Linear Partial Differential Equations
1.6 The Method of Separation of Variables
We next note that
∂2
∂x∂y
(x−y)u
= ∂
∂x (x−y)∂u
∂y −u
= (x−y)uxy+ (uy−ux). (1.5.52) Whenm= 1, equation (1.5.51) becomes
(x−y)uxy=ux−uy so that (1.5.52) reduces to
∂2
∂x∂y
(x−y)u
= 0. (1.5.53)
This shows that the solution of (1.5.53) is(x−y)u = φ(x) +ψ(y). Hence, the solution of (1.5.51) withm= 1is
u(x, y) = φ(x) +ψ(y)
x−y , (1.5.54)
whereφandψare arbitrary functions.
We multiply (1.5.51) by(x−y), and apply the derivative ∂x∂y∂2 , so that the result is, due to (1.5.52),
(x−y) ∂2
∂x∂y(uxy) + ∂
∂y − ∂
∂x
(uxy) =m ∂
∂x− ∂
∂y
uxy. Or
(x−y) ∂2
∂x∂y(uxy) = (m+ 1) ∂
∂x− ∂
∂y
(uxy). (1.5.55) Hence, ifuis a solution of (1.5.51), thenuxyis a solution of (1.5.51) withmreplaced bym+ 1. Whenm= 1, the solution is given by (1.5.54), and hence, the solution of (1.5.51) takes the form
u(x, y) = ∂2m−2
∂xm−1∂ym−1
φ(x) +ψ(y) x−y
, (1.5.56)
whereφandψare arbitrary functions.
variables. However, the question of separability of a partial differential equation into two or more ordinary differential equations is by no means a trivial one. In spite of this question, the method is widely used in finding solutions of a large class of initial boundary-value problems. This method of solution is known as the Fourier method (or the method of eigenfunction expansion). Thus, the procedure outlined above leads to the important ideas of eigenvalues, eigenfunctions, and orthogonality, all of which are very general and powerful for dealing with linear problems. The following ex- amples illustrate the general nature of this method.
Example 1.6.1 (Transverse Vibration of a String). We consider the one-dimensional linear wave equation
utt=c2uxx, 0< x < , t >0, (1.6.1) wherec2 =T∗/ρ,T∗is a constant tension, andρis the constant line density of the string. The initial and boundary conditions are
u(x,0) =f(x), ut(x,0) =g(x), 0≤x≤, (1.6.2ab) u(0, t) = 0 =u(, t), t >0, (1.6.3ab) wherefandgare the initial displacement and initial velocity, respectively.
According to the method of separation of variables, we assume a separable solu- tion of the form
u(x, t) =X(x)T(t)= 0, (1.6.4) whereXis a function ofxonly, andTis a function oftonly.
Substituting this solution into equation (1.6.1) yields 1
X d2X
dx2 = 1 c2T
d2T
dt2. (1.6.5)
Since the left side of this equation is a function ofxonly and the right-hand side is a function oftonly, it follows that (1.6.5) can be true only if both sides are equal to the same constant value. We then write
1 X
d2X dx2 = 1
c2T d2T
dt2 =λ, (1.6.6)
whereλis an arbitrary separation constant. Thus, this leads to the pair of ordinary differential equations
d2X
dx2 =λX, (1.6.7a)
d2T
dt2 =λc2T. (1.6.7b)
We solve this pair of equations by using the boundary conditions which are obtained from (1.6.3ab) as
u(0, t) =X(0)T(t) = 0 fort >0, (1.6.8) u(, t) =X()T(t) = 0 fort >0. (1.6.9) Hence, we takeT(t)= 0to obtain
X(0) = 0 =X(). (1.6.10ab)
There are three possible cases: (i)λ >0, (ii)λ= 0, (iii)λ <0.
For case (i),λ=α2>0. The solution of (1.6.7a) is
X(x) =Aeαx+Be−αx, (1.6.11) which together with (1.6.10ab) leads toA=B = 0. This leads to a trivial solution u(x, t) = 0.
