0. Thus u is given by
2
∫ px
t
s
u(x, t ) = C1
exp −
d s.
(7.5)
4
0
Note that
u(x, 0) =
p
− π
p
π
if x < 0,
if x > 0,
(7.6)
where u(x, 0) is interpreted as lim u(x, t ).
t →0+
Differentiating u w.r.t. x using the formula (7.5) gives
2
C
x
u x (x, t ) = p1 exp −
.
(7.7)
4t
t
∫
1
Choose C1 so that R u x (x, t ) d x = 1, and this gives C1 = 2pπ . We define fundamental
solution, using (7.7), by
2
1
x
K(x, t ) = p exp −
(7.8)
.
4t
2 πt
Remark 7.1 (Reasons behind the choice of fundamental solution).
(i) Fundamental
solution is expected to pick-up effects of impulses which are usually concentrated at
a point. Note that for a function u(x, 0) as in (7.6), the corresponding ‘derivative’
function represents an impulse at x = 0.
(ii) Of course, we need to verify that fundamental solution defined by (7.8) has the
properties that are expected of a fundamental solution, which will be done in the
next few results.
7.1.1 Cauchy problem for homogeneous equation
Let K1 : R × R × (0, ∞) be defined by
|x − y|2
1
K1 (x, y, t ) = p exp −
4t
2 πt
(7.9)
In the following result, we collect some properties of K1 .
Lemma 7.2. The function K1 has the following properties:
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IIT Bombay
7.1. Fundamental solution and its applications
189
(i) K1 ∈ C ∞ (R × R × (0, ∞)).
(ii) K1 (x, y, t ) > 0 for t > 0.
€∂
Š
∂2
(iii) ∂ t − ∂ x 2 K1 (x, y, t ) = 0 for t > 0.
∫
(iv) R K1 (x, y, t ) d y = 1 for x ∈ R, t > 0.
(v) For any δ > 0,
∫
lim
t →0+
|y−x|>δ
K1 (x, y, t ) d y = 0
(7.10)
uniformly for x ∈ R.
Proof. Step 1: proof of (i), (ii), (iii): Properties (i), (ii), and (iii) follow from the definition
of K1 as given by (7.9).
Step 2: proof of (iv): Setting z =
y−x
p ,
2 t
we have
∫
1
K1 (x, y, t ) d y = p
π
R
∫
R
exp −z 2 d z = 1.
Step 3: proof of (v): Introducing a change of variable as in Step 2, we get
∫
∫
1
K1 (x, y, t ) d y = p
exp −z 2 d z.
π |z|> pδ
|y−x|>δ
(7.11)
(7.12)
2 t
Since
∫
R
exp −z
2
p
d z = π, we have
∫
lim
t →0
δ
|z|> 2p t
exp −z 2 d z = 0.
(7.13)
This completes the proof of (v).
Solution to Cauchy problem for homogeneous heat equation
u t − u x x = 0,
u(x, 0) = φ(x),
x ∈ R, t > 0,
(7.14a)
x ∈R
(7.14b)
for
can be expressed using fundamental solution, and is the content of the following result.
Theorem 7.3. Let φ : R → R be a continuous and bounded function. Let u be defined by
∫
∫
|x − y|2
1
u(x, t ) = K1 (x, y, t )φ(y) d y = p
φ(y) d y.
(7.15)
exp −
4t
4πt R
R
Then
(i) u ∈ C ∞ (R × (0, ∞)).
(ii) u is a solution of heat equation u t = k u x x for t > 0.
(iii) If we extend the function to R × [0, ∞) by setting u(x, 0) = φ(x), then u ∈ C (R ×
[0, ∞)).
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Chapter 7. Heat Equation
Proof. Step 1: proof of (i), (ii): Assertions (i) and (ii) follow from Lemma 7.2.
