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2 Heat Equation
2.1 Derivation
Ref: Strauss, Section 1.3. Below we provide two derivations of the heat equation,
ut − kuxx = 0 k > 0. (2.1)
This equation is also known as the diffusion equation.
2.1.1 Diffusion
Consider a liquid in which a dye is being diffused through the liquid. The dye will move from higher concentration to lower concentration. Let u(x, t) be the concentration (mass per unit length) of the dye at position x in the pipe at time t. The total mass of dye in the pipe from x 0 to x 1 at time t is given by
M (t) =
∫ (^) x 1
x 0
u(x, t) dx.
Therefore, dM dt
∫ (^) x 1
x 0
ut(x, t) dx.
By Fick’s Law, dM dt = flow in − flow out = kux(x 1 , t) − kux(x 0 , t),
where k > 0 is a proportionality constant. That is, the flow rate is proportional to the concentration gradient. Therefore, ∫ (^) x 1
x 0
ut(x, t) dx = kux(x 1 , t) − kux(x 0 , t).
Now differentiating with respect to x 1 , we have
ut(x 1 , t) = kuxx(x 1 , t).
Or, ut = kuxx.
This is known as the diffusion equation.
2.1.2 Heat Flow
We now give an alternate derivation of (2.1) from the study of heat flow. Let D be a region in Rn. Let x = [x 1 ,... , xn]T^ be a vector in Rn. Let u(x, t) be the temperature at point x,
time t, and let H(t) be the total amount of heat (in calories) contained in D. Let c be the specific heat of the material and ρ its density (mass per unit volume). Then
H(t) =
D
cρu(x, t) dx.
Therefore, the change in heat is given by
dH dt
D
cρut(x, t) dx.
Fourier’s Law says that heat flows from hot to cold regions at a rate κ > 0 proportional to the temperature gradient. The only way heat will leave D is through the boundary. That is, dH dt
∂D
κ∇u · n dS.
where ∂D is the boundary of D, n is the outward unit normal vector to ∂D and dS is the surface measure over ∂D. Therefore, we have ∫
D
cρut(x, t) dx =
∂D
κ∇u · n dS.
Recall that for a vector field F , the Divergence Theorem says ∫
∂D
F · n dS =
D
∇ · F dx.
(Ref: See Strauss, Appendix A.3.) Therefore, we have ∫
D
cρut(x, t) dx =
D
∇ · (κ∇u) dx.
This leads us to the partial differential equation
cρut = ∇ · (κ∇u).
If c, ρ and κ are constants, we are led to the heat equation
ut = k∆u,
where k = κ/cρ > 0 and ∆u =
∑n i=1 uxixi^.
2.2 Heat Equation on an Interval in R
2.2.1 Separation of Variables
Consider the initial/boundary value problem on an interval I in R,
ut = kuxx x ∈ I, t > 0 u(x, 0) = φ(x) x ∈ I u satisfies certain BCs.
In practice, the most common boundary conditions are the following:
Any positive eigenvalues? First, we check if we have any positive eigenvalues. That is, we check if there exists any λ = β^2 > 0. Our eigenvalue problem (2.4) becomes { X′′^ + β^2 X = 0 0 < x < l X(0) = 0 = X(l).
The solutions of this ODE are given by
X(x) = C cos(βx) + D sin(βx).
The boundary condition X(0) = 0 =⇒ C = 0.
The boundary condition
X(l) = 0 =⇒ sin(βl) = 0 =⇒ β = nπ l n = 1, 2 ,....
Therefore, we have a sequence of positive eigenvalues
λn =
(nπ l
with corresponding eigenfunctions
Xn(x) = Dn sin
(nπ l x
Is zero an eigenvalue? Next, we look to see if zero is an eigenvalue. If zero is an eigenvalue, our eigenvalue problem (2.4) becomes { X′′^ = 0 0 < x < l X(0) = 0 = X(l).
The general solution of the ODE is given by
X(x) = C + Dx.
The boundary condition X(0) = 0 =⇒ C = 0.
The boundary condition X(l) = 0 =⇒ D = 0.
Therefore, the only solution of the eigenvalue problem for λ = 0 is X(x) = 0. By definition, the zero function is not an eigenfunction. Therefore, λ = 0 is not an eigenvalue.