For case (ii),λ= 0. In this case, the solution of (1.6.7a) is
X(x) =Ax+B. (1.6.12)
Then. we use (1.6.10ab) to obtainA=B = 0. This also leads to the trivial solution u(x, t) = 0.
For case (iii),λ < 0, and hence, we writeλ = −α2 so that the solution of equation (1.6.7a) gives
X =Acosαx+Bsinαx, (1.6.13)
whence, using (1.6.10ab), we derive the nontrivial solution
X(x) =Bsinαx, (1.6.14)
whereBis an arbitrary nonzero constant. Clearly, sinceB= 0andX() = 0,
sinα= 0, (1.6.15)
which gives the solution for the parameterα α=αn=
nπ
, n= 1,2,3, . . . . (1.6.16) Note thatn= 0 (α= 0)leads to the trivial solutionu(x, t) = 0, and hence, the case n= 0is to be excluded.
Evidently, we see from (1.6.16) that there exists an infinite set of discrete values ofαfor which the problem has a nontrivial solution. These valuesαnare called the eigenvalues, and the corresponding solutions are
Xn(x) =Bnsin nπx
. (1.6.17)
We next solve (1.6.7a) withλ=−α2nto find the solution forTn(t)as
Tn(t) =Cncos(αnct) +Dnsin(αnct), (1.6.18)
where Cn andDn are constants of integration. Combining (1.6.17) with (1.6.18) yields the solution from (1.6.4) as
un(x, t) = ancos nπct
+bnsin nπct
sin nπx
, (1.6.19)
wherean = CnBn,bn =BnDn are new arbitrary constants andn = 1,2,3, . . .. These solutionsun(x, t), corresponding to eigenvaluesαn = (nπ ), are called the eigenfunctions. Finally, since the problem is linear, the most general solution is ob- tained by the principle of superposition in the form
u(x, t) = ∞ n=1
ancosnπct
+bnsinnπct
sin
nπx
, (1.6.20) provided the series converges and it is twice continuously differentiable with respect toxandt. The arbitrary coefficientsanandbnare determined from the initial con- ditions (1.6.2ab) which give
u(x,0) =f(x) = ∞ n=1
ansin nπx
, (1.6.21)
ut(x,0) =g(x) = πc
∞
n=1
nbnsin nπx
. (1.6.22)
Either by a Fourier series method (see Appendix B) or by direct multiplication of (1.6.21) and (1.6.22) bysin(mπx )and integrating from0to, we can find an andbnas
an =2
0
f(x) sin nπx
dx, bn= 2 nπc
0
g(x) sin nπx
dx, (1.6.23ab) in which we have used the result
0
sin mπx
sin nπx
dx=
2δmn, (1.6.24)
whereδmnare Kronecker deltas. Thus, (1.6.20) represents the solution whereanand bnare given by (1.6.23ab). Hence, the problem is completely solved.
We examine the physical significance of the solution (1.6.19) in the context of the free vibration of a string of length. The eigenfunctions
un(x, t) = (ancosωnt+bnsinωnt) sin nπx
,
ωn =nπc
, (1.6.25) are called thenth normal modes of vibration or thenth harmonic, andωnrepresents the discrete spectrum of circular (or radian) frequency orνn= ω2πn =nc2, which are called the angular frequencies. The first harmonic(n= 1)is called the fundamental
Fig. 1.4 Several modes of vibration in a string.
harmonic and all other harmonics(n > 1)are called overtones. The frequency of the fundamental mode is given by
ω1=πc
, ν1= 1 2
T∗
ρ . (1.6.26ab)
Result (1.6.26ab) is considered the fundamental law (or Mersenne law) of a stringed musical instrument. The angular frequency of the fundamental mode of transverse vibration of a string varies as the square root of the tension, inversely as the length, and inversely as the square root of the density. The period of the fundamental mode is T1= 2πω
1 = 2c, which is called the fundamental period. Finally, the solution (1.6.20) describes the motion of a plucked string as a superposition of all normal modes of vibration with frequencies which are all integral multiples(ωn = nω1 or νn = nν1)of the fundamental frequency. This is the main reason for the fact that stringed instruments produce more sweet musical sounds (or tones) than drum instruments.