∫
Step 2: proof of (iii): For δ > 0 (to be chosen later), in view of R K1 (x, y, t ) d y = 1,
we have
∫
|u(x, t ) − φ(ξ )| = K1 (x, y, t ) (φ(y) − φ(ξ )) d y ∫R
∫
≤
K1 (x, y, t ) |φ(y) − φ(ξ )| d y +
K1 (x, y, t ) |φ(y) − φ(ξ )| d y
|y−x|<δ
|y−x|>δ
In the last inequality, there are two integrals: the first integral can be made small using
continuity of φ, and the second integral goes to zero as t → 0 by Lemma 7.2. We will
ε
choose δ > 0 so that |φ(y) − φ(ξ )| < 2 whenever |y − ξ | < 2δ. Thus we have
∫
∫
K1 (x, y, t ) |φ(y) − φ(ξ )| d y ≤
K1 (x, y, t ) |φ(y) − φ(ξ )| d y
|y−x|<δ
|y−ξ |<2δ
ε
<
2
∫
|y−ξ |<2δ
K1 (x, y, t ) d y
∫
ε
K (x, y, t ) d y
<
2 R 1
ε
< .
2
(7.16)
By (7.10), there exists t0 so that for 0 < t < t0 , we have
∫
ε
K1 (x, y, t ) |φ(y) − φ(ξ )| d y < .
2
|y−x|>δ
(7.17)
Combining the inequalities (7.16) and (7.17), we get for (x, t ) ∈ R × (0, ∞) such that
|x − y| < δ and t < t0 , the following inequality
|u(x, t ) − φ(ξ )| <
ε ε
+ = ε,
2 2
thus proving that u is continuous at points of x-axis, and that u(x, 0) = φ(x) for each
x ∈ R.
Remark 7.4. The Cauchy data φ need not be a bounded function for the formula (7.15)
to represent a solution to Cauchy problem. If there exists M , a such that φ satisfies for
every x ∈ R
2
|φ(x)| ≤ M e a x ,
then the formula (7.15) represents a solution to Cauchy problem for 0 < t <
1
u ∈ C ∞ (R × (0, 4a ).
1
,
4a
and
Remark 7.5 (infinite speed of propagation).
(i) The solution having the form (7.15)
has the following domain of dependence property. Solution at any point (x, t ) ∈
R × (0, ∞) depends on the values of initial data φ(x) at all x ∈ R, and thus domain
of dependence of solution at any point (x, t ) for t > 0 is the entire real line. The
domain of influence of any point on x-axis is R × (0, ∞). This shows that information from Cauchy data reaches all points instantly. This suggests that heat equation
may not be suitable to study physical phenomenon.
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IIT Bombay
7.1. Fundamental solution and its applications
191
(ii) In addition to the hypotheses of Theorem 7.3 assume that the Cauchy data φ ̸= 0
is such that φ(x) ≥ 0 for all x ∈ R. It immediately follows from the formula (7.15)
that u(x, t ) > 0 for all x ∈ R and t > 0. This is another instance illustrating infinite
speed of propagation property of heat equation.
Note that Theorem 7.3 asserts only the existence of a solution. In general uniqueness
is not expected for Cauchy problem for heat equation, posed for x ∈ R. However we can
prove that Cauchy problem admits only one solution when we are looking for solutions
belonging to special classes of functions, like solutions having a controlled growth of the
2
type |u(x, t )| ≤ M e ax [20].
The following example of Tychonoff illustrates non-uniqueness of solutions.
Example 7.6 (Tychonoff’s example). Let g : R → R be given by
€ 1Š
¨
exp − t 2
if t > 0,
g (t ) =
0
if t ≤ 0.
Then the series
u(x, t ) :=
∞
∑
g (k) (t ) 2k
x
(2k)!
k=0
converges and is a solution to homogeneous heat equation, and u(x, 0) = 0. For further
details, see [20, 23]. However v(x, t ) ≡ 0 is also a solution to the same Cauchy problem.
Thus a Cauchy problem may have more than one solution.
7.1.2 Duhamel’s principle for nonhomogeneous equation
As explained in Subsection 4.1.4, Duhamel principle gives us a way to solve nonhomogeneous problems corresponding to a linear differential operator, by superposition of solutions of a family of corresponding homogeneous problems.
The Cauchy problem for nonhomogeneous heat equation is given by
u t − u x x = f (x, t ),
u(x, 0) = φ(x),
x ∈ R, t > 0,
(7.18a)
x ∈ R.
(7.18b)
for
Let S(t )ϕ denote the solution of the Cauchy problem
u t − u x x = 0, x ∈ R, t > 0,
u(x, 0) = φ(x) for x ∈ R.
Taking cue from (4.76), we expect the following formula to yield a solution of the
Cauchy problem (7.18)
∫t
(7.19)
u(x, t ) = (S(t )φ) (x) +
(S(t − τ) fτ ) (x) d τ,
0
where fτ (x) := f (x, τ).
