7 Separation of Variables

7 Separation of Variables Chapter 5, An Introduction to Partial Differential Equations, Pichover and Rubinstein In this section we introduce the tec...
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7

Separation of Variables

Chapter 5, An Introduction to Partial Differential Equations, Pichover and Rubinstein In this section we introduce the technique, called the method of separations of variables, for solving initial boundary value-problems.

7.1

Heat Equation

We consider the heat equation satisfying the initial conditions ( ut = kuxx , x ∈ [0, L], t > 0 u(x, 0) = φ(x), x ∈ [0, L]

(7.1)

We seek a solution u satisfying certain boundary conditions. The boundary conditions could be as follows: (a) Dirichlet

u(0, t) = u(L, t) = 0.

(b) Neumann

ux (0, t) = ux (L, t).

(c) Periodic

u(−L, t) = u(L, t) and ux (−L, t) = ux (L, t).

We look for solutions of the form u(x, t) = X(x)T (t) where X and T are function which have to be determined. Substituting u(x, t) = X(x)T (t) into the equation, we obtain X(x)T ′ (t) = kX ′′ (x)T (t) from which, after dividing by kX(x)T (t), we get T ′ (t) X ′′ (x) = . kT (t) X(x) The left side depends only on t whereas the right hand side depends only on x. Since they are equal, they must be equal to some constant −λ. Thus T ′ + λkT = 0

(7.2)

X ′′ + λX = 0.

(7.3)

and

53

The general solution of the first equation is given T (t) = Be−λkt for an arbitrary constant B. The general solutions of the second equation are as follows. √ √ (1) If λ < 0, then X(x) = α cosh −λx + β sinh −λx. (2)

(3)

If λ = 0, then X(x) = αx + β. √ √ If λ > 0, then X(x) = α cos λx + β sin λx.

(7.4)

In addition, the function X which solves the second equation will satisfy boundary conditions depending on the boundary condition imposed on u. The problem ( X ′′ + λX = 0 (7.5) X satisfies boundary conditions is called the eigenvalue problem, a nontrivial solution is called an eigenfunction associated with the eigenvalue λ. 7.1.1

Heat equation with Dirichlet boundary conditions

We consider (7.1) with the Dirichlet condition u(0, t) = u(L, t) = 0

for all t ≥ 0.

In this case the eigenvalue problem (7.5) becomes ( X ′′ + λX = 0 X(0) = X(L) = 0.

(7.6)

We have to find nontrivial solutions X of the eigenvalue problem (7.6). If λ = 0, then X(x) = αx + β and 0 = X(0) = α · 0 + β implies that β = 0 and 0 = X(L) = αL implies that α = 0. If λ < 0. Then 0 = X(0) = α cosh 0 + β sinh 0 = α and 0 = X(L) = β sinh L shows that also β = 0. We conclude that λ ≤ 0 is not an eigenvalue √ of the problem (7.5). Finally, √ consider 0 = X(L) = β sin λL. Since λ > 0.Then 0 = X(0) = α cos λ · 0 = α and √ X is nontrivial solution, β 6= 0 and hence sin λL = 0. Consequently, λ=

 nπ 2 L

,

54

n≥1

and the corresponding eigenfunction is given by nπx Xn (x) = sin x L After substituting λ = (nπ/L)2 to (7.2), we get the family of solutions nπ 2

Tn (t) = Bn e−k( L ) t . Thus we have obtained the following sequence of solutions nπx −k( nπ )2 t L un (x, t) = Xn (x)Tn (x) = Bn sin e . L We obtain more solutions by taking linear combinations of the un ’s ( recall the superposition principle) u(x, t) =

N X

un (x, t) =

N X

Bn sin

n=1

n=1

nπx −k( nπ )2 t L e , L

and then by passing to the limit N → ∞, u(x, t) =

∞ X

Bn sin

n=1

nπx −k( nπ )2 t L e . L

(7.7)

Finally, we consider the initial condition. At t = 0, we must have u(x, 0) =

∞ X

Bn sin

n=1

nπx = φ(x). L

(7.8)