Any negative eigenvalues? Last, we check for negative eigenvalues. That is, we look for an eigenvalue λ = −γ^2. In this case, our eigenvalue problem (2.4) becomes { X′′^ − γ^2 X = 0 0 < x < l X(0) = 0 = X(l).
The solutions of this ODE are given by
X(x) = C cosh(γx) + D sinh(γx).
The boundary condition X(0) = 0 =⇒ C = 0.
The boundary condition X(l) = 0 =⇒ D = 0.
Therefore, there are no negative eigenvalues. Consequently, all the solutions of (2.4) are given by
λn =
(nπ l
Xn(x) = Dn sin
(nπ l x
n = 1, 2 ,....
¶
Example 2. (Periodic Boundary Conditions) Find all solutions to the eigenvalue problem { −X′′^ = λX −l < x < l X(−l) = X(l), X′(−l) = X′(l). (2.5)
Any positive eigenvalues? First, we check if we have any positive eigenvalues. That is, we check if there exists any λ = β^2 > 0. Our eigenvalue problem (2.5) becomes { X′′^ + β^2 X = 0 −l < x < l X(−l) = X(l), X′(−l) = X′(l).
The solutions of this ODE are given by
X(x) = C cos(βx) + D sin(βx).
The boundary condition
X(−l) = X(l) =⇒ D sin(βl) = 0 =⇒ D = 0 or β = nπ l
The boundary condition
X′(−l) = X′(l) =⇒ Cβ sin(βl) = 0 =⇒ C = 0 or β = nπ l
Therefore, we have a sequence of positive eigenvalues
λn =
(nπ l
with corresponding eigenfunctions
Xn(x) = Cn cos
(nπ l x
(nπ l x
Now that we have done a couple of examples of solving eigenvalue problems, we return to using the method of separation of variables to solve (2.2). Recall that in order for a function of the form u(x, t) = X(x)T (t) to be a solution of the heat equation on an interval I ⊂ R which satisfies given boundary conditions, we need X to be a solution of the eigenvalue problem, (^) { X′′^ = −λX x ∈ I X satisfies certain BCs
for some scalar λ and T to be a solution of the ODE
−T ′^ = kλT.
We have given some examples above of how to solve the eigenvalue problem. Once we have solved the eigenvalue problem, we need to solve our equation for T. In particular, for any scalar λ, the solution of the ODE for T is given by
T (t) = Ae−kλt
for an arbitrary constant A. Therefore, for each eigenfunction Xn with corresponding eigen- value λn, we have a solution Tn such that the function
un(x, t) = Tn(t)Xn(x)
is a solution of the heat equation on the interval I which satisfies our boundary conditions. Note that we have not yet accounted for our initial condition u(x, 0) = φ(x). We will look at that next. First, we remark that if {un} is a sequence of solutions of the heat equation on I which satisfy our boundary conditions, than any finite linear combination of these solutions will also give us a solution. That is,
u(x, t) ≡
∑^ N
n=
un(x, t)
will be a solution of the heat equation on I which satisfies our boundary conditions, assuming each un is such a solution. In fact, one can show that an infinite series of the form
u(x, t) ≡
∑^ ∞
n=
un(x, t)
will also be a solution of the heat equation, under proper convergence assumptions of this series. We will omit discussion of this issue here.
2.2.2 Satisfying our Initial Conditions
We return to trying to satisfy our initial conditions. Assume we have found all solutions of our eigenvalue problem. We let {Xn} denote our sequence of eigenfunctions and {λn} denote our sequence of eigenvalues. Then for each λn, we have a solution Tn of our equation for T. Let u(x, t) =
n
Xn(x)Tn(t) =
n
AnXn(x)e−kλnt.
Our goal is to choose An appropriately such that our initial condition is satisfied. In partic- ular, we need to choose An such that
u(x, 0) =
n
AnXn(x) = φ(x).
In order to find An satisfying this condition, we use the following orthogonality property of eigenfunctions. First, we make some definitions. For two real-valued functions f and g defined on Ω,
〈f, g〉 =
Ω
f (x)g(x) dx
is defined as the L^2 inner product of f and g on Ω. The L^2 norm of f on Ω is defined as
||f ||^2 L (^2) (Ω) = 〈f, f 〉 =
Ω
|f (x)|^2 dx.
We say functions f and g are orthogonal on Ω ⊂ Rn^ if
〈f, g〉 =
Ω
f (x)g(x) dx = 0.