We will now check that the second term on the right hand side of (7.19) solves the
nonhomogeneous heat equation, as the first term corresponds to solution of the homogeneous wave equation satisfying the given Cauchy data.
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Chapter 7. Heat Equation
Indeed, differentiating the equation (7.19) w.r.t. t gives
∫ t
∫t
∂
∂
(S(t − τ) fτ ) (x) d τ = S(0) f t +
(S(t − τ) fτ ) (x) d τ
∂t
0
0 ∂ t
∫t
∂
= f (x, t ) +
(S(t − τ) fτ ) (x) d τ.
0 ∂ t
Since S(t − τ) fτ are solutions to homogeneous heat equation, we have
∂
∂2
(S(t − τ) fτ ) =
(S(t − τ) fτ ) .
∂t
∂ x2
Thus we get
∂
∂2
−
∂ t ∂ x2
∫
t
(S(t − τ) fτ ) (x) d τ = f (x, t ).
0
Using the definition of the operator S(t ) in the formula (7.19) for the solution of Cauchy
problem for nonhomogeneous heat equation, we get the following expression for solution.
∫
|x − y|2
1
exp −
φ(y) d y
u(x, t ) = p
4t
4πt R
∫
∫t
|x − y|2
1
+
exp −
f (y, τ) d y d τ.
(7.20)
p
4(t − τ)
4π(t − τ) R
0
Remark 7.7. Note that w(x, t ; τ) := S(t − τ) fτ solves the following problem
w t − w x x = 0,
x ∈ R, t > τ
(7.21a)
and satisfies the conditions
w(x, τ; τ) = f (x, τ) for
Thus the integral
∫
0
x ∈ R.
(7.21b)
t
(S(t − τ) fτ ) (x) d τ
in the Duhamel’s formula has the form
∫t
w(x, t ; τ) d τ
0
7.2 Initial boundary value problem for Heat equation
Consider the Initial Boundary Value Problem (IBVP) for heat equation given by
u t = k u x x , 0 < x < π, 0 < t < T
u(0, t ) = g1 (t ), 0 ≤ t ≤ T
u(π, t ) = g3 (t ), 0 ≤ t ≤ T
u(x, 0) = g2 (x),
Sivaji
0≤x ≤π
(7.22a)
(7.22b)
(7.22c)
(7.22d)
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193
where k > 0.
Definition 7.8 (solution to IBVP). Let R be the the rectangle (0, π)×(0, T ). Let CH denote
the collection of all functions φ : R → R such that the functions φ, φ x , φ x x , φ t belong to the
space C (R). A function v ∈ CH is said to be a solution of the IBVP (7.22) if v satisfies all the
equations (7.22a) - (7.22d)
Since the equation (7.22a) is linear and homogeneous, principle of superposition holds
for its solutions. Thus a solution to the IBVP (7.22) may be obtained by a superposition
of three solutions of IBVP, where each one of these solves IBVP with exactly one of the
functions g1 , g2 , g3 being non-zero.
We illustrate how to obtain a solution of the IBVP (7.22) where g1 = g3 = 0, and other
two cases are left as an exercise. Thus, we consider the Initial Boundary Value Problem
for heat equation given by
u t = k u x x , 0 < x < π, 0 < t < T ,
u(0, t ) = 0,
0 ≤ t ≤ T,
u(π, t ) = 0,
u(x, 0) = f (x),
0 ≤ t ≤ T,
0 ≤ x ≤ π,
(7.23a)
(7.23b)
(7.23c)
(7.23d)
where k > 0.
We show the existence of solutions to the IBVP (7.23) using the method of separation
of variables. Substituting
u(x, t ) = X (x)T (t )
in the equation (7.23a), and re-arranging the terms, we get
X ′′ (x)
T ′ (t )
=
X (x)
kT (t )
(7.24)
Since the LHS and RHS of the equation (7.24) are functions of the variables x and t respectively, each of them must be a constant function. Thus we have, for some constant
λ,
X ′′ (x)
T ′ (t )
=
=λ
X (x)
kT (t )
(7.25)
From the equation (7.25), we get the following two ODEs
X ′′ (x) − λX (x) = 0,
T ′ (t ) − λT (t ) = 0.