In order to describe waves produced in the plucked string with zero initial veloc- ity(ut(x,0) = 0), we write the solution (1.6.25) in the form
un(x, t) =ansin nπx
cos nπct
, n= 1,2,3, . . . . (1.6.27) These solutions are called standing waves with amplitudeansin(nπx ), which van- ishes at
x= 0, n,2
n, . . . , .
These are called the nodes of the nth harmonic. The string displaysnloops separated by the nodes as shown in Figure1.4.
It follows from elementary trigonometry that (1.6.27) takes the form un(x, t) = 1
2an sinnπ
(x−ct) + sinnπ
(x+ct)
. (1.6.28)
This shows that a standing wave is expressed as a sum of two progressive waves of equal amplitude traveling in opposite directions. This result is in agreement with the d’Alembert solution.
Finally, we can rewrite the solution (1.6.19) of the nth normal modes in the form un(x, t) =cnsin
nπx
cos
nπct −εn
, (1.6.29)
wherecn= (a2n+b2n)12 andtanεn= (ban
n).
This solution represents transverse vibrations of the string at any pointxand at any timet with amplitudecnsin(nπx )and circular frequencyωn = nπc . This form of the solution enables us to calculate the kinetic and potential energies of the transverse vibrations. The total kinetic energy (K.E.) is obtained by integrating with respect toxfrom0to, that is,
Kn=K.E.=
0
1 2ρ
∂un
∂t 2
dx, (1.6.30)
whereρis the line density of the string. Similarly, the total potential energy (P.E.) is given by
Vn=P.E.= 1 2T∗
0
∂un
∂x 2
dx. (1.6.31)
Substituting (1.6.29) in (1.6.30) and (1.6.31) gives Kn= 1
2ρ nπc
cn
2
sin2 nπct
−εn
0
sin2 nπx
dx
= ρc2π2
4 (ncn)2sin2 nπct
−εn
= 1
4ρωn2c2nsin2(ωnt−εn), (1.6.32) whereωn= nπc .
Similarly, Vn = 1
2T∗ nπcn
2
cos2 nπct
−εn 0
cos2 nπx
dx
= π2T∗
4 (ncn)2cos2 nπct
−εn
= 1
4ρωn2c2ncos2(ωnt−εn). (1.6.33) Thus, the total energy of the nth normal modes of vibrations is given by
En=Kn+Vn= 1
4ρ(ωncn)2=const. (1.6.34) For a given string oscillating in a normal mode, the total energy is proportional to the square of the circular frequency and to the square of the amplitude.
Finally, the total energy of the system is given by E=
∞ n=1
En=1 4ρ
∞ n=1
ω2nc2n, (1.6.35) which is constant becauseEn=const.
Example 1.6.2 (One-Dimensional Diffusion Equation). The temperature distribution u(x, t)in a homogeneous rod of lengthsatisfies the diffusion equation
ut=κuxx, 0< x < , t >0, (1.6.36) with the boundary and initial conditions
u(0, t) = 0 =u(, t), t≥0, (1.6.37ab)
u(x,0) =f(x), 0≤x≤, (1.6.38)
whereκis a diffusivity constant.
We assume a separable solution of (1.6.36) in the form
u(x, t) =X(x)T(t)= 0. (1.6.39) Substituting (1.6.39) in (1.6.36) gives
1 X
d2X dx2 = 1
κT dT
dt. (1.6.40)
Since the left-hand side depends only onxand the right-hand side is a function of timetonly, result (1.6.40) can be true only if both sides are equal to the same constantλ. Thus, we obtain two ordinary differential equations
d2X
dx2 −λX= 0, dT
dt −λκT = 0. (1.6.41ab)
Forλ≥0, the only solution of the form (1.6.39) consistent with the given boundary conditions isu(x, t)≡0. Hence, for negativeλ=−α2,
d2X
dx2 +α2X = 0, dT
dt +κα2T = 0, (1.6.42ab) which admit solutions as
X(x) =Acosαx+Bsinαx (1.6.43) and
T(t) =Cexp
−κα2t
, (1.6.44)
whereA,B, andCare constants of integration.