We say boundary conditions are symmetric if
[f ′(x)g(x) − f (x)g′(x)]|xx==ba = 0
for all functions f and g satisfying the boundary conditions.
Lemma 3. Consider the eigenvalue problem (2.3) with symmetric boundary conditions. If Xn, Xm are two eigenfunctions of (2.3) with distinct eigenvalues, then Xn and Xm are or- thogonal.
Proof. Let I = [a, b].
λn
∫ (^) b
a
Xn(x)Xm(x) dx = −
∫ (^) b
a
X n′′ (x)Xm(x) dx
∫ (^) b
a
X n′(x)X m′(x) dx − X n′(x)Xm(x)|xx==ba
∫ (^) b
a
Xn(x)X m′′(x) dx + [Xn(x)X m′(x) − X n′(x)Xm(x)]|xx==ba
= −λm
∫ (^) b
a
Xn(x)Xm(x) dx,
using the fact that the boundary conditions are symmetric. Therefore,
(λn − λm)
∫ (^) b
a
Xn(x)Xm(x) dx = 0,
Example 5. (Periodic Boundary Conditions) In the case of periodic boundary conditions on the interval [−l, l], we showed earlier that our eigenvalues and eigenfunctions are given by
λn =
(nπ l
, Xn(x) =
cos
(nπ l x
sin
(nπ l x
) n = 1, 2 ,...
λ 0 = 0, X 0 (x) = C 0.
Therefore, our solutions for Tn are given by
Tn(t) = Ane−kλnt^ =
Ane−k(nπ/l)^2 t^ n = 1, 2 ,... A 0 n = 0.
Now using the fact that for any integer n ≥ 0, un(x, t) = Xn(x)Tn(t) is a solution of the heat equation which satisfies our periodic boundary conditions, we define
u(x, t) =
n
Xn(x)Tn(t) = A 0 +
∑^ ∞
n=
[
An cos
(nπ l x
(nπ l x
)]
e−k(nπ/l) (^2) t .
Now, taking into account our initial condition, we want
u(x, 0) = A 0 +
∑^ ∞
n=
[
An cos
(nπ l x
(nπ l x
)]
= φ(x).
It remains only to find coefficients satisfying this equation. From our earlier discus- sion, using the fact that periodic boundary conditions are symmetric (check this!), we know that eigenfunctions corresponding to distinct eigenvalues will be orthogonal. For peri- odic boundary conditions, however, we have two eigenfunctions for each positive eigenvalue. Specifically, cos
(nπ l x
and sin
(nπ l x
are both eigenfunctions corresponding to the eigenvalue λn = (nπ/l)^2. Could we be so lucky that these eigenfunctions would also be orthogonal? By a straightforward calculation, one can show that
∫ (^) l
−l
cos
(nπ l x
sin
(nπ l x
dx = 0.
They are orthogonal! This is not merely coincidental. In fact, for any eigenvalue λ of (2.3) with multiplicty m (meaning it has m linearly independent eigenfunctions), the eigenfunc- tions may always be chosen to be orthogonal. This process is known as the Gram-Schmidt orthogonalization method. The fact that all our eigenfunctions are mutually orthogonal will allow us to calculate coefficients An, Bn so that our initial condition is satisfied. Using the technique described
above, and letting 〈f, g〉 denote the L^2 inner product on [−l, l], we see that
A 0 =
〈 1 , φ〉 〈 1 , 1 〉
2 l
∫ (^) l
−l
φ(x) dx
An = 〈cos(nπx/l), φ〉 〈cos(nπx/l), cos(nπx/l)〉
l
∫ (^) l
−l
cos
(nπ l
x
φ(x) dx
Bn =
〈sin(nπx/l), φ〉 〈sin(nπx/l), sin(nπx/l)〉
l
∫ (^) l
−l
sin
(nπ l x
φ(x) dx.
Therefore, the solution of (2.2) on the interval I = [−l, l] with periodic boundary condi- tions is given by
u(x, t) = A 0 +
∑^ ∞
n=
[
An cos
(nπ l x
(nπ l x
)]
e−k(nπ/l) (^2) t
where
A 0 =
2 l
∫ (^) l
−l
φ(x) dx
An =
l
∫ (^) l
−l
cos
(nπ l x
φ(x) dx
Bn =
l
∫ (^) l
−l
sin
(nπ l x
φ(x) dx.
