Ecuaciones diferenciales parciales de Logan - Chapter 4 Solutions

8/10/2019 Ecuaciones diferenciales parciales de Logan - Chapter 4 Solutions

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CHAPTER 1

Partial Differential Equations on BoundedDomains

1. Separation of Variables

Exercise 1. The solution is

u(x, t ) =

n =1an en

2 t sin nx

where

an = 2

/ 2 sin nxdx = 2n ((1)n cos(n/ 2))

Thus

u(x, t ) = 2

et sin x 2

e4t sin2x + 23

e9t sin3x + 25

e25 t sin5x +

Exercise 2. The solution is given by formula (4.14) in the text, where the coeffi-cients are given by (4.15) and (4.16). Since G(x) = 0 we have cn = 0. Then

dn = 2

/ 2

0x sin nx dx +

2

/ 2( x)sin nx dx

Using the antiderivative formula

x sin nx dx = (1 /n 2)sin nx

(x/n )cos nx we

integrate to get

dn = 4n 2

sin n

2

Exercise 3. Substituting u(x, y ) = (x)(y) we obtain the Sturm-Liouville prob-lem

= , x(0, l); (0) = (l) = 0and the differential equation

= 0The SLP has eigenvalues and eigenfunctions

n = n2

2

/l2

, n (x) = sin( nx/l )and the solution to the equation is

n (y) = an cosh(ny/l ) + bn sinh(ny/l )

1


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2 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS

Therefore

u(x, y ) =

n =1(an cosh(ny/l ) + bn sinh( ny/l )) sin( nx/l )

Now we apply the boundary conditions:

u(x, 0) = F (x) =

n =1an sin(nx/l )

and

u(x, 1) = G(x) =

n =1(an cosh(n/l ) + bn sinh(n/l )) sin( nx/l )

Thusan =

2l

0F (x) sin( nx/l )dx

andan cosh(n/l ) + bn sinh( n/l ) =

2l

0G(x) sin( nx/l )dx

which gives the coefficients an and bn .

Exercise 4. Substituting u = y(x)g(t) into the PDE and boundary conditionsgives the SLP

y = y, y(0) = y(1) = 0and, for g, the equation

g + kg + c2g = 0The SLP has eigenvalues and eigenfunctions

n = n22 , yn (x) = sin nx, n = 1 , 2, . . .

The g equation is a linear equation with constant coefficients; the characteristicequations is

m2 + km + c2 = 0which has roots

m = 12

(k k2 4c2n22)

By assumption k < 2c, and therefore the roots are complex for all n. Thus thesolution to the equation is (see the Appendix in the text on ordinary differentialequations)

gn (t) = ekt (an cos(mn t) + bn sin(mn t))where

mn = 12 4c2n22 k2)Then we form the linear combination

u(x, t ) =

n =1ekt (an cos(mn t) + bn sin(mn t))sin( nx )

Now apply the initial conditions. We have

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

n =1an sin(nx )


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2. FLUX AND RADIATION CONDITIONS 3

and thus

an = 1

0f (x)sin nx dx

The initial condition u t = 0 at t = 0 yields

u t (x, 0) = 0 =

n =1(bn mn ka n )sin( nx )

Thereforebn mn ka n = 0

or

bn = kanmn

= kmn

1

0f (x)sin nx dx

2. Flux and Radiation Conditions

Exercise 1. This problem models the transverse vibrations of a string of length l

when the left end is xed (attached) and the right end experience no force; however,the right end can move vertically. Initially the string is displaced by f (x) and it isnot given an initial velocity.

