Figure 2.10:The space-time domain for the heat distribution in a ring.
Step 1 (Separation of variables):We use separation of variables again and look for special solutions of the form
u(θ,t)=A(θ)B(t).
By inserting this into the heat equation we get A(θ)B0(t)−A00(θ)B(t)=0⇐⇒B0(t)
B(t) =A00(θ) A(θ).
The variables are separated in the sense that the two sides of the equation above depend on different variables and, as in Section2.11, both have to be constant, which is denoted byλ. Thus
A00(θ)=λA(θ), B0(t)=λB(t).
Step 2 (Solution to the separated equations):The initial condition u(θ, 0)=A(θ)B(0)=g(θ)
is a 2π-periodic function. We do a similar case study on the possible values ofλ, as we did in the case of the Laplace equation.
λ>0 Thenλ=µ2,µ>0, and thus we haveA00(θ)−µ2A(θ)=0. This ODE has the general solution
A(θ)=c1eµθ+c2e−µθ,
which is periodic only ifc1=c2=0. This gives the trivial solutionu=0 and thus we exclude this alternative.
λ=0 ThenA00(θ)=0, which implies A(θ)=c1θ+c2. Here the only possibility that is compatible with the periodicity ofAis the solutionA(θ)=c2is constant.
In this case we also get thatB(t)=c3so that this gives only the constant solution.
λ<0 Thenλ= −µ2,µ>0 and the ODE forA(θ) is A00(θ)+µ2A(θ)=0. The general complex solution of this ODE is
A(θ)=c1eiµθ+c2e−iµθ
for some constantsc1andc2. Again forAto be periodic we needµto be an integer which gives thus, considering all cases above, that the only admissible values for λare of the formλ= −j2,j∈Z. For these values we haveB0(t)+j2B(t)=0, which has the general solution
B(t)=ce−j2t.
Thus we have special solutions, called the normal modes, of the form u(θ,t)=e−j2tei jθ, j∈Z.
By the construction, these functions are solutions of the one-dimensional heat equation for every j∈Z.
Step 3 (Fourier series solution of the entire problem): Since the heat equation is linear, any linear combination of the special solutions above will give again a solution. Motivated by the superposition principle, we define
u(θ,t)= X∞ j=−∞
aje−j2tei jθ, −πÉθ<π, t>0.
The next step is to determine coefficientsaj. By the initial conditionu(θ, 0)=g(θ), we have
X∞ j=−∞
ajei jθ=g(θ)
which identifies the coefficientsaj as the Fourier coefficientsg(b j) of the initial datag. Thus
u(θ,t)= X∞ j=−∞
g(j)eb −j2tei jθ.
This infinite series converges absolutely, since the Fourier coefficients g(j) areb bounded if g∈L1([−π,π]) (see Lemma 2.20(2)) and the term e−j2t decays ex-tremely fast ast>0 and jis large. Thus
X∞ j=−∞
¯
¯
¯ bg(j)e−j2tei jθ¯
¯
¯É X∞ j=−∞
|g(j)b |e−j2tÉ kgkL1([−π,π])
X∞ j=−∞
e−j2t
| {z }
<∞
.
The last series converges, which shows that the series defining the solution to the heat equation converges absolutely.
Step 4 (Explicit representation formula):Next goal is to derive an inte-gral representation for the solution similar to the convolution with the Poisson kernel for the Dirichlet problem for the Laplace equation in the unit disc. Using the definition of the Fourier coefficients we obtain
u(θ,t)= X∞ j=−∞
g(j)eb −j2tei jθ
= lim
n→∞
n
X
j=−n
µ 1 2π
Z π
−πg(s)e−i jsds
¶
e−j2tei jθ
= lim
n→∞
1 2π
Z π
−πg(s) Ã n
X
j=−n
e−j2tei j(θ−s)
! ds.
We have the uniform estimate
¯
¯
¯
¯
¯
n
X
j=−n
e−j2tei j(θ−s)
¯
¯
¯
¯
¯ É
X∞ j=−∞
e−j2t< ∞.
Here it is crucial thatt>0, so these estimates do not generalize to negative times.
By switching the order of the limit and the integral, we obtain u(θ,t)= 1
2π Z π
−πg(s) lim
n→∞
à n X
j=−n
e−j2tei j(θ−s)
! ds
= 1 2π
Z π
−πg(s) Ã ∞
X
j=−∞
e−j2tei j(θ−s)
! ds.
By defining the heat kernel for the circle as Ht(θ)=
X∞ j=−∞
e−j2tei jθ, −πÉθ<π, t>0,
we see that the solution of the heat equation on the circle is given by the convolu-tion
u(θ,t)=(g∗Ht)(θ)= 1 2π
Zπ
−πg(s)Ht(θ−s)ds, −πÉθ<π, t>0.
TH E M O R A L: The solution of the initial value problem for the heat equation on the circle is a convolution of the initial temperature distribution with the heat kernel.
