Powers D.L. Boundary Value Problems and Partial Differential Equations

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38 Chapter 0 Ordinary Differential Equations

(Here, E is Young’s modulus and I is the second moment of the cross sec-

tion.) Solve this problem if w(x) = w

, constant.

19. If the beam of Exercise 18 is built into a wall at the left end and is unsup-

ported at the right end, the boundary conditions become

u(0) = 0, u



(0) = 0, u



(a) =0, u



(a) =0.

Solve the same differential equation subject to these conditions.

0.4 Singular Boundary Value Problems

A boundary value problem can be singular in two different ways. In one case,

an endpoint of the interval of interest is a singular point of the differential

equation. In the other, the interval is inﬁnitely long.

Regular Singular Point

Recall that a point x

is a (regular) singular point of the differential equation



+k(x)u



+p(x )u =f (x)

if the products

(x − x

)k(x), (x − x

)

p(x)

both have Taylor series expansions centered at x

but either k(x) or p(x) or

both become inﬁnite as x → x

. For example, the point x

= 1 is a regular

singular point of the differential equation

(1 −x)u





+xu = 0.

In standard form, the equation is



1 −x



1 −x

u =0.

Since both

k(x) =

1 −x

and p(x) =

1 −x

become inﬁnite at x = 1, but (x − 1)k(x) and (x − 1)

p(x) both have Taylor

series expansions about the center x =1, the point x

=1 is a regular singular

point. Another convenient example is provided by the Cauchy–Euler equation

of Section 1, which has a regular singular point at the origin.

This situation typically arises when a boundary point is a mathematical

boundary without being a physical boundary. For instance, a circular disk of

0.4 Singular Boundary Value Problems 39

radius c may be described in polar (r,θ)coordinates as occupying the region

0 ≤r ≤ c. The origin, at r =0, is a mathematical boundary, yet physically this

point is in the interior of the disk.

At a singular point, one cannot specify a value for u(x

), the solution of

the differential equation, or for its derivative. However, it is usually necessary

to require that both u(x

) and u



) be ﬁnite, or bounded. Tacitly, we always

require that the solution and its derivative be ﬁnite at every point of the interval

where we are solving a differential equation. But when a singular point is a

boundary point of that interval, we enforce the condition explicitly. In the

example that follows we shall see how these conditions act so as to make the

solution of a boundary value problem unique.

Example: Radial Heat Flow.

Suppose a long cylindrical bar, surrounded by a medium at temperature T,

carries an electrical current. If heat ﬂows in the radial direction much faster

than in the axial direction, the temperature u(r) in the rod may be described

by the problem





=−H, 0 ≤r < c, (1)

u(c) =T. (2)

Here, c is the radius of the rod, r is a polar coordinate, and H (constant) is

proportional to the electrical power being converted into heat.

In this problem, only the physical boundary condition has been noted. The

mathematical boundary r =0 is a singular point, as is clear from the differen-

tial equation in the form

=−H.

Thus, at this point we will require that u and du/dr be ﬁnite:

u(0), u



(0) ﬁnite. (3)

Now the differential equation (1) is easy to solve. Multiply through by r and

integrate once to ﬁnd that

=−H

Divide through this equation by r and integrate once more to determine that

u(r) =−H

ln(r) +c

40 Chapter 0 Ordinary Differential Equations

Application of the special condition, that u(0) and u



(0) be ﬁnite, immediately

tellsusthatc

= 0; for both, ln(r) and its derivative 1/r become inﬁnite as r

approaches 0.

The physical boundary condition, Eq. (2), says that

u(c) =−H

=T.

Hence, c

=Hc

/4 +T, and the complete solution is

u(r) = H

−r

)

+T. (4)



From this example, it is clear that the “artiﬁcial” boundary condition,

boundedness of u(r) at the singular point r = 0, works just the way an or-

dinary boundary condition works at an ordinary (not singular) point. It gives

one condition to be fulﬁlled by the unknown constants c

and c

,whichare

then completely determined by the second boundary condition.