Substituting u = y(x)g(t) into the PDE and boundary conditions gives the SLP

y = y, y(0) = y(l) = 0and, for g, the equation

g + c2g = 0The SLP has eigenvalues and eigenfunctions

n = ((2 n + 1) /l )2 , yn (x) = sin((2 n + 1) x/l ), n = 0 , 1, 2, . . .

and the equation for g has general solution

gn

(t) = an

sin((2 n + 1) ct/l ) + bn

cos((2n + 1) ct/l )

Then we form

u(x, t ) =

n =0(an sin((2 n + 1) ct/l ) + bn cos((2n + 1) ct/l )) sin((2 n + 1) x/l )

Applying the initial conditions,

u(x, t ) = f (x) =

n =0bn sin((2 n + 1) x/l )

which yields

bn = 1

||sin((2 n + 1) x/l )

||2

l

0f (x) sin((2 n + 1) x/l )

And

u t (x, 0) = 0 =

n =0an cn sin((2 n + 1) x/l )


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which gives an = 0. Therefore the solution is

u(x, t ) =

n =0bn cos((2n + 1) ct/l )) sin((2 n + 1) x/l )

Exercise 2. This problem models the steady state temperatures in a rectangular

plate that is insulated on both sides, whose temperature is zero on the top, andwhose temperature is f (x) along the bottom. Letting u = g(y)(x) and substitutinginto the PDE and boundary conditions gives the Sturm-Liouville problem

= , (0) = (a) = 0and the differential equation

g g = 0The eigenvalues and eigenfunctions are 0 = 0 , (x) = 1 and

n = n22 /a 2 , n (x) = cos( nx/a ), n = 1 , 2, 3, . . .

The solution to the g equation is, corresponding to the zero eigenvalue, g0(y) =c0y + d0 , and corresponding to the positive eigenvalues,

gn (y) = cn sinh(ny/a ) + dn cosh(ny/a )Thus we form the linear combination

u(x, y ) = c0y + d0 +

n =1(cn sinh( ny/a ) + dn cosh(ny/a )) cos( nx/a )

Now apply the boundary conditions on y to compute the coefficients:

u(x, 0) = f (x) = d0 +

n =1dn cos(nx/a )

which gives

d0 = 1a

a

0f (x)dx, dn =

2a

a

0f (x) cos(nx/a )dx

Nextu(x, b) = 0 = c0b + d0 +

n =1(cn sinh( nb/a ) + dn cosh(nb/a )) cos( nx/a )

Therefore

c0 = d0 /b, c n = cosh(nb/a )sinh( nb/a )

dn

Exercise 3. The ux at x = 0 is (0, t ) ux (0, t ) = a0u(0, t ) > 0, so thereis heat ow into the bar and therefore adsorption. At x = 1 we have (1, t ) =ux (1, t ) = a1u(1, t ) > 0, and therefore heat is owing out of the bar, which isradiation. The right side of the inequality a0 + a1 > a0a1 is positive, so thepositive constant a1 , which measures radiation, must greatly exceed the negativeconstant a0 , which measures adsorption.

In this problem substituting u = y(x)g(t) leads to the Sturm-Liouville problem

y = y, y(0) a0y(0) = 0 , y(1) + a1y(1) = 0


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2. FLUX AND RADIATION CONDITIONS 5

and the differential equationg = g

There are no nonpositive eigenvalues. If we take = k2 > 0 then the solutions are

y(x) = a cos kx + bsin kx

Applying the two boundary conditions leads to the nonlinear equation

tan k = (a0 + a1)kk2 a0a1To determine the roots k, and thus the eigenvalues = k2 , we can graph both sidesof this equation to observe that there are innitely many intersections occuring atkn , and thus there are innitely many eigenvalues n = k2n . The eigenfunctions are

yn (x) = cos kn x + a0kn

sin kn x

So the solution has the form

u(x, t ) =

n =1cn e

2n t (cos kn x +

a0kn

sin kn x)

The cn are then the Fourier coefficients

cn = ( f, y n )/ ||yn ||2

If a0 = 1/ 4 and a1 = 4 thentan k =

3.75kk2 + 1

From a graphing calculator, the rst four roots are approximately k1 = 1 .08, k2 =3.85, k3 = 6 .81, k4 = 9 .82.