The boundary conditions forX(x)are
X(0) = 0 =X(), (1.6.45)
which are used to find AandB in solution (1.6.43). It turns out thatA = 0and B = 0. Hence,
sinα= 0, (1.6.46)
which gives the eigenvalues
α=αn =nπ
, n= 1,2,3, . . . . (1.6.47) The valuen= 0is excluded because it leads to a trivial solution. Thus, the eigen- functions are given by
Xn(x) =Bnsin nπx
, (1.6.48)
whereBnare nonzero constants.
Withα=αn= nπ , we combine (1.6.44) with (1.6.48) to obtain the solution for un(x, t)as
un(x, t) =anexp − nπ
2
κt
sin nπx
, (1.6.49)
wherean = BnCnis a new constant. Thus, (1.6.47) and (1.6.49) constitute an in- finite set of eigenvalues and eigenfunctions. Thus, the most general solution is ob- tained by the principle of superposition in the form
u(x, t) = ∞ n=1
anexp − nπ
2
κt
sin nπx
. (1.6.50)
Now, the initial condition implies that f(x) =
∞ n=1
ansin nπx
, (1.6.51)
which determinesan, in view of (1.6.24), as an= 2
0
f(x) sin nπx
dx. (1.6.52)
Thus, the final form of the solution is given by u(x, t) =
∞ n=1
2
0
f x
sin nπx
dx
exp − nπ
2
κt
sin nπx
. (1.6.53)
It follows from the series solution (1.6.53) that the series satisfies the given boundary and initial conditions. It also satisfies equation (1.6.36) because the series is conver- gent for allx(0 ≤x≤)andt ≥0and can be differentiated term by term. Phys- ically, the temperature distribution decays exponentially with timet. This shows a striking contrast to the wave equation, whose solution oscillates in timet. The time scale of decay for the nth mode isTd∼ 1κ(nπ )2which is directly proportional to2 and inversely proportional to the thermal diffusivity.
The method of separation of variables is applicable to the wave equation and the diffusion equation, and also to problems involving Laplace’s equation and other equations in two or three dimensions with a wide variety of initial and boundary conditions. We consider the following examples.
Example 1.6.3 (Two-Dimensional Diffusion Equation). We consider
ut=κ(uxx+uyy), 0< x < a, 0< y < b, t >0, (1.6.54)
u(x, y, t) =f(x, y) att= 0, (1.6.55)
u(x, y, t) = 0 on∂D, (1.6.56)
where∂Dis the boundary of the rectangle defined by0≤x≤a,0≤y≤b.
The method here is precisely the same as in the previous examples except that we seek a solution of (1.6.54) in the form
u(x, y, z) =S(x, y)T(t)= 0, (1.6.57) so thatSandTsatisfy the equations
∂2S
∂x2 +∂2S
∂y2 +λS= 0, (1.6.58)
∂T
∂t +κλT = 0. (1.6.59)
Forλ≤0, the separable solution (1.6.57) with the given boundary data leads only to a trivial solutionu(x, y, t)≡0. Hence, for positiveλ, we solve (1.6.58), (1.6.59) subject to the given boundary and initial conditions. Equation (1.6.58) is an elliptic equation, and here we seek a solutionS(x, y)which satisfies the boundary conditions S(0, y) = 0 =S(a, y) for0≤y≤b, (1.6.60) S(x,0) = 0 =S(x, b) for0≤x≤a. (1.6.61) We also seek a separable solution of (1.6.58) in the form
S(x, y) =X(x)Y(y)= 0 (1.6.62) and find thatX(x)andY(y)satisfy the equation
X X +Y
Y +λ= 0,
that is,
X
X =−μ=− Y
Y +λ
. (1.6.63)
Or
