Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers

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10.10 The Schr¨odinger Equation and the Hydrogen Atom 391

(x)=0, for x ≥ a. (10.10.44)

From a physical point of view, the solution of the Schr¨odinger equation

must b e continuous everywhere including at the boundaries. Thus, match-

ing of solutions at x =+

a is required so that

(a)=C sin ak + D cos ak =0=ψ

(a) , (10.10.45)

(−a)=−C sin ak + D cos ak =0=ψ

(−a) . (10.10.46)

This system of linear homogeneous equations h as nontrivial solutions

for C and D only if the determinant of the coeﬃcient matrix vanishes. This

means that

sin ak cos ak =0. (10.10.47)

There are two possible nontrivial solutions for the set of conditions

(10.10.47).

Case 1. Even solution: cos ak =0.

In this case, it follows from (10.10.45)–(10.10.46) that C = 0. Hence,

ak =(2n +1)

,nis an integer,

or,

= k

(2n +1)

. (10.10.48)

Consequently, (10.10.42) gives the energy levels E = E

(2n +1)



8Ma

. (10.10.49)

In this case, the nontrivial solution in region 2 takes t he form

(x)=D cos kx, −a ≤ x ≤ a. (10.10.50)

Case 2. Odd solution: sin ak =0.

It follows from (10.10.45)–(10.10.46) that D =0andsinak =0holds.

Consequently,

ak = nπ, n is an integer, n =0,

or,



nπ



. (10.10.51)

392 10 Higher-Dimensional Boundary-Value Problems

Thus, the energy levels are given by



(nπ)

2Ma

. (10.10.52)

The nontrivial solution in region 2 is

(x)=C sin kx, −a ≤ x ≤ a. (10.10.53)

Thus, it tur ns out that, corresponding to every value of E

given by

(10.10.49) or (10.10.52), there exists a physically acceptable solution. Hence,

the general solution of the Schr¨odinger equation is obtained from (10.10.7)

in the form

ψ (x, t)=



(x)exp



−

itE





, (10.10.54)

where C

are constants.

In classical mechanics, the motion of the particle is al lowed for E>0.

In quantum mechanics, it follows from (10.10.49) or (10.10.52) that particle

motion is allowed for discrete values of energy, that is, the energy for this

system is quantized. This is a remarkable contrast between the results of

the classical mechanics and quantum mechanics.

Finally, it follows from the above analysis is that

lim

|x|→∞

ψ (x)=0. (10.10.55)

Such a system, where the wave function vanishes beyond range or asymp-

totically, is called a bound state, and energy is quantized. A very common

example is the hydrogen atom which was discussed in this section. In the

present system ψ (x)=0forx

≥ a

. Therefore, this system is also referred

to as a particle in a box of length 2a. The probability for ﬁnding the particle

outside this region is zero.

10.11 Method of Eigenfunctions and Vibration of

Membrane

Consider the nonhomogeneous initial boundary-value problem

L [u]=ρu

− G in D (10.11.1)

with prescribed homogeneous boundary conditions on the boundary B of

D, and the initial conditions

u (x

,...,x

, 0) = f (x

,...,x

) , (10.11.2)

,...,x

, 0) = g (x

,...,x

) . (10.11.3)

10.11 Method of Eigenfunctions and Vibration of Membrane 393

Here ρ ≡ ρ (x

,...,x

) is a real-valued positive continuous function and

G ≡ G (x

,...,x

) is a real-valued continuous function.

We assume that the only solution of the associated homogeneous prob-

lem

L [u]=ρu

(10.11.4)

with the prescribed boundary conditi ons is th e trivi al solution. Then, if

there exists a solution of the given prob lem (10.11.1)–(10.11.3), it can be

represented by a series of eigenfunctions of the associated eigenvalue prob-

lem

L [ϕ]+λρϕ = 0 (10.11.5)

with ϕ satisfying the boundary conditions given for u. For problems with

one space variable, see Section 7.8.

As a speciﬁc example, we shall determine the solution of the problem of

forced vibration of a rectangular membrane of length a and width b.The

problem is

− c

∇

u = F (x, y, t)inD (10.11. 6)

u (x , y, 0) = f (x, y) , 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.11.7)

(x, y, 0) = g (x, y) , 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.11.8)

u (0,y,t)=0,u(a, y, t)=0, (10.11. 9)

u (x , 0,t)=0,u(x, b, t)=0. (10.11.10)

The associated eigenvalue problem is

∇

ϕ + λϕ =0 in D,

ϕ = 0 on the boundary B of D.