Exercise 4. Letting c(x, t ) = y(x)g(t) leads to the periodic boundary value prob-lem

y = y, y(0) = y(2l), y(0) = y(2l)and the differential equation

g = Dgwhich has solution

g(t) = eDt

The eigenvalues and eigenfunctions are found exactly as in the solution of Exercise4, Section 3.4, with l replaced by 2 l. They are

0 = 0 , n = n22 /l 2 , n = 1 , 2, . . .

andy0(x) = 1 , yn (x) = an cos(nx/l ) + bn sin(nx/l ), n = 1 , 2, . . .

Thus we form

c(x, t ) = a0/ 2 +

n =1en

2 2 Dt/l 2 (an cos(nx/l ) + bn sin(nx/l )

Now the initial condition gives

c(x, 0) = f (x) = a0/ 2 +

n =1(an cos(nx/l ) + bn sin(nx/l )


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which is the Fourier series for f . Thus the coefficients are given by

an = 1l

2

0lf (x) cos(nx/l )dx, n = 0 , 1, 2, /ldots

and

bn = 1l

2

0lf (x) sin( nx/l )dx, n = 1 , 2, /ldots

3. Laplaces Equation

Exercise 2. Substituting u(r, ) = g()y(r ) into the PDE and boundary conditionsgives the Sturm-Liouville problem

g = g, g(0) = g(/ 2) = 0and the differential equation

r 2y + ry + y = 0This SLP has been solved many times in the text and in the problems. The eigen-

values and eigenfunctions aren = 4 n2 , gn () = sin(2 n), n = 1 , 2, . . .

The y equation is a Cauchy-Euler equation and has bounded solution

yn (r ) = r 2n

Form

u(r, ) =

n =1bn r 2n sin(2n)

Then the boundary condition at r = R gives

u(R, ) = f () =

n =1bn R2n sin(2n)

Hence the coefficients are

bn = 1R 2n

/ 2

0f () sin(2 n)d

Exercise 3. Substituting u(r, ) = g()y(r ) into the PDE and boundary conditionsgives the Sturm-Liouville problem

g = g, g(0) = g(/ 2) = 0and the differential equation

r 2y + ry + y = 0The eigenvalues and eigenfunctions are

n = (2 n + 1) 2 , gn () = sin((2 n + 1) ), n = 0 , 1, 2, . . .The y equation is a Cauchy-Euler equation and has bounded solution

yn (r ) = r 2n +1


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3. LAPLACES EQUATION 7

Form

u(r, ) =

n =1bn r 2n +1 sin((2 n + 1) )

Then the boundary condition at r = R gives

u(R, ) = f () =

n =1bn R2n +1 sin((2 n + 1) )

Hence the coefficients are

bn = 1

R 2n +1 / 2

0f () sin((2 n + 1) )d

Exercise 4. Substituting r = 0 into Poissons integral formula (4.34) of the textwe instantly get the temperature at the origin as

u(0, 0) = 12

2

0f ()d

The right side is the average of the function f () over the interval [0 , 2].

Exercise 5. Let w = u + v where u satises the Neumann problem and v satisesthe boundary condition n v = 0. ThenE (w) = E (u + v)

= 1

2 (u u + 2u v + v v)dV (hu hv)dA= E (u) + u v dV + 12 v vdV hv dA= E (u) + vu n dA v u dV + 12 v vdV hv dA= E (u) + vh dA +

12 v vdV hv dA= E (u) + 1

2 v vdV So E (u) E (w).Exercise 6. Multiply both sides of the PDE by u and integrate over . We obtain

u u dV = c u2dV Now use Greens rst identity to obtain

u

u

ndA

u

u dV = c

u2dV

or

au 2dA u u dV = c u2dV


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The left side is negative and the right side is positive. Then both must be zero, or

u2dV = 0Hence u = 0 in .