X+μX = 0, (1.6.64a)
Y+ (λ−μ)Y = 0. (1.6.64b)
These equations have to be solved with the boundary conditions
X(0) = 0 =X(a), (1.6.65a)
Y(0) = 0 =Y(b). (1.6.65b) The eigenvalue problem (1.6.64a), (1.6.64b) with (1.6.65a), (1.6.65b) gives the eigenvalues
μm= mπ
a 2
, (1.6.66)
and the corresponding eigenfunctions Xm(x) =Amsin
mπx a
, (1.6.67)
whenm= 1,2,3, . . .. Thus, equation (1.6.64b) becomes
Y+ (λ−μm)Y = 0, (1.6.68)
which has to be solved with (1.6.65b). This is another eigenvalue problem similar to that already considered and leads to the eigenvalues
λn−μm= nπ
b 2
(1.6.69) and the corresponding eigenfunctions
Yn(y) =Bnsin nπy
b
, (1.6.70)
wheren= 1,2,3, . . .. In other words, the solution of equation (1.6.58) becomes Smn(x, y) =Xm(x)Yn(y) =Amnsin
mπx a
sin
nπy b
, (1.6.71) whereAmn=AmBnare constants together with the eigenvalues
λmn=μm+ nπ
b 2
= m2
a2 +n2 b2
π2, (1.6.72)
wherem= 1,2,3, . . .andn= 1,2,3, . . ..
Withλmnas eigenvalues, we solve (1.6.59) to obtain
Tmn(t) =Bmnexp(−λmnκt), (1.6.73) whereBmnare integrating constants.
Finally, the solution (1.6.57) can be expressed as a double series u(x, y, t) =
∞ m=1
∞ n=1
amnsin mπx
a
sin nπy
b
exp(−λmnκt), (1.6.74) whereamnare constants to be determined from the initial condition so that
f(x, y) =u(x, y,0) = ∞ m=1
∞ n=1
amnsin mπx
a
sin nπy
b
. (1.6.75) To find constantsamn, we multiply (1.6.75) bysin(rπxa )and integrate the result with respect toxfrom0toawith fixedy, so that
a 2
∞ n=1
amnsin nπy
b
= a
0
f(x, y) sin mπx
a
dx. (1.6.76) The right-hand side is a function ofyand is set equal tog(y), so that
a 0
f(x, y) sin mπx
a
dx=g(y). (1.6.77)
Then, the coefficients amn (m fixed) in (1.6.76) are found by multiplying it by sin(nπyb )and integrating with respect toyfrom0tob, so that
ab 4
amn=
b 0
g(y) sin nπy
b
dy, (1.6.78)
whence
amn= 4
ab
a 0
b 0
f(x, y) sin mπx
a
sin nπy
b
dx dy. (1.6.79) Thus, the solution of the problem is given by (1.6.74) where amn is represented by (1.6.79). The method of construction of the solution shows that the initial and boundary conditions are satisfied by the solution. Moreover, the uniform convergence of the double series justifies differentiation of the series, and this, in turn, permits us to verify the solution by direct substitution in the original diffusion equation (1.6.54).
Example 1.6.4 (Dirichlet’s Problem for a Circle). We consider the Laplace equation in cylindrical polar coordinates(r, θ, z)as
urr+1 rur+ 1
r2uθθ= 0, 0< r < a, 0≤θ <2π, (1.6.80)
with the boundary condition
u(a, θ) =f(θ) for allθ. (1.6.81) According to the method of separation of variables, we seek a solution in the form
u(r, θ) =R(r)Θ(θ)= 0. (1.6.82) Substituting this solution in equation (1.6.80) gives
r2R R +rR
R =−Θ Θ =λ.
Hence,
r2R+rR−λR= 0 and (1.6.83a)
Θ+λΘ= 0. (1.6.83b)
ForΘ(θ), we naturally require periodic boundary conditions
Θ(θ+ 2π) =Θ(θ) for − ∞< z <∞. (1.6.84) Due to the periodicity condition, forλ < 0, the solution (1.6.82) leads to a trivial solution. So, there are two cases: (i)λ= 0and (ii)λ >0.