Semi-Inﬁnite and Inﬁnite Intervals

Another type of singular boundary value problem is one for which the inter-

val of interest is inﬁnite. (Of course, this is always a mathematical abstraction

that cannot be realized physically.) For instance, on the interval 0 < x < ∞,

sometimes called a semi-inﬁnite interval, as it does have one ﬁnite endpoint, a

boundary condition would normally be imposed at x = 0. At the other “end,”

no boundary condition is imposed, because no boundary exists. However, we

normally require that both u(x) and u



(x) remain bounded as x increases. In

precise terms, we require that there exist constants M and M



for which



u(x)



≤M and





(x)



≤M



are both satisﬁed for all x, no matter how large. We never identify M or M



and the entire condition is usually written

u(x) and u



(x) bounded as x →∞.

Example: Cooling Fin.

A long cooling ﬁn has one end held at a constant temperature T

and ex-

changesheatwithamediumattemperatureT through convection. The tem-

perature u(x) in the ﬁn satisﬁes the requirements

κA

(u −T), 0 < x, (5)

u(0) = T

(6)

0.4 Singular Boundary Value Problems 41

(see Section 3). As the problem has been posed for a semi-inﬁnite interval

(because the ﬁn is very long and, perhaps, to mask our ignorance of what is

happening at the other physical end), we must also impose the condition

u(x), u



(x) bounded as x →∞. (7)

Now, the general solution of the differential equation (5) is

u(x) = T +c

cosh(µx) +c

sinh(µx),

where µ =



hC/κA. The boundary condition at x =0requiresthat

u(0) = T

: T + c

The boundedness condition, Eq. (7), requires that

=−c

The reason for this is that of all the linear combinations of cosh and sinh, the

only one that is bounded as x →∞is

cosh(µx) −sinh(µx) = e

−µx

and its constant multiples. The ﬁnal solution is easily found to be

u(x) = T +(T

−T)



cosh(µx) −sinh(µx)





Satisfying the boundedness condition in the example would have been sim-

pler if we had expressed the general solution of the differential equation (5)

u(x) = T +c



µx



−µx

We would have seen immediately that choosing c



=0 is the only way to satisfy

the boundedness condition. We summarize the observation as a rule of thumb:

the solution of

−µ

u =0

on an interval I is best expressed as

u(x) =



cosh(µx) +c

sinh(µx), if I is ﬁnite,

µx

−µx

, if I is inﬁnite.

EXERCISES

1. Put each of the following equations in the form



+ku



+pu = f

42 Chapter 0 Ordinary Differential Equations

and identify the singular point(s).





=u;

dφ



sin(φ)

dφ



=sin(φ)u;





1 −x





=0;

dρ



dρ



=−λ

2. The temperature u inalargeobjecthavingaholeofradiusc in the middle

may be said to obey the equations





=0, r > c,

u(c) = T.

Solve the problem, adding the appropriate boundedness condition.

3. Compact kryptonite produces heat at a rate of H cal/s cm

.Ifasphere

(radius c) of this material transfers heat by convection to a surrounding

medium at temperature T,thetemperatureu(ρ) in the sphere satisﬁes the

boundary value problem

dρ



dρ



−H

, 0 <ρ<c,

−κ

dρ



u(c) −T



Supply the proper boundedness condition and solve. What is the tempera-

ture at the center of the sphere?

4. (Critical radius) The neutron ﬂux u in a sphere of uranium obeys the dif-

ferential equation

dρ



dρ



+(k −1) Au =0

in the range 0 <ρ<a,whereλ is the effective distance traveled by a neu-

tron between collisions, A is called the absorption cross section, and k is

the number of neutrons produced by a collision during ﬁssion. In addition,

the neutron ﬂux at the boundary of the sphere is 0. Make the substitution

u = v/ρ and 3(k − 1)A/λ = µ

, and determine the differential equation

satisﬁed by v(ρ). See Section 0.1, Exercise 19.

5. Solve the equation found in Exercise 4 and then ﬁnd u(ρ) that satisﬁes

the boundary value problem (with boundedness condition) stated in Ex-

ercise 4. For what radius a is the solution not identically 0?