The eigenvalues for this problem according to Section 10.6 are given by





,m,n=1, 2, 3 ...

and the corresponding eigenfunctions are

(x, y)=sin



mπx



sin



nπy



Thus, we assume the solution

u (x , y, t)=

∞



m=1

∞



n=1

(t)sin



mπx



sin



nπy



and the forcing function

394 10 Higher-Dimensional Boundary-Value Problems

F (x, y, t)=

∞



m=1

∞



n=1

(t)sin



mπx



sin



nπy



Here F

(t) are given by

(t)=



F (x, y, t)sin



mπx



sin



nπy



dx dy.

Note that u automatically satisﬁes the homogeneous boundary conditions.

Now inserting u (x, y, t)andF (x, y, t) in equation (10.11.6), we obtain

¨u

+ c

= F

where α

=(mπ/a)

+(nπ/b)

.Wehaveassumedthatu is twice contin-

uously diﬀerentiable with respect to t. Thus, the solution of the preceding

ordinary diﬀerential equation takes the form

(t)=A

cos (α

ct)+B

sin (α

ct)

(α



(τ)sin[α

c (t − τ)] dτ.

The ﬁrst initial condition gives

u (x , y, 0) = f (x, y)=

∞



m=1

∞



n=1

sin



mπx



sin



nπy



Assuming that f (x, y) is continuous in x and y, the coeﬃcient A

of the

double Fourier series is given by



f (x, y)sin



mπx



sin



nπy



dx dy.

Similarly, from the remaining initial condition, we have

(x, y, 0) = g (x, y)=

∞



m=1

∞



n=1

(α

c)sin



mπx



sin



nπy



and hence, for continuous g (x, y),

(ab α



g (x, y)sin



mπx



sin



nπy



dx dy.

The solution of the given initial boundary-value problem is therefore given

u (x , y, t)=

∞



m=1

∞



n=1

(t)sin



mπx



sin



nπy



10.12 Time-Dependent Boundary-Value Problems 395

provided the series for u and i ts ﬁrst and second derivatives converge uni-

formly.

If F (x, y, t)=e

x+y

cos ωt,then

(t)=

4mnπ

+ a

)(n

+ b

)

1+(−1)

m+1

1+(−1)

n+1

cos ωt

= C

cos ωt.

Hence, we have

(t)=

(α



cos ωt sin [α

c (t − τ)] dτ

(α

− ω

)

(cos ωt − cos α

ct)

provided ω =(α

c). Thus, the solution may be written in the form

u (x , y, t)=

∞



m=1

∞



n=1

(α

− ω

)

(cos ωt − cos α

ct)

×sin



mπx



sin



nπy



10.12 Time-Dependent Boundary-Value Problems

The preceding chapters have been devoted to problems with homogeneous

boundary conditions. Due to the frequ ent occurrence of problems with time

dependent boundary condition s in practice, we consider the forced vibration

of a rectangular membrane with moving boundaries. The p roblem here is

to determine the displacement function u which satisﬁes

− c

∇

u = F (x, y, t) , 0 <x<a, 0 <y<b, (10.12.1)

u (x , y, 0) = f (x, y) , 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.12.2)

(x, y, 0) = g (x, y) , 0 ≤ x ≤ a, 0 ≤ y ≤ b, (10.12.3)

u (0,y,t)=p

(y, t) , 0 ≤ y ≤ b, t ≥ 0, (10.12.4)

u (a , y, t)=p

(y, t) , 0 ≤ y ≤ b, t ≥ 0, (10.12.5)

u (x , 0,t)=q

(x, t) , 0 ≤ x ≤ a, t ≥ 0, (10.12.6)

u (x , b, t)=q

(x, t) , 0 ≤ x ≤ a, t ≥ 0. (10.12.7)

For such problems, we seek a solution in the form

u (x , y, t)=U (x, y, t)+v (x, y, t) , (10.12.8)