The coefficients, Bn can be computed as follows. Fix m ∈ N. Multiplying the above equality by sin mπx L and then integrating over [0, L], we get Z L Z LX ∞ mπx nπx mπx φ(x) sin Bn sin dx = sin dx L L L 0 0 n=1 Z L ∞ X mπx nπx sin dx. Bn sin = L L 0 n=1 RL P (above we ignore the question about interchanging ∞ n=1 with 0 .) Since ( Z L 0 n 6= m nπx mπx sin sin dx = L (7.9) L L n = m, 0 2 we find that Bm =

2 L

Z

L

φ(x) sin 0

55

mπx dx. L

(7.10)

Example 7.1. Consider the problem, ut − uxx = 0,

0 < x < π, t > 0

u(0, t) = u(π, t) = 0, t ≥ 0 ( x 0 ≤ x ≤ π/2 u(x, 0) = φ(x) = π − x π/2 ≤ x ≤ π.

(7.11)

Here k = 1 and [0, L] = [0, π]. In view of (7.8), the solution u(x, t) is given by ∞ X 2 Bn (sin nx)e−n t . u(x, t) = n=1

Using (7.10), the coefficients Bn are equal to Z 2 π Bn = φ(x) sin nx dx π 0 Z Z 2 π 2 π/2 x sin nx dx + (π − x) sin nx dx. = π 0 π π/2

Integrating by parts we find that the right-hand side is equal to     4 nπ 2 −x cos nx sin x π/2 2 −(π − x) cos nx sin x π + = + 2 − 2 sin . 2 π n n π n n πn 2 0 π/2 Since nπ sin = 2

( 0 (−1)k+1

n = 2k n = 2k − 1

for k ≥ 1, we get u(x, t) =

∞ X

Bn (sin nx)e−n

n=1 ∞ X

4 = π

n=1

2t

(−1)n+1 2 sin[(2n − 1)x]e−(2n−1) t . 2 (2n − 1)

(7.12)

We claim that u is indeed the classical solution of the equation (7.11). To see this, denote by un (x, t) =

(−1)n+1 2 sin[(2n − 1)x]e−(2n−1) t (2n − 1)2 56

so that u(x, t) =

4 π

P

n un (x, t). Since (−1)n+1 1 −(2n−1)2 t ≤ sin[(2n − 1)x]e |un (x, t)| = 2 (2n − 1) (2n − 1)2 P 1 and the series n (2n−1) 2 converges. Hence, in view of the Weierstrass Mtest (see Review), the series on the right hand side converges uniformly to a continuous function u in the region D = {(x, t)| 0 ≤ x ≤ π, t ≥ 0}. It remain to show that u is differentiable with respect to t and twice differentiable with respect to x, and satisfies the heat equation. Take ε > 0, and let

Dε = {(x, t)| 0 < x < π,

t > ε}.

Differentiating un with respect to t we get (2n − 1)2 2 t −(2n−1) ≤ e−(2n−1)2 ε . |un (x, t)t | = sin[(2n − 1)x]e (2n − 1)2 P 2 The series n e−(2n−1) ε converges (apply the ratio test). Hence applying the Weierstrass M-test again we find that the series of un (x, t)t ’ converges uniformly to a continuous function on Dε . Similarly, differentiating with respect to x twice we get (2n − 1)2 2 t −(2n−1) ≤ e−(2n−1)2 ε |un (x, t)xx | = sin[(2n − 1)x]e 2 (2n − 1) P and again n un (x, t)xx converges uniformly on Dε . Hence, using that (un )t − (un )xx = 0, X X X ut − uxx = (un )t − (un )xx = [(un )t − (un )xx ] = 0. n

n

n

So, u is a solution on Dε and since ε was arbitrary, u is a classical solution on D. 7.1.2

Heat equation with Neumann boundary conditions

We consider the heat equation (7.1) but with Neumann boundary conditions ux (0, t) = ux (0, t) = 0

for all t ≥ 0.