The uniqueness is standard. Let u and v be two solutions to the boundaryvalue problem

u cu = f, x n u + au = g, x Then the difference w = u v satises the homogeneous problem

w cw = 0 , x n w + aw = 0 , x By the rst part of the problem we know w = 0 and therefore u = v.

Exercise 7. The solution is given by equation (4.31) in the text. Here

f () = 4 + 3 sin

The right side is its Fourier series, so the Fourier coefficients are given bya02

= 4 , Rb1 = 3

with all the other Fourier coefficients identically zero. So the solution isu(r, ) = 4 +

3rR

sin

Exercise 8. Multiplying the equation u = 0 by u, integrating over , and thenusing Greens identity gives

u u dV = uu ndA u udV = 0Thus

u udV = 0which implies

u = 0Thus u =constant.

4. Cooling of a Sphere

Exercise 1. The problem is

y 2

y = y, y(0) bounded , y() = 0

Making the transformation Y = y we get Y = Y . If = k2 < 0 thenY = a sinh k + bcosh k

or y = 1(a sinh k + bcosh k)For boundedness at = 0 we set b = 0. Then y() = 0) forces sinh k = 0. Thusk = 0. Consequently, there are no negative eigenvalues.


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4. COOLING OF A SPHERE 9

Figure 1. Temperature at the center of the sphere in Exercise 2.

Exercise 2. From the formula developed in the text the temperature at = 0 is

u(0, t ) = 74

n =1(1)n +1

1n

en2 kt

where k = 5 .58 inches-squared per hour. A graph of an approximation using ftyterms is shown in the gure.

Exercise 3. The boundary value problem is

u t = k(u + 2

u )

u (R, t ) = hu (R, t ), t > 0u(, 0) = f (), 0 R

Assume u = y()g(t). Then the PDE and boundary conditions separate into theboundary value problem

y + (2 / )y + y = 0 , y(R) = hy(R), y boundedand the differential equation

g= kgThe latter has solution g(t) = exp(kt ). One can show that the eigenvalues arepositive. So let = p2 and make the substitution Y = y, as in the text, to obtain

Y + p2Y = 0This has solution

Y () = a cos p + bsin p


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But Y (0) = 0 forces a = 0 (because y is bounded). Then the other boundarycondition forces p to satisfy the nonlinear equation

tan Rp = Rp1 Rh

If we graph both sides of this equation against p we note that there are innitelymany intersections, giving innitely many roots pn , n = 1 , 2, . . . , and therefore

innitely many eigenvalues n = p2n . The corresponding eigenfunctions are

yn () = 1 sin( n )Thus we haveu(, t ) =

n =1cn e n kt 1 sin( n )

The cn are found from the initial condition. We have

u(, 0) = f () =

n =1cn 1 sin( n )

Thus

cn =

R

0 f () sin( n )d

R0 sin2( n )d

Exercise 4. Representing the Laplacian in spherical coordinates, the boundaryvalue problem for u = u(, ), where (0, 1) and (0, ), is

u = u + 2

u + 1

2 sin (sin u ) = 0

u(1, ) = f (), 0 Observe, by symmetry of the boundary condition, u cannot depend on the angle .Now assume u = R()Y (). The PDE separates into two equations,

2R + 2 R R = 0and 1

sin (sin Y ) = Y

We transform the Y equation by changing the independent variable to x = cos .Then we get, using the chain rule,

1sin

dd

= ddx

So the Y equation becomes

ddx

(1 x2)dydx

= y 1 < x < 1By the given facts, this equation has bounded, orthogonal, solutions yn (x) = P n (x)on [

1, 1] when = n = n(n + 1), n = 0 , 1, 2, . . . . Here P n (x) are the Legendre

polynomials.Now, the Requation then becomes

2R + 2 R n(n + 1) R = 0


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4. COOLING OF A SPHERE 11

This is a Cauchy-Euler equation (see the Appendix in the text) with characteristicequation

m(m 1) + m n(n + 1) = 0The roots are m = n, (n + 1). Thus

Rn () = an n

are the bounded solutions (the other root gives the solution n

1

, which is un-bounded at zero). Therefore we form

u(, ) =

n =0an n P n (cos )

or, equivalently

u(, x ) =

n =0an n P n (x)