¶
2.2.3 Fourier Series
In the case of Dirichlet boundary conditions above, we looked for coefficients so that
φ(x) =
∑^ ∞
n=
An sin
(nπ l x
We showed that if we could write our function φ in terms of this infinite series, our coefficients would be given by the formula
An =
l
∫ (^) l
0
sin
(nπ l x
φ(x) dx.
For a given function φ defined on (0, l) the infinite series
φ ∼
∑^ ∞
n=
An sin
(nπ l x
where An ≡
l
∫ (^) l
0
sin
(nπ l x
φ(x) dx
is called the Fourier sine series of φ. Note: The notation ‘∼’ just means the series associated with φ. It doesn’t imply that the series necessarily converges to φ.
- The Fourier sine series of φ converges to φ (in some sense).
- The infinite series actually satisfies the heat equation.
We will not discuss these issues here, but refer the reader to convergence results in Strauss as well as the notes from 220a. Later, in the course, we will prove an L^2 convergence result of eigenfunctions.
Complex Form of Full Fourier Series. It is sometimes useful to write the full Fourier series in complex form. We do so as follows. The eigenfunctions associated with the full Fourier series are given by (^) { cos
(nπ l x
, sin
(nπ l x
for n = 0, 1 , 2 ,.. .. Using deMoivre’s formula,
eiθ^ = cos θ + i sin θ,
we can write
cos
(nπ l x
einπx/l^ + e−inπx/l 2 sin
(nπ l x
einπx/l^ − e−inπx/l 2 i
Now, of course, any linear combination of eigenfunctions is also an eigenfunction. Therefore, we see that
cos
(nπ l x
(nπ l x
= einπx/l
cos
(nπ l x
− i sin
(nπ l x
= e−inπx/l
are also eigenfunctions. Therefore, the eigenfunctions associated with the full Fourier series can be written as (^) { einπx/l
n =... , − 2 , − 1 , 0 , 1 , 2 ,....
Now, let’s suppose we can represent a given function φ as an infinite series expansion in terms of these eigenfunctions. That is, we want to find coefficients Cn such that
φ(x) =
∑^ ∞
n=−∞
Cneinπx/l.
As described earlier, eigenfunctions corresponding to distinct eigenvalues will be orthogonal, as periodic boundary conditions are symmetric. Therefore, we have
∫ (^) l
−l
einπx/leimπx/l^ dx = 0 for m 6 = n.
For eigenfunctions corresponding to the same eigenvalue, we need to check the L^2 inner product. In particular, for the eigenvalue λn = (nπ/l)^2 , we have two eigenfunctions: einπx/l and e−inπx/l. By a straightforward calculation, we see that ∫ (^) l
−l
einπx/leinπx/l^ dx = 0 for n 6 = 0 ∫ (^) l
−l
einπx/le−inπx/l^ dx = 2l.
Therefore, our coefficients Cn would need to be given by
Cn =
2 l
∫ (^) l
−l
φ(x)e−inπx/l^ dx.
Consequently, the complex form of the full Fourier series for a function φ defined on (−l, l) is given by
φ ∼
∑^ ∞
n=−∞
Cneinπx/l^ where Cn =
2 l
∫ (^) l
−l
φ(x)e−inπx/l^ dx.
2.3 Fourier Transforms
2.3.1 Motivation
Ref: Strauss, Section 12. We would now like to turn to studying the heat equation on the whole real line. Consider the initial-value problem, (^) { ut = kuxx, −∞ < x < ∞ u(x, 0) = φ(x).
In the case of the heat equation on an interval, we found a solution u using Fourier series. For the case of the heat equation on the whole real line, the Fourier series will be replaced by the Fourier transform. Above, we discussed the complex form of the full Fourier series for a given function φ. In particular, for a function φ defined on the interval [−l, l] we define its full Fourier series as
φ ∼
∑^ ∞
n=−∞
Cneinπx/l^ where Cn =
2 l
∫ (^) l
−l
φ(x)e−inπx/l^ dx.
Plugging the coefficients Cn into the infinite series, we see that
φ ∼
∑^ ∞
n=−∞
[
2 l
∫ (^) l
−l
φ(y)e−inπy/l^ dy
]
einπx/l.
Now, letting k = nπ/l, we can write this as
φ ∼
∑^ ∞
n=−∞
[
2 π
∫ (^) l
−l
φ(y)e−i(y−x)k^ dy
]
π l
Therefore, we just need to look at ∫ (^) ∞
−∞
e−ixξe−≤x^2 dx.