In this case the eigenvalue problem (7.5) becomes ( X ′′ + λX = 0 X ′ (0) = X ′ (L) = 0 57

(7.13)

As before the problem doesn’t have negative eigenvalues. If λ = 0, the general solution is X(x) = αx + β so that 0 = X ′ (0) = β, implies that λ0 = 0 is an eigenvalue with the unique (up to multiplication by a constant) eigenfunction X√ λ > 0, then the general solution of the problem 0 (x) ≡ 1. If√ is X(x) = α cos λx+β sin λx form which we conclude that 0 = X ′ (0) = β √ √ and 0 = X ′ (L) = − λα sin λL implies that λ > 0 is an eigenvalue if and only if  nπ 2 , n≥1 λ = λn = L and the corresponding eigenfunction Xn (x) is given by Xn (x) = cos

nπx . L

Then the corresponding solutions of T ′ + λkT = 0 are T0 (t) = B0 nπ 2

Tn (t) = Bn e−k( L ) t ,

n ≥ 1.

Thus we obtain a sequence of solutions −k

un (x, t) = Bn e

„ nπ «2

L

t

cos

nπx , L

n≥0

which we combine to form the series X nπ 2 nπ u(x, t) = Bn e−k( L ) t cos x. L n≥0

At t = 0, we have φ(x) = u(x, 0) =

X

Bn cos

n≥0

nπx . L

To compute Bm ’s, multiply this equality by cos mπx L and integrate over [0, L]. Then Z L Z L X nπx mπx mπx cos φ(x) cos dx = Bn cos dx. L L L 0 0 n≥0

Since Z

0

L

mπx dx = cos L 58

(

L 0

m=0 m ≥ 1,

(7.14)

and Z

L 0

nπx mπx cos cos dx = L L

it follows that Z 1 L φ(x) dx B0 = L 0 7.1.3

and

Bm

2 = L

Z

(

0 L 2

n 6= m n = m,

L

φ(x) cos 0

(7.15)

mπx dx, m ≥ 1. L

Heat equation with periodic boundary conditions

Next we consider (7.1) with the periodic boundary conditions u(−L, t) = u(L, t)

and ux (−L, t) = ux (L, t)

for all t ≥ 0.

In this case the eigenvalue problem takes the form ( X ′′ + λX = 0 X(0) = X(L), X ′ (0) = X ′ (L)

(7.16)

This follows from X(−L)T (t) = X(L)T (t) and X ′ (−L)T (t) = X(L)T (t). To find eigenvalues, we first consider λ < 0. From (7.4) and since sinh is odd and cosh is even, the condition X(−L) = X(L) implies that √ β sinh −λL = 0 so that β = 0. The condition X ′ (−L) = X ′ (L) implies that √ α sinh −λL = 0 so that α = 0. If λ = 0, then X(−L) = −αL + β = αL + β = X(L) so that α = 0. So λ0 = 0 is an eigenvalue with the corresponding eigenfunction X ≡ 1. Finally, let λ > 0. Then X(−L) = X(L) gives either√β = 0 or √0 (x) nπ λ = L and the condition X ′ (−L) = X ′ (L) gives either α = 0 or λ = nπ L . Hence the positive eigenvalues are λn =

 nπ 2 L

,

n≥1

and the corresponding eigenfunctions are  nπx   nπx  + Cn sin . Xn (x) = Bn cos L L 59

Thus the product solutions of the periodic boundary problem are u0 (x, t) = A0   nπx   nπx  nπ 2 un (x, t) = Bn cos + Cn sin e−( L ) t L L

which can be combined to form the series  nπx   nπx  X nπ 2 + Cn sin e−( L ) t . u(x, t) = A0 + Bn cos L L n≥1

At t = 0, we have φ(x) = u(x, 0) = A0 +

 nπx   nπx  X + Cn sin . Bn cos L L

n≥1

Integrating over [−L, L], we get 1 A0 = 2L

Z

L

φ(x) −L

 RL  RL nπ dx = sin dx = 0. Next multiplying both sides since −L cos nπx −L L  L by cos mπx and integrating over [−L, L] leads to L Bm since

L

Z

cos

−L L

Z

sin

−L

1 = L

 nπx  L

 nπx  L

L

Z

φ(x) cos

−L

cos cos

 mπx  L

 mπx  L

 mπx 

dx =

L

(

0 L

dx = 0,

dx

n 6= m n=m all n, m ≥ 1.