396 10 Higher-Dimensional Boundary-Value Problems

where v is the new dependent variable to be determined. Before ﬁnding v,

we must ﬁrst determine U. If we substitute equation (10.12.8) into equations

(10.12.1)–(10.12.7), we respectively obtain

− c

+ v

)=F − U

+ c

+ U

F (x, y, t)

and

v (x, y, 0) = f (x, y) − U (x, y, 0) =

f (x, y) ,

(x, y, 0) = g (x, y) − U

(x, y, 0) = 5g (x, y) ,

v (0,y,t)=p

(y, t) − U (0,y,t)=5p

(y, t) ,

v (a, y, t)=p

(y, t) − U (a, y, t)=5p

(y, t) ,

v (x, 0,t)=q

(x, t) − U (x, 0,t)=5q

(x, t) ,

v (x, b, t)=q

(x, t) − U (x, b, t)=5q

(x, t) .

In order to make the conditions on v homogeneous, we set

= 5p

= 5q

=0,

so that

U (0,y,t)=p

(y, t) ,U(a, y, t)=p

(y, t) , (10.12.9)

U (x, 0,t)=q

(x, t) ,U(x, b, t)=q

(x, t) . (10.12.10)

In order that the boundary conditions be compatible, we assume that the

prescribed functions take th e forms

(y, t)=ϕ (y) p

∗

(y, t) ,p

(y, t)=ϕ (y) p

∗

(y, t) ,

(x, t)=ψ (x) q

∗

(x, t) ,q

(x, t)=ψ (x) q

∗

(x, t) ,

where the function ϕ must vanish at the end points y =0,y = b and the

function ψ must vanish at x =0,x = a.Thus,U (x, y, t) which satisﬁes

equations (10.12.9)–(10.12.10) takes the form

U (x, y, t)=ϕ (y)

∗

+ p

∗

)

+ ψ (x)

∗

+ q

∗

)

The problem then is to ﬁnd the fun ction v (x, y, t) which satisﬁes

− c

+ v

F (x, y, t) ,

v (x, y, 0) =

f (x, y) ,v

(x, y, 0) = 5g (x, y) ,

v (0,y,t)=0,v(a, y, t)=0,

v (x, 0,t)=0,v(x, b, t)=0.

This is an initial boundary-value problem with homogeneous boundary con-

ditions, which has already been solved.

10.12 Time-Dependent Boundary-Value Problems 397

As a particular case, consider the following problem

− c

+ u

)=0,

u (x , y, 0) = 0,u

(x, y, 0) =

sin



πx



u (0,y,t)=0,u(a, y, t)=0,

u (x , 0,t)=0,u(x, b, t)=sin



πx



sin t.

We assume a solution in the form

u (x , y, t)=v (x, y, t)+U (x, y, t) .

The function U (x, y, t) which satisﬁes

U (0,y,t)=0,U(a, y, t)=0,

U (x, 0,t)=0,U(x, b, t)=sin



πx



sin t

U (x, y, t)=sin



πx



sin t



Thus, the new p r oblem to be solved is

− c

+ v



1 −



sin



πx



sin t,

v (x, y, 0) = 0,v

(x, y, 0) = 0,

v (0,y,t)=0,v(a, y, t)=0,

v (x, 0,t)=0,v(x, b, t)=0.

Then, we ﬁnd F

from

(t)=



F (x, y, t)sin



mπx



sin



nπy



dx dy,

where

F (x, y, t)=



1 −



sin



πx



sin t,

and obtain

(t)=

2(−1)

n+1



1 −



sin t.

Now we determine v

(t)whicharegivenby

398 10 Higher-Dimensional Boundary-Value Problems

(t)=A

cos (α

ct)+B

sin (α

ct)



(τ)sin[α

c (t − τ)] dτ.

Since v (x, y, 0) = 0, A

= 0, but

ab α





−

sin

πx



sin



mπx



sin



nπy



dx dy

2(−1)

nac

Thus, we have

(t)=

2(−1)

nac

sin (α

ct)

2(−1)

n (1 − α

)



− c



(sin α

ct − αc sin t) .

The solution is therefore given by

u (x , y, t)=

sin



πx



sin t +

∞



m=1

∞



n=1

(t)sin



mπx



sin



nπy



10.13 Exercises

1. Solve the Dirichlet problem

∇

u =0, 0 <x<a, 0 <y<b, 0 <z<c,

u (0,y,z)=sin



πy



sin



πz



,u(a, y, z)=0,

u (x , 0,z)=0,u(x, b, z)=0,

u (x , y, 0) = 0,u(x, y, c)=0.