Completing the square, we have
−≤x^2 − ixξ = −≤
[
x^2 + iξx ≤
iξ 2 ≤
iξ 2 ≤
) 2 ]
[
x +
iξ 2 ≤
)] 2
iξ 2 ≤
Therefore, we have ∫ (^) ∞
−∞
e−ixξe−≤x^2 dx =
−∞
e−≤[x+(iξ/^2 ≤)]^2 e−ξ^2 /^4 ≤^ dx.
Now making the change of variables, z = [x + (iξ/ 2 ≤)], we have ∫ (^) ∞
−∞
e−≤[x+(iξ/^2 ≤)] 2 e−ξ (^2) / 4 ≤ dx = e−ξ (^2) / 4 ≤
Γ
e−≤z 2 dz
where Γ is the line in the complex plane given by
Γ ≡
z ∈ C : y = x + iξ 2 ≤ , x ∈ R
Without loss of generality, we assume ξ > 0. A similar analysis works if ξ < 0. Now ∫
Γ
e−≤z 2 dz = (^) R→lim+∞
ΓR
e−≤z 2 dz
where ΓR is the line segment in the complex plane given by
ΓR ≡
z ∈ C : z = x + iξ 2 ≤ , |x| ≤ R
Now define Λ^1 R, Λ^2 R and Λ^3 R as shown in the picture below. That is,
Λ^1 R ≡ {x ∈ R : |x| ≤ R}
Λ^2 R ≡
z ∈ C : z = x + iy, x, y ∈ R, x = R, 0 ≤ y ≤ ξ 2 ≤
Λ^3 R ≡
z ∈ C : z = x + iy; x, y ∈ R; x = −R, 0 ≤ y ≤ ξ 2 ≤
− R R
Γ R
Λ^1
(^32) R R
R
Re z
Im z (
ξ/2ε
From complex analysis, we know that ∫
C
e−≤z 2 dz = 0
where C is the closed curve given by C = ΓR ∪Λ^1 R ∪Λ^2 R ∪Λ^3 R traversed in the counter-clockwise direction. Therefore, we have ∫
ΓR
e−≤z 2 dz =
ΛR
e−≤z 2 dz
where the integral on the right-hand side is the line integral given by ΛR = Λ^3 R ∪ Λ^1 R ∪ Λ^2 R traversed in the direction shown. Therefore, ∫
Γ
e−≤z 2 dz = (^) R→lim+∞
ΛR
e−≤z 2 dz.
But, as R → +∞, ∫
ΛjR
e−≤z 2 dz → 0 forj = 2, 3 ∫
Λ^1 R
e−≤z^2 dz →
−∞
e−≤x^2 dx.
Therefore, (^) ∫
Γ
e−≤z 2 dz =
−∞
e−≤x 2 dx.
Consequently, we have ∫ (^) ∞
−∞
e−ixξe−≤x 2 dx = e−ξ (^2) / 4 ≤
Γ
e−≤z 2 dz
= e−ξ (^2) / 4 ≤
−∞
e−≤x 2 dx
= e−ξ (^2) / 4 ≤
−∞
e−xe 2 d˜x √ ≤
= e− √ξ^2 /^4 ≤ ≤
π.
Therefore, we have
f̂ (ξ) = 1 (2π)n/^2
e−ξ^12 /^4 ≤ √ ≤
π
e−ξ^2 n/^4 ≤ √ ≤
π
(2≤)n/^2 e−|ξ| (^2) / 4 ≤ ,
as claimed.
Next, we use the fact that if f, g are in L^1 (Rn), then f ,̂ ̂g are in L∞(Rn), and, moreover, ∫
Rn
f (x)̂g(x) dx =
Rn
f (ξ)g(ξ) dx. (2.9)
This fact can be seen by direct substitution, as shown below, ∫
Rn
f (x)̂g(x) dx =
Rn
f (x)
[
(2π)n/^2
Rn
e−ix·ξg(ξ) dξ
]
dx
Rn
[
(2π)n/^2
Rn
e−ix·ξf (x) dx
]
g(ξ) dξ
=
Rn
f (ξ)g(ξ) dξ.