(7.26a)
(7.26b)
Since we are interested in finding a non-trivial solution, the boundary conditions (7.23b)(7.23c) give rise to the following conditions on the function X :
X (0) = 0,
X (π) = 0,
(7.27)
The ODE (7.26a) has non-trivial solutions satisfying the boundary conditions (7.27)
if and only if λ = −n 2 for some n ∈ N. Thus λn = −n 2 are eigenvalues and the corresponding eigenfunctions are given by Xn (x) = sin nx.
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Chapter 7. Heat Equation
For each n ∈ N, the solution of ODE (7.26b) with λ = λn = −n 2 (upto a constant
2
multiple) is given by Tn (t ) = e −n k t .
Since for each n ∈ N, the function un (x, y) = Xn (x)Tn (t ) is a solution (7.23a), we
propose a formal solution of (7.23a) by superposition of the sequence of solutions (un ) as
follows:
u(x, t ) =
∞
∑
bn e −n
2
kt
sin nx.
(7.28)
n=1
The coefficients bn in the formula (7.28) will be determined using the condition (7.23d).
Thus we get
f (x) =
∞
∑
bn sin nx.
(7.29)
n=1
Thus bn are the Fourier sine coefficients of f , which are given by
bn =
2
π
∫
π
f (s) sin ns d s .
(7.30)
0
Thus the formal solution of (7.23) is given by
u(x, t ) ≈
∫
∞ ∑
2 π
2
f (s ) sin n s d s e −n k t sin n x.
π
0
n=1
(7.31)
7.2.1 Uniqueness for IBVP via energy method
In this subsection, we establish uniqueness of solutions to most general IBVP for nonhomogeneous heat equation by energy considerations.
Multiplying the equation (7.23a) with u and integrating over the interval [0, π] gives
∫
π
0
∫
u ut d x = k
π
u u x x d x.
0
(7.32)
Note that the LHS of the equation (7.32) can be written as
d
dt
∫
π
0
1 2
u dx =
2
∫
π
0
u u t d x,
and integrating by parts on RHS of the equation (7.32) yields
∫
π
0
∫
u ux x d x = −
Thus the energy integral E(t ) =
1 ∫π
2
0
π
0
π
u x u x d x + u u x 0 .
π
u 2 (x, t ) d x is non-increasing if k u u x 0 ≤ 0. For
example if u satisfies u(0, t ) = 0 and u x (π, t ) = 0, we get
we get E(t ) ≤ E(0) for t > 0.
Sivaji
d
dt
E(t ) ≤ 0. As a consequence,
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7.3. Maximum principle for Heat equation and its consequences
195
Using energy method, we can establish that solutions to an IBVP for heat equation
are unique. Consider the IBVP
u t = k u x x + f (x, t ), 0 < x < π, 0 < t
u(0, t ) = g1 (t ), 0 ≤ t
(7.33a)
(7.33b)
u(π, t ) = g3 (t ), 0 ≤ t
u(x, 0) = g2 (x), 0 ≤ x ≤ π
(7.33c)
(7.33d)
where k > 0, f , g1 , g2 , g3 are known functions.
Theorem 7.9. Let u and v be solutions to the IBVP (7.33) on the strip S := (0, π) × (0, ∞).
Then u = v on S .
Proof. Define w := u − v. Then w solves homogeneous heat equation with zero initialboundary conditions on the strip S . Note that E(t ) is a non-increasing function, and
thus E(t ) ≤ E(0). But E(0) = 0. Thus E(t ) = 0 and hence w(x, t ) = 0 on S .
7.3 Maximum principle for Heat equation and its consequences
We begin this section with the notion of parabolic boundary which plays an important
role in the study of maximum principle for heat equation.
Definition 7.10 (Parabolic Boundary). Let R be the rectangle (0, π) × (0, T ). The boundary ∂ R of the rectangle is the union of lines Li (i = 1, 2, 3, 4) where
L1 = {(0, t ) : 0 ≤ t ≤ T } ,
L2 = {(x, 0) : 0 ≤ x ≤ π} ,
L3 = {(π, t ) : 0 ≤ t ≤ T } ,
L4 = {(x, T ) : 0 ≤ x ≤ π} .
The parabolic boundary of the rectangle R, which is denoted by ∂P R, is defined by
∂P R = L1 ∪ L2 ∪ L3 .
The following maximum principle holds for solutions of heat equation.
Theorem 7.11. Let u ∈ CH be a solution of the heat equation. Then the maximum value of
u on R is achieved on the parabolic boundary ∂P R.