2. Solve the Neumann problem

∇

u =0, 0 <x<1, 0 <y<1, 0 <z<1,

(0,y,z)=0,u

(1,y,z)=0,

(x, 0,z)=0,u

(x, 1,z)=0,

(x, y, 0) = cos πx cos πy, u

(x, y, 1) = 0.

3. Solve the Robin boundary-value problem

∇

u =0, 0 <x<π, 0 <y<π, 0 <z<π,

u (0,y,z)=f (y, z ) ,u(π, y, z)=0,

(x, 0,z)=0,u

(x, π, z)=0,

(x, y, 0) + hu(x, y, 0)

(x, y, π)+hu(x, y, π)

⎫

⎬

⎭

h = constant.

10.13 Exercises 399

4. Determine the solution of each of the following problems for a cylinder:

(a) ∇

u =0,r<a,0 <θ<2π, 0 <z<l,

u (a , θ, z)=0,u(r, θ, l)=0,u(r, θ, 0) = f (r, θ) .

(b) ∇

u =0,r<a,0 <θ<2π, 0 <z<l,

u (a , θ, z)=f (θ, z ) ,u

(r, θ, 0) = 0,u

(r, θ, l)=0.

5. Find the solution of the Dirichlet problem for a sphere

∇

u =0,r<a,0 <θ<π, 0 <ϕ<2π,

u (a , θ, ϕ)=cos

θ.

6. Solve the Dirichlet problem in a region bounded by two concentric

spheres

∇

u =0, a<r<b, 0 <θ<π, 0 <φ<2π,

u (a , θ, φ)=f (θ, φ) ,u(b, θ, φ)=g (θ, φ) .

7. Find the steady-state temperature distribution in a cylinder of radiu s

a if a constant ﬂow of heat T is supplied at the end z = 0, and the

surface r = a and the end z = l are maintained at zero temperature.

8. Find the potential of the electrostatic ﬁeld inside a cylinder of length l

and radius a, if each end of the cylinder is grounded, and the surface is

charged to a potential u

9. Determine the p otential of the electric ﬁeld inside a sphere of radius

a, if the upper half of the sphere is charged to a potential u

and the

lower half to a potential u

10. Solve the Dirichlet problem for a half cylinder

∇

u =0,r<1, 0 <θ<π, 0 <z<1,

u (1,θ,z)=0,u(r, 0,z)=0,u(r, π, z)=0,

u (r, θ, 0) = 0,u(r, θ, 1) = f (r, θ) .

11. Solve the Neumann problem for a sph ere

400 10 Higher-Dimensional Boundary-Value Problems

∇

u =0,r<1, 0 <θ<π, 0 <ϕ<2π,

(1,θ,ϕ)=f (θ, ϕ) ,

where



2π



f (θ, ϕ)sinθdθdϕ=0.

12. Find the solution of the initial boundary-value problem

= c

∇

u, 0 <x<1, 0 <y<1,t>0,

u (x , y, 0) = sin

πx sin πy, u

(x, y, 0) = 0, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,

u (0,y,t)=0,u(1,y,t)=0, 0 ≤ y ≤ 1,t>0,

u (x , 0,t)=0,u(x, 1,t)=0, 0 ≤ x ≤ 1,t>0.

13. Obtain the solution of the problem

= c

∇

u, r < a, 0 <θ<2π, t > 0,

u (r, θ, 0) = f (r, θ) ,u

(r, θ, 0) = g (r, θ) ,u(a, θ, t)=0.

14. Determine the temperature distribution in a rectangular plate with ra-

diation from its surface. The temperature distribution is described by

= k (u

+ u

) − h (u − u

) , 0 <x<a, 0 <y<b, t>0,

u (x , y, 0) = f (x, y) ,

u (0,y,t)=0,u(a, y, t)=0,

u (x , 0,t)=0,u(x, b, t)=0,

where k, h and u

are constants.

15. Solve the heat conduction problem in a circular plate

= k



θθ



,r<1, 0 <θ<2π, t > 0,

u (r, θ, 0) = f (r, θ) ,u(1,θ,t)=0.

16. Solve the initial boundary-value problem

= c

∇

u, 0 <x<1, 0 <y<1, 0 <z<1,t>0,

u (x , y, z, 0) = sin πx sin πy sin πz ,