0.5 Green’s Functions 43

6. Inside a nuclear fuel rod, heat is constantly produced by nuclear reaction.

Atypicalrodisabout3mlongandabout1cmindiameter,sotemperature

variationalongthelengthismuchlessthanalongaradius.Thus,wetreat

the temperature in such a rod as a function of the radial variable alone. Find

this temperature u(r), which is the solution of the boundary value problem





=−

, 0 < r < a,

u(a) =T

7. For the problem of Exercise 6, ﬁnd the temperature at the center of the

rod, u(0), using these values for the parameters: a = 0.5cm,thepower

density g = 418 W/cm

=100 cal/s cm

,conductivityκ =0.01 cal/s cm

◦

and the surface temperature T

=325

◦

8. A model for microwave heating of food uses this equation for the tempera-

ture u(x) in a large solid object:

=−Ae

−x/L

, 0 < x.

Here, A is a constant representing the strength of the radiation and prop-

erties of the object, and L is a characteristic length, known as penetration

depth, that depends on frequency of the radiation and properties of the ob-

ject. (Typically, L is about 12 cm in frozen raw beef or 2 cm thawed.) Show

that the boundary condition u



(0) = 0 is incompatible with the condition

that u(x) be bounded as x goes to inﬁnity. [See C.J. Coleman, The mi-

crowave heating of frozen substances, Applied Math. Modeling, 14 (1990):

439–443.]

9. Solve the differential equation in Exercise 8 subject to the conditions

u(0) =T

, u(x) bounded.

0.5 Green’s Functions

The most important features of the solution of the boundary value problem,

+k(x)

+p(x )u =f (x), l < x < r, (1)

αu(l) − α



(l) = 0, (2)

βu(r) + β



(r) = 0, (3)

The primes on the constants α



,β



are not to indicate differentiation, of course, but to

show that they are coefﬁcients of derivatives.

44 Chapter 0 Ordinary Differential Equations

can be developed by using the variation-of-parameters solution of the differ-

ential equation (1), as presented in Section 2. To begin, we need to have two

independent solutions of the homogeneous equation

+k(x)

+p(x )u =0, l < x < r. (4)

Let us designate these two solutions as u

(x) and u

(x). It will simplify algebra

later if we require that u

satisfy the boundary condition at x = l and u

the

condition at x = r;

αu

(l) − α



(l) = 0, (5)

βu

(r) +β



(r) = 0. (6)

According to Theorem 3 of Section 2, the general solution of the differential

equation (1) can be written as

u(x) = c

(x) +c

(x) +





(z)u

(x) −u

(z)u

(x)



f (z)

W(z)

dz. (7)

Recall that in the denominator of the integrand, we have the Wronskian of u

and u

W(z) =



(z) u

(z)



(z) u



(z)



, (8)

which is nonzero because u

and u

are independent. We will need to know

the following derivative of the function in Eq. (7):



(x) +c



(x) +





(z)u



(x) −u

(z)u



(x)



f (z)

W(z)

dz.

(See Leibniz’s rule in the Appendix.)

Now let us apply the boundary condition,Eq.(2),tothegeneralsolution

u(x).First,atx =l we have

αu(l) − α



(l) = c



αu

(l) − α



(l)





αu

(l) − α



(l)



=0. (9)

Note that the integrals in u and u



are both 0 at x =l. Because of the boundary

condition (5) imposed on u

,Eq.(9)reducesto



αu

(l) − α



(l)



=0, (10)

and we conclude that c

=0.

0.5 Green’s Functions 45

Second, the boundary condition at x = r becomes

βu(r) +β



(r) = c



βu

(r) +β



(r)







(z)



βu

(r) +β



(r)



−u

(z)



βu

(r) +β



(r)



f (z)

W(z)

dz =0. (11)

Now, the boundary condition (6) on u

at x = r eliminates one term of the

integrand, leaving



βu

(r) +β



(r)



−



(z)



βu

(r) +β



(r)



f (z)

W(z)

dz =0. (12)

The common factor of βu

(r) +β



(r) can be canceled from both terms, and

we then ﬁnd



(z)

f (z)

W(z)

dz. (13)

Now we have found c

and c

so that u(x) in Eq. (7) satisﬁes both boundary

conditions. If we use the values of c

and c

as found, we have

u(x) = u

(x)



(z)

f (z)