Proof. Since u ∈ C (R), the maximum value of u is attained somewhere in R. We would
like to show that this maximum is also attained on the parabolic boundary ∂P R. Let M
and m be defined by
M := max u
R
m := max u
∂P R
(7.34)
Clearly m ≤ M . The proof of the theorem will be complete if we prove that m < M is
not possible.
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Chapter 7. Heat Equation
t
L4
(0, T )
(π, T )
R
L1
L3
.
(0, 0)
(π, 0)
L2
x
Figure 7.1. Rectangle for Heat equation
Let L∗4 denote L4 without the end-points, i.e.,
L∗4 = L4 \ {(0, T ), (π, T )}.
Assume that m < M holds. Let (x1 , t1 ) ∈ R ∪ L∗4 be such that u(x1 , t1 ) = M .
Define the function v : R → R by
v(x, t ) = u(x, t ) +
M −m
(x − x1 )2
4π2
(7.35)
For (x, t ) ∈ ∂P R, we have
v(x, t ) ≤ m +
M −m 2
M −m
π =m+
<M
2
4π
4
(7.36)
Further, v(x1 , t1 ) = u(x1 , t1 ) = M . Thus the function v assumes its maximum value,
namely M , on R ∪ L∗4 . Let (x2 , t2 ) ∈ R ∪ L∗4 be such that v(x2 , t2 ) = M . Note that
0 < x2 < π. Note that
(a) if (x2 , t2 ) ∈ R, then we must have v t (x2 , t2 ) = 0, and
(b) if (x2 , t2 ) ∈ L∗4 , then we must have v t (x2 , t2 ) ≥ 0.
Thus we have v t (x2 , t2 ) ≥ 0. In view of the relations
•
˜
M −m
v t (x2 , t2 ) = u t (x2 , t2 ) = k u x x (x2 , t2 ) = k v x x (x2 , t2 ) −
,
2π2
we get
0 ≤ v t (x2 , t2 ) < kv x x (x2 , t2 ).
(7.37)
However, v x x (x2 , t2 ) ≤ 0 since v attains a maximum at (x2 , t2 ), which contradicts (7.37).
Thus m < M is not possible, which completes the proof of the theorem.
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7.3. Maximum principle for Heat equation and its consequences
197
Remark 7.12. Note that the maximum principle proved here is the weak maximum principle. A strong maximum principle which is similar to that of laplace equation also holds
for heat equation, and a proof may be found in [12].
Corollary 7.13. Let u ∈ CH be a solution of the heat equation. Then the minimum value of
u on R is achieved on the parabolic boundary ∂P R.
Proof. The assertion of the corollary follows by applying Theorem 7.11 to the function
v := −u.
Corollary 7.14 (uniqueness of solutions to IBVP). The initial boundary value problem
(7.22) for the heat equation has at most one solution.
Proof. Let u1 , u2 be solutions of the initial boundary value problem (7.22). We need to
show that u1 = u2 . Consider w defined by w := u1 − u2 . Then w satisfies the following
IBVP
wt = k wx x ,
0 < x < π, 0 < t < T
w(0, t ) = 0,
w(π, t ) = 0,
0≤t ≤T
0≤t ≤T
w(x, 0) = 0,
0≤x ≤π
By Theorem 7.11, and Corollary 7.13, we conclude that w attains both its maximum and
minimum on the parabolic boundary, but w = 0 there. Hence w ≡ 0.
Theorem 7.15. Consider the IBVP (7.23), where f is such that the fourier series of f converges
uniformly to f . Then the formal solution given by (7.31) is indeed a solution of the IBVP (7.23).
Proof.
Let us now check the validity of the formal solution given by (7.31).
(i) The series in (7.31) converges uniformly for (x, t ) ∈ [0, π] × (0, T ], when f is integrable on the interval [0, π]. For,
∫
∫
2 π
2 π
| f (s)|| sin ns | d s ≤
| f (s)| d s ≤ c < ∞,
|bn | ≤
(7.39)
π 0
π 0
∑∞
|bn e −n
2
kt
sin nx| ≤ c e −n
2
kt
≤ c e −n
2
k t0
(7.40)
Since the series n=1 e −n k t0 is convergent (follows from ratio test), we conclude
that the series in (7.31) converges uniformly for (x, t ) ∈ [0, π] × [t0 , T ], and hence
u is a continuous function. Since t0 is arbitrary, we conclude that u is continuous
on [0, π] × (0, T ].