Therefore, letting f (x) = e−≤|x|^2 and letting g(x) = w(x) as defined above, substituting f and g into (2.9) and using Claim 7 to calculate the Fourier transform of f , we have ∫
Rn
e−≤|ξ| 2 ̂ w(ξ) dξ =
Rn
(2≤)n/^2 e−|x| (^2) / 4 ≤ w(x) dx. (2.10)
Now we take the limit of both sides above as ≤ → 0 +. First,
lim ≤→ 0 +
Rn
e−≤|ξ| 2 ̂ w(ξ) dξ =
Rn
w(ξ) dξ. (2.11)
Second, we claim
lim ≤→ 0 +
(2≤)n/^2
Rn
e−|x| (^2) / 4 ≤ w(x) dx = (2π)n/^2 w(0). (2.12)
We prove this claim as follows. In particular, we will prove that
1 (4π≤)n/^2
Rn
e−|x| (^2) / 4 ≤ w(x) dx → w(0) as ≤ → 0 +.
First, we note that 1 (4π≤)n/^2
Rn
e−|x|^2 /^4 ≤^ dx = 1. (2.13)
This follows directly from the fact that ∫ (^) ∞
−∞
e−z 2 dz =
π.
Therefore,
1 (4π≤)n/^2
Rn
e−|x| (^2) / 4 ≤ w(x) dx − w(0) =
(4π≤)n/^2
Rn
e−|x| (^2) / 4 ≤ [w(x) − w(0)] dx.
Now, we will show that for all γ > 0 there exists an ˜≤ > 0 such that ∣∣ ∣ ∣
(4π≤)n/^2
Rn
e−|x|^2 /^4 ≤[w(x) − w(0)] dx
∣ < γ
for 0 < ≤ < ˜≤, thus, proving (2.12). Let B(0, δ) be the ball of radius δ about 0. (We will choose δ sufficiently small below.) Now break up the integral above into two pieces, as follows, ∣ ∣∣ ∣
(4π≤)n/^2
Rn
e−|x| (^2) / 4 ≤ [w(x) − w(0)] dx
(4π≤)n/^2
B(0,δ)
e−|x| (^2) / 4 ≤ [w(x) − w(0)] dx
∣∣^1
(4π≤)n/^2
Rn−B(0,δ)
e−|x| (^2) / 4 ≤ [w(x) − w(0)] dx
≡ I + J.
First, for term I, we have ∣ ∣∣ ∣
(4π≤)n/^2
B(0,δ)
e−|x| (^2) / 4 ≤ [w(x) − w(0)] dx
∣ ≤ |w(x)^ −^ w(0)|L∞(B(0,δ))
(4π≤)n/^2
Rn
e−|x| (^2) / 4 ≤ dx
< γ 2
for δ sufficiently small, using the fact that w ∈ C(Rn) and (2.13). Now for δ fixed small, we look at term J, ∣∣ ∣∣^1 (4π≤)n/^2
Rn−B(0,δ)
e−|x| (^2) / 4 ≤ [w(x) − w(0)] dx
(4π≤)n/^2
Rn−B(0,δ)
e−|x| (^2) / 4 ≤ |w(x)| dx
(4π≤)n/^2
Rn−B(0,δ)
e−|x| (^2) / 4 ≤ |w(0)| dx
∣∣^1
(4π≤)n/^2 e−|x| (^2) / 4 ≤
L∞(Rn−B(0,δ))
Rn
|w(x)| dx
- e−δ^2 /^8 ≤|w(0)| 2 n/^2 (8π≤)n/^2
Rn−B(0,δ)
e−|x|^2 /^8 ≤^ dx
≤ C
(4π≤)n/^2
e−|δ|^2 /^4 ≤
∣ +^ Ce
−δ^2 / 8 ≤ 1 (8π≤)n/^2
Rn
e−|x|^2 /^8 ≤^ dx
≤ C
∣∣^1
(4π≤)n/^2 e−|δ| (^2) / 4 ≤
∣∣ + Ce−δ^2 /^8 ≤
< γ 2
for ≤ sufficiently small, using the fact that for a fixed δ 6 = 0,
lim ≤→ 0 +
(4π≤)n/^2 e−|δ| (^2) / 4 ≤ = 0 = lim ≤→ 0 +^ e−δ (^2) / 8 ≤ .
Therefore, we conclude that I + J < γ
for ≤ chosen sufficiently small, and, thus, (2.12) is proven.