For case (i), we have the solution
u(r, θ) = (A+Blogr)(Cθ+D). (1.6.85) Sincelogris singular atr= 0, hence,B = 0. Foruto be periodic with period 2π,C= 0. Hence, the solutionumust be constant forλ= 0.
Forλ >0, the solution of equation (1.6.83b) is Θ(θ) =Acos√
λθ+Bsin√
λθ. (1.6.86)
Since Θ(θ)is periodic with period2π,√
λmust be an integernso thatλ = n2, n= 1,2,3, . . .. Thus, solution (1.6.86) becomes
Θ(θ) =Acosnθ+Bsinnθ. (1.6.87) The equation (1.6.83a) is the Euler equation withλ= n2. It gives solutions of the formR(r) =rα= 0so that (1.6.83a) gives
α(α−1) +α−n2 rα= 0, whenceα=±n. Thus, the solution forR(r)is given by
R(r) =Crn+Dr−n. (1.6.88)
Since R(r) → ∞as r → 0 because of the termr−n, we getD = 0. Thus, the solution (1.6.82) reduces to
u(r, θ) =Crn(Acosnθ+Bsinnθ). (1.6.89) By the superposition principle, the solution of the Laplace equation within a circular region including the originr= 0is
u(r, θ) =1 2a0+
∞ n=1
rn(ancosnθ+bnsinnθ), (1.6.90) wherea0,an, andbnare constants to be determined from the boundary conditions, and the first term12a0represents the constant solution forλ= 0 (n= 0).
Finally, using the boundary condition (1.6.81), we derive f(θ) =u(a, θ) = 1
2a0+ ∞ n=1
an(ancosnθ+bnsinnθ). (1.6.91) This is exactly the Fourier series representation forf(θ), and hence, the coefficients are given by
an = 1 πan
2π 0
f(φ) cosnφ dφ, n= 0,1,2,3, . . . , bn = 1
πan 2π
0
f(φ) sinnφ dφ, n= 1,2,3, . . . . Substituting the values foranandbninto (1.6.91) yields the solution
u(r, θ) = 1 2π
2π 0
f(φ)dφ +1
π ∞ n=1
r a
n 2π 0
(cosnθcosnφ+ sinnθsinnφ)f(φ)dφ
= 1 2π
2π 0
f(φ)
1 + 2 ∞ n=1
r a
n
cosn(θ−φ)
dφ, (1.6.92)
where the term inside the set of braces in the above integral can be summed by writing it as a geometric series, that is,
1 + ∞ n=1
r a
n
exp
in(θ−φ) +
∞ n=1
r a
n
exp
−in(θ−φ)
= 1 + rexp{i(θ−φ)}
a−rexp{i(θ−φ)}+ rexp{−i(θ−φ)} a−rexp{−i(θ−φ)}
= (a2−r2)
a2−2arcos(θ−φ) +r2. Thus, the final form of the solution is
u(r, θ) = 1 2π
2π 0
(a2−r2)f(φ)dφ
a2−2arcos(θ−φ) +r2. (1.6.93) This formula is known as Poisson’s integral formula representing the solution of the Laplace equation within the circle of radiusain terms of values prescribed on the circle. It has several important consequences. First, we set r = 0 andθ = 0 in (1.6.93) to obtain
u(0,0) = 1 2π
2π 0
f(φ)dφ. (1.6.94)
This states that the value of uat the center is equal to the mean value ofuon the boundary of the circle. This is called the mean value property.
We rewrite (1.6.93) in the form u(r, θ) =
2π 0
P(r, θ−φ)f(φ)dφ, (1.6.95) whereP(r, θ−φ)is called the Poisson kernel given by
P(r, θ−φ) = 1
2π· (a2−r2)
a2−2arcos(θ−φ) +r2, (1.6.96) which is zero forr=abutθ=φ. Further,
f(θ) = lim
r→a−u(r, θ) = 2π
0
rlim→a−P(r, θ−φ)
f(φ)dφ, which implies that
rlim→a−P(r, θ−φ) =δ(θ−φ), (1.6.97) whereδ(x)is the Dirac delta function.