Theorem 2.46 (Solution to the heat equation on the unit circle). The solu-tion of the (periodic) initial value problem (2.18) is
u(θ,t)=(g∗Ht)(θ,t)= 1 2π
Z π
−πg(s)Ht(θ−s)ds, −πÉθ<π, t>0, whereHt(θ)=P∞
j=−∞e−j2tei jθ. The initial values are obtained in the sense that limt→0u(θ,t)=g(θ) −πÉθÉπ.
Remark 2.47. Another way to derive the solution to a PDE problem is start with the Fourier series expansion, insert it in the PDE and try to determine the coefficients. This approach works for the Laplace equation, the heat equation and the wave equation, but we shall discuss only the initial value problem (2.18) in detail. For a fixedt>0, consider the Fourier series of the functionθ7→u(θ,t). This gives
u(θ,t)= X∞ j=−∞
cj(t)ei jθ, where the Fourier coefficients are
cj(t)= 1 2π
Z π
−πu(s,t)e−i jsds, j∈Z.
Observe that the Fourier coefficients depend ont. We claim that the PDE foru implies that the Fourier coefficients satisfy an ODE as a function oft. To see this, we switch the order of integral and derivative to have
c0j(t)=Çu Çt
µ 1 2π
Zπ
−πu(s,t)e−i jsds
¶
= 1 2π
Zπ
−π
Çu
Çt(s,t)e−i jsds, j∈Z. Then we use the PDE and Lemma2.31twice to obtain
c0j(t)= 1 2π
Z π
−π
Çu
Çt(s,t)e−i jsds
= 1 2π
Z π
−π
Ç2u
Çs2(s,t)e−i jsds
=i j 1 2π
Z π
−π
Çu
Çs(s,t)e−i jsds
=(i j)2 1 2π
Z π
−πu(s,t)e−i jsds
= −j2cj(t), j∈Z.
TH E M O R A L: The PDE becomes an ODE on the Fourier side.
The general solution ofc0j(t)+j2c(t)=0 iscj(t)=aje−j2t, whereajis a constant.
By the Fourier series representation u(θ,t)=
X∞ j=−∞
cj(t)ei jθ= X∞ j=−∞
aje−j2tei jθ, −πÉθ<π, t>0.
By the initial conditionu(θ, 0)=g(θ), we have u(θ, 0)=
X∞ j=−∞
ajei jθ=g(θ)
which identifies the coefficientsaj as the Fourier coefficientsg(b j) of the initial datag. Thus
u(θ,t)= X∞ j=−∞
g(j)eb −j2tei jθ.
TH E M O R A L: In the this approach we inserted the Fourier series represen-tation of the function in the PDE and determined the Fourier coefficients by the initial condition. In the beginning of this section we derived the same formula for the solution by separation of variables and this lead to the Fourier series.
Although the result is same, the main difference is that in the original argument we did not have to assume in the beginning that the solution is given by the Fourier series. However, theory of Fourier series is needed in both approaches.
Let us consider another problem for the one-dimensional heat equation. We study the temperature distribution in a thin uniform bar of given lengthL>0 with insulated lateral surface and no internal sources of heat, subject to certain boundary and initial conditions. To describe the problem, letu(x,t), 0<x<L,t>0, be the temperature at the pointxat timet. Assume that the initial temperature distribution of the bar isu(x, 0)=g(x) and assume, that the ends of the bar are held at constant temperature 0◦C, for example. This is related to the previously considered problem of the heat distribution on a circle, since we can cut the circle and unfold it so that it becomes a bar.
Figure 2.11:The heat distribution in a bar.
The following theorem gives a solution of the heat distribution in the bar of lengthL>0. We shall prove this in the exercises using the method of separation of variables.
Theorem 2.48 (Solution of the heat equation on a bar). The solution of
Çu
Çt −a2Ç2u
Çx2=0, 0<x<L, t>0, u(0,t)=u(L,t)=0, tÊ0,
u(x, 0)=g(x), 0ÉxÉL.
is
u(x,t)= X∞ j=1
aje−λ2jtsin jπx L , where
aj=2 L
Z L 0
g(x) sinjπx
L dx and λj=ajπ
L, j=1, 2, . . . .
Figure 2.12:The space-time domain for the heat distribution in a bar.
The method of separation of variables can also be used to solve heat conduction problems with other boundary conditions than those given above.
Example 2.49. Consider a bar of lenghtπis in boiling water. After reaching the temperature 100◦C throughout, the bar is taken out and immersed in a medium with constant freezing temperature 0◦C. The the bar are kept insulated and we assume thata2=1. Then the temperature distribution is
u(x,t)= X∞ j=1
aje−j2tsin(jx),
where
aj=2 π
Z π
0
100 sin(jx)dx=200
jπ(1−cos(jπ)).
After some simplifications (exercise) u(x,t)=400
π X∞ j=0
e−(2j+1)2t
2j+1 sin((2j+1)x).
Here we used the fact that 1−cos(jπ)=0 if jis even and 2 if jis odd.