Applying the boundary condition gives the coefficients. We have

u(1, x) = f (arccos x) =

n =0an P n (x)

By orthogonality we get

an = 1

||P n ||2 1

1f (arccos x)P n (x)dx

or

an = 1

||P n ||2

0f ()P n (cos )sin d

By direct differentiation we get

P 0(x) = 1 , P 1(x) = x, P 2(x) = 12

(3x2 1), P 3(x) = 52

x3 32

x

Also the norms are given by

||P 0||2

= 2 , ||P 1||2

= 23, ||P 2||

2=

25

When f () = sin the rst few Fourier coefficients are given by

a0 = 4

, a1 = a3 = 0 , a2 = 0.49Therefore a two-term approximation is given by

u(, ) 4

0.492

2(3cos2 1)

Exercise 5. To determine the temperature of the earth we must derive the tem-perature formula for any radius R (the calculation in the text uses R = ). Themethod is exactly the same, but now the eigenvalues are n = n22 /R 2 and the

eigenfunctions are yn = 1 sin(n/R ), for n = 1 , 2, . . . . Then the temperature is

u(, t ) = 2RT 0

n =1

(1)n +1n

elam n kt 1 sin(n/R )


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Now we compute the geothermal gradient at the surface, which is u (, t ) at = R.We obtain

u (R, t ) = 2T 0R

n =1en

2 2 kt/R 2

If G is the value of the geothermal gradient at the current time t = tc , then

RG2T 0 =

n =1e

n 2 2 kt c /R 2

We must solve for t c . Notice that the sum has the form

n =1ean

2

where a = 2kt c /R 2 . We can make an approximation by noting that the sumrepresents a Riemann sum approximation to the integral

0 eax 2 dx = 12 /aSo we use this value to approximate the sum, i.e.,

n =1en

2

2

kt c /R2

R

2 kt cSolving for t c gives

tc = T 20G2k

Substituting the numbers in from Exercise 5 in Section 2.4 gives tc = 5 .15(10)8years. This is the same approximation we found earlier.

5. Diffusion in a Disk

Exercise 1. The differential equation is

(ry ) = ry . Multiply both sides by y

and integrate over [0 , R] to get

R

0 (ry )ydr = R

0ry 2dr

Integrating the left hand side by parts gives

ryy |R0 R

0r (y)2dr =

R

0ry 2dr

But, since y and y are assumed to be bounded, the boundary term vanishes. Theremaining integrals are nonnegative and so 0.Exercise 2. Let y, and w, be two eigenpairs. Then (ry ) = ry and

(rw ) = rw . Multiply the rst of these equations by w and the second by

y, and then subtract and integrate to get

R

0[(ry )w + ( rw )y]dr = ( )

R

0ruwdr


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5. DIFFUSION IN A DISK 13

Figure 2. The Bessel functions J 0(zn r ).

Now integrate both terms in the rst integral on the left hand side by parts to get

(ry w + rw y) |R0 + R

0(ry w rw y)dr = ( )

R

0ruwdr

The left side of the equation is zero and so y and w are orthogonal with respect tothe weight function r .

Exercise 3. We have

u(r, t ) =

n =1cn e0.25 n t J 0(zn r )

where

cn = 1

0 5r4(1 r )J 0(zn r )dr

1

0 J 0(zn r )2rdr

We have z1 = 2 .405, z2 = 5 .520, z3 = 8 .654. Use a computer algebra program tocalculate a three-term approximation.

Exercise 4. The eigenfunctions J 0(zn r ), n = 1 , 2, 3, 4 are sketched in the gure.For larger n the number of oscillations increases.

Exercise 5. The Maple worksheet follows.