 Finally, multiplying both sides by sin mπx and integrating over [−L, L] L leads to Z  mπx  1 L dx Cm = φ(x) sin L −L L since

Z

L

−L

sin

 nπx  L

sin

 mπx  L

60

( 0 dx = L

n 6= m n = m.

7.2

Separation of variable for the wave equation

The above techniques can be used to solve the wave equation. We focus on the wave equation satisfying Dirichlet boundary condition (Neumann boundary conditions are treated in the book). We consider the initial boundary problem utt − c2 uxx = 0, 0 < x < L, t > 0 u(0, t) = u(L, t) = 0,

u(x, 0) = φ,

t≥0

0 0    v (0, t) = 0, v (L, t) = 0 t≥0 x x  v(x, 0) = 0 x ∈ [0, L].    v (x, 0) = 0 x ∈ [0, L]. t Define the energy

1 E(t) = 2

Z

L

vt (x, t)2 + c2 vx (x, t)2

0



dx.

Differentiating with respect to t, we get Z L  ′ E (t) = vt (x, t)vtt (x, t) + c2 vx (x, t)vxt (x, t) dx.

(7.21)

0

Note that

c2 vx vxt = c2 [(vx vt )x − vxx vt ] = c2 (vx vt )x − vtt vt where we have used vtt − c2 vxx = 0. Substituting this into (7.21) and using the boundary conditions, we get Z L ∂ (vx (x, t)v(x, t))dx = c2 [vx (x, t)vt (x, t)]L E ′ (t) = c2 0 = 0. ∂x 0 Hence E(t) = constant for all t. At t − 0, we have v(x, 0) = 0 so that vx (x, 0) = 0. Since vt (x, 0) = 0, we conclude that E(0) = 0 and so E(t) = 0 for all t ≥ 0. This implies that vt (x, t)2 + c2 vx (x, t)2 = 0 for all x ∈ [0, T ] and t ≥ 0. Thus, vt (x, t) = vx (x, t) = 0 for all x ∈ [0, T ] and t ≥ 0. Consequently, v ≡ constant. But then v(x, 0) = 0, implies that v ≡ 0. 

Review–Infinite series of functions P Formally, an infinite series n≥1 an is a pair (an ) P consisting of a sequence P and a sequence (sn ) of partial sums, sn = nk=1 ak . The series an converges, provided that a sequence (sn ) converges, that is, there is s such that 64

P lim sn = s. The number s is called the an and is deP∞ P value of the series noted by s = a . The series a is said to converge absolutely, if n n n=1 P P P |an | converges. If |a | converges, then a converges. However, the n n P P convergence of an does notP guarantee convergence |an |. For P examP P of ple, (−1)n /n converges but 1/n diverges. If a converges but |an | n P diverges, then series an is said to converge conditionally. Let (fn ) be a P sequence of functions defined on some interval [a, b].PWe can form a series fn (x) for every x ∈ [a, b]. One fn P says that the series converges pointwise on [a, b] if for the series fn (x) converges for every point x ∈ [a, b]. That is, for each x ∈ [a, b], n X as n → ∞. fk (x) → 0 f (x) − k=1

If

n X max f (x) − fk (x) → 0 as n → ∞, x∈[a,b] k=1 P then, the series fn is said to converge uniformly on [a, b].

Theorem 7.5 (Weierstrass M-test). Let (fn ) be a sequence of functions on [a, b] and let (Mn ) be a sequence of positive constants satisfying: (a) |fn (x)| ≤ Mn for all x ∈ [a, b] and all n, P (b) the series Mn converges. P The the series fn converges absolutely and uniformly on [a, b].

Theorem 7.6. Let (fn ) be a sequence Pof functions on [a, b] converging pointwise to the function f , i.e., f (x) = ∞ n=1 fn (x). P (a) If the fn ’s are continuous and the series fn converges uniformly on [a, b], then its sum f is a continuous function on [a, b], and Z b Z bX ∞ Z b ∞ X f (x) dx. fn (x) dx = fn (x) dx = n=1 a

a

a n=1

P ′ (b) If the fn ’s are of class C 1 and the series fn of derivatives converges uniformly on [a, b], then its sum f is of class C 1 and !′ ∞ X X ′ f (x) = fn (x) = fn′ (x). n=1

65

n=1