W(z)





(z)u

(x) −u

(z)u

(x)



f (z)

W(z)

dz. (14)

The solution becomes more compact if we break the interval of integration at

x in the ﬁrst integral, making it



(z)

f (z)

W(z)

dz =



(z)

f (z)

W(z)

dz +



(z)

f (z)

W(z)

dz. (15)

When the integrals on the range l to x are combined, there is some cancella-

tion, and our solution becomes

u(x) =



(z)u

(x)

f (z)

W(z)

dz +



(x)u

(z)

f (z)

W(z)

dz. (16)

Finally, these two integrals can be combined into one. We ﬁrst deﬁne the

Green’s function for the problem (1), (2), (3) as

G(x, z) =











(z)u

(x)

W(z)

, l < z ≤ x,

(x)u

(z)

W(z)

, x ≤ z < r.

(17)

46 Chapter 0 Ordinary Differential Equations

Then the formula given in Eq. (16) for u simpliﬁes to

u(x) =



G(x, z)f (z) dz. (18)

Example.

Solve the problem that follows by constructing the Green’s function.

−u =−1, 0 < x < 1,

u(0) =0, u(1) = 0.

First, we must ﬁnd two independent solutions of the homogeneous differential

equation u



− u = 0 that satisfy the boundary conditions as required. The

generalsolutionofthehomogeneous differential equation is

u(x) = c

cosh(x) +c

sinh(x).

As u

(x) is required to satisfy the condition at the left, u

(0) = 0, we take

= 0, c

= 1 and conclude u

(x) = sinh(x). The second solution is to sat-

isfy u

(1) = 0. We may take

(x) = sinh(1) cosh(x) −cosh(1) sinh(x) = sinh(1 −x).

The Wronskian of the two solutions is

W(x) =



sinh(x) sinh(1 −x)

cosh(x) −cosh(1 −x)



=−sinh(1).

Now, by Eq. (17), the Green’s function for this problem is

G(x, z) =











sinh(z) sinh(1 −x)

−sinh(1)

, 0 < z ≤ x,

sinh(x) sinh(1 −z)

−sinh(1)

, x ≤ z < 1.

Furthermore, since f (x) =−1, the solution, by Eq. (18), is the integral

u(x) =



−G(x, z) dz.

To actually carry out the integration, we must break the interval of integration

at x,thusrevertingineffecttoEq.(16).Theresult:

0.5 Green’s Functions 47

u(x) =



sinh(z) sinh(1 −x)

sinh(1)

dz +



sinh(x) sinh(1 −z)

sinh(1)

sinh(1 −x)

sinh(1)

cosh(z)



sinh(x)

sinh(1)



−cosh(1 − z)





sinh(1 −x)

sinh(1)



cosh(x) −1



sinh(x)

sinh(1)



cosh(1 − x) −1



sinh(1 −x) cosh(x) +sinh(x) cosh(1 −x)

sinh(1)

−

sinh(1 −x) +sinh(x)

sinh(1)

= 1 −

sinh(1 −x) +sinh(x)

sinh(1)

This, ﬁnally, is easily seen to be the correct solution. In this instance, there

are much quicker ways to arrive at the same result. The advantage of the

Green’s function is that it shows how the solution of the problem depends

on the inhomogeneity f (x). It is an efﬁcient way to obtain the solution in

some cases.



Now let us look back over the calculations and see if there is some place they

might fail. Aside from the possibility that the coefﬁcients k(x) or p(x) in the

differential equation might not be continuous, it seems that division by 0 is

the only possibility of failure. Quantities canceled or divided by were

W(x) =



(x) u

(x)



(x) u



(x)



αu

(l) − α



(l),

βu

(r) +β



(r)

in Eqs. (7), (10), and (12), respectively. It can be shown that all three of these

are 0 if any one of them is 0, and, in that case, u

(x) and u

(x) are proportional.

We summarize in a theorem.

Theorem. Let k(x),p(x),andf(x) be continuous, l ≤ x ≤r. The b oundary value

problem

+k(x)

+p(x )u =f (x), l < x < r,

αu(l) − α



(l) = 0, (i)

βu(r) + β



(r) = 0, (ii)