(ii) The derivatives u t , u x , u x x exist as continuous functions on [0, π] × (0, T ]. This
follows from the fact that the series in (7.31) can be differentiated term-by-term
once w.r.t. t and twice w.r.t. x. Since proofs of assertions regarding u x , u x x are on
similar lines, we present the proof for the case of u t . Note that the series
2
∞
∑
(−n 2 k)bn e −n
2
kt
sin nx
(7.41)
n=1
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198
Chapter 7. Heat Equation
is uniformly convergent for (x, t ) ∈ [0, π] × [t0 , T ], which follows from ratio test,
∑
2
−n 2 k t0
the convergence of the series ∞
, and the inequalities
n=1 n k e
|bn n 2 k e −n
2
kt
sin nx| ≤ c n 2 e −n
2
kt
≤ c e −n
2
k t0
.
(7.42)
Thus u t is a continuous function on [0, π] × [t0 , T ]. Since t0 > 0 is arbitrary, it
follows that u t is a continuous function on [0, π] × (0, T ].
Thus it follows that u satisfies heat equation on the domain [0, π] × (0, T ]. It remains
to show that u satisfies the initial-boundary conditions (7.23b)-(7.23d).
In order to show that u is continuous on [0, π] × [0, T ], we show that the sequence
of partial sums of the series in (7.31) is uniformly Cauchy on [0, π] × [0, T ]. Let the N th
partial sum be denoted by SN (x, t ) which is given by
SN (x, t ) =
N
∑
bn e −n
2
kt
sin nx.
(7.43)
n=1
For m ≥ l , let w l ,m be defined by w l ,m (x, t ) = S m (x, t ) − S l (x, t ). Thus
w l ,m (x, t ) =
m
∑
bn e −n
2
kt
sin nx.
(7.44)
n=l +1
Note that w l ,m is a solution of (7.23), and satisfies
w l ,m (x, 0) =
m
∑
bn sin n x.
(7.45)
n=l +1
Applying maximum principle to the functions w l ,m we get
|w l ,m (x, t )| ≤ |w l ,m (x, 0)|
(7.46)
∑m
Note that n=l +1 bn sin nx is a uniformly Cauchy sequence, since the Fourier sine series
for f is assumed to converge uniformly to f on the interval [0, π]. Thus the function u
is continuous on [0, π] × [0, T ], and satisfies the initial condition u(x, 0) = f (x).
Remark 7.16. Note from the above proof that the series (7.31) was proved to be differentiable w.r.t. t by proving that series resulting from term-by-term differentiation of (7.31) is
uniformly convergent, which followed from the presence of exponentially decaying term
2
e −n k t . By a similar argument, it follows that the function defined by the series (7.31) is infinitely differentiable w.r.t. x and t . Thus a solution of heat equation belongs to the space
C ∞ (R), even when u(x, 0) = f is not. This is described as the regularizing (smoothing)
effect of heat equation.
Sivaji
IIT Bombay
Exercises
199
Exercises
General
7.1. Show that the heat equation u t − u x x = 0 under the change of coordinates z = ax
and τ = a 2 t takes the form wτ − w z z = 0.
7.2. [23] Goal of this exercise is to prove Weierstrass’s approximation theorem using
formula (7.15). Let f ∈ C [a, b ].
(i) Extend the function to φ : R → R by

 f (a)
f (x)
φ(x) =

f (b )
if x < a,
if a ≤ x ≤ b ,
if x > b .
Note that φ is continuous and bounded on R.
(ii) Consider Cauchy problem for homogeneous heat equation with Cauchy
data φ as defined in (i) above. Write down the formula (7.15). Show that
u(x, t ) → f (x) uniformly for a ≤ x ≤ b , as t → 0.
(iii) Approximate K1 (x, y, t ) by its truncated power series w.r.t. x − y, and conclude Weierstrass approximation theorem.
(iv) Further show that any f ∈ C0m (R) and each of its derivatives upto order m
can be approximated by polynomials uniformly in any bounded set containing the support of u.
7.3. [40] Solve the heat equation with constant dissipation:
u t − u x x + b u = 0 for x ∈ R,
u(x, 0) = φ(x),
where b > 0 is a constant. (Hint: Try a solution having the form u(x, t ) =
e −b t v(x, t )).