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6. Sources on Bounded Domains

Exercise 1. Use Duhamels principle to solve the problem

u tt c2uxx = f (x, t ), 0 < x < , t > 0u(0, t ) = u(, t ) = 0 , t > 0

u(x, 0) = u t (x, 0) = 0 , 0 < x < 1Consider the problem for w = w(x,t, ), where is a parameter:

wtt c2wxx = 0 , 0 < x < , t > 0w(0, t , ) = w(,t, ) = 0 , t > 0

w(x, 0, ) = 0 , wt (x, 0, ) = 0 , 0 < x < 1

This problem was solved in Section 4.1 (see (4.14)(4.14)). The solution is

w(x,t, ) =

n =1cn ( )sin nct sin nx

wherecn ( ) =

2

nc

0f (x, )sin nx dx

So the solution to the original problem is

u(x, t ) = t

0w(x, t , )d

Exercise 2. If f = f (x), and does not depend on t, then the solution can bewritten

u(x, t ) = 2

n =1

0f (r )sin nr dr (

t

0en

2 k ( t ) d sin nx

But a straightforward integration gives

t

0 en2

k( t ) d = 1kn 2 (1 en

2

kt )

Therefore

u(x, t ) = 2

n =1

1kn 2

(1 en2 kt )

0f (r )sin nr dr sin nx

Taking the limit as t givesU (x) lim u(x, t ) = u(x, t ) =

2

n =1

1kn 2


Now consider the steady state problem

kv = x( x), v(0) = v() = 0This can be solved directly by integrating twice and using the boundary conditionsto determine the constants of integration. One obtains

v(x) = 112k

(2x 2 x4 3x)


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6. SOURCES ON BOUNDED DOMAINS 15

To observe that the solution v(x) is the same as the limiting solution U (x) weexpand the right side of the vequation in its Fourier sine series on [0 , ]. Then

kv =

n =1cn sin nx

where

cn = 2

0r ( r )sin nrdr

Integrating the differential equation twice gives

kv(x) + kv(0)x =

n =1cn n2(sin nx x)

Evaluating at x = gives

kv(0) =

n =1cn n2

Whence

kv(x) =

n =1

cn n2 sin nx

or

v(x) = 2k

n =1

1n2


Hence v(x) = U (x).

Exercise 3. Following the hint in the text we have

u(x, y ) =

n =1gn (y)sin nx

where we ndg

n n2gn = f n (y), gn (0) = gn (1) = 0

where f n (y) are the Fourier coefficients of f (x, y ). From the variation of parametersformula

gn (y) = aeny + beny 2n

y

0sinh( n( y))f n ( )d

Now gn (0) = 0 implies b = a. So we can writegn (y) = 2 a sinh ny

2n

y


which gives, using gn (1) = 0,

a = 1

n sinh n 1

0sinh( n( 1))f n ( )d

Thus the gn (y) are given by

gn (y) = 2 sinh ny

n sinh n 1

0sinh( n( 1))f n ( )d

2n

y



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u t = u + f (r, t ) 0 r < R, t > 0u(R, t ) = 0 , t > 0

u(r, 0) = 0 , 0 < r < R

For w = w(r,t, ) we consider the problem

wt = w 0 r < R, t > 0w(R,t ) = 0 , t > 0w(r, 0, ) = f (r, ), 0 < r < R

This is the model for heat ow in a disk of radius R; the solution is given byequation (4.53) in Section 4.5 of the text. It is

w(r,t, ) = cn ( )e n kt J 0(zn r/R )

where zn are the zeros of the Bessel function J 0 , n = z2n /R 2 and

cn ( ) = 1

||J 0(zn r/R )||2 R

0f (r, )J 0(zn r/R )rdr

Then

u(r, t ) = R

0w(r, t , )d

7. Parameter Identication Problems

Exercise 1. We have p1 = p0er 1 t

, p2 = p0er 2 t

Dividing the two equations and then taking natural logarithms gives

r 1 r 2 = 1t

(