7.4. [40] If u(x, t ) = f (x − at ) is a solution of homogeneous heat equation, then find
f and show that the speed a is arbitrary.
7.5. [40] Let u be a solution of 2u t = u x x . Let v be defined by
2 
‹
1
x
x 1
v(x, t ) = p exp
u
,
.
2t
t t
t
Show that v satisfies the ‘backward heat equation’ 2v t = −v x x for t > 0.
IBVP
7.6. Solve the following initial boundary value problem:
ut = k ux x
x > 0, t > 0,
u(0, t ) = 1
u(x, 0) = 0
t ≥ 0,
x ≥ 0.
(Ans: u(x, t ) = 0 for x > t and u(x, t ) = 1 for 0 < x < t .)
October 21, 2015
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200
Chapter 7. Heat Equation
7.7. [40] Consider heat equation posed for x ∈ (0, l ) with Robin boundary conditions
u x (0, t )−a0 u(0, t ) = 0 and u x (l , t )+a l u(l , t ) = 0, where a0 > 0 and a l > 0 are constants. Use energy method to show that the endpoints contribute to the decrease
∫l
of 0 u 2 (x, t ) d x. These boundary conditions are called ‘radiating’, ‘dissipative’
boundary conditions as energy is lost at the boundary.
Maximum principles
7.8. [16] The initial boundary value problem for Heat equation is well-posed. Existence was proved by separation of variables method and uniqueness using maximum principle. Now prove the stability part of the well-posedness.
7.9. State and prove a maximum principle for solutions of an IBVP for u t = k∆u,
where ∆ is the laplacian in Rd .
7.10. Consider the following initial boundary value problem:
u t = k u x x , 0 < x < π, 0 < t < T
u(0, t ) = g (t ), 0 ≤ t ≤ T
u(π, t ) = g (t ), 0 ≤ t ≤ T
u(x, 0) = f (x),
0≤x ≤π
(a) Using maximum principle, show that the above initial boundary value problem has a unique solution.
(b) Using maximum principle, show that the solutions to IBVP are stable in
the sense that for each ε > 0, there exists a δ > 0 such that for functions
f1 , f2 , g1 , g2 , h1 , h2 satisfying
max | f1 (x) − f2 (x)| < δ, max | g1 (t ) − g2 (t )| < δ, max |h1 (t ) − h2 (t )| < δ,
x∈[0,π]
t ∈[0,T ]
t ∈[0,T ]
the corresponding solutions u1 , u2 of the IBVP corresponding to the data
( f , g , h) = ( f1 , g1 , h1 ) and ( f , g , h) = ( f2 , g2 , h2 ) respectively satisfy
max |u1 (x, t ) − u2 (x, t )| ≤ ε,
R
where R := (0, π) × (0, T ).
7.11. [40] Let u be a solution of u t = u x x , 0 ≤ x ≤ l , 0 ≤ t < ∞.
(a) Let M (T ) be the maximum of the function u on the rectangle {(x, t ) : 0 ≤
x ≤ l , 0 ≤ t ≤ T }. Is M (T ) (strictly) decreasing or increasing as a function
of T ?
(b) Let m(T ) be the minimum of the function u on the rectangle {(x, t ) : 0 ≤
x ≤ l , 0 ≤ t ≤ T }. Is m(T ) (strictly) decreasing or increasing as a function
of T ?
7.12. [40] The aim of this exercise is to show that the maximum principle does not hold
for the equation u t = x u x x which has a variable coefficient.
(a) Verify that the function u(x, t ) = −2x t − x 2 is a solution. Find its maximum
on the rectangle {(x, t ) : −2 ≤ x ≤ 2, 0 ≤ t ≤ 1}.
(b) Where exactly does our proof of the maximum principle fail in the case of
this equation?
Sivaji
IIT Bombay
Exercises
201
7.13. [33] For T > 0, let R be defined by
R = {(x, t ) : 0 < x < π, 0 < t < T }.
Let u be a solution to the problem
ut − ux x = 0
u(0, t ) = u(π, t ) = 0
u(x, 0) = sin2 x
on R,
for 0 ≤ t ≤ T ,
for 0 ≤ x ≤ π.
Using maximum principle, show that 0 ≤ u(x, t ) ≤ e −t sin x holds for (x, t ) ∈ R.
October 21, 2015
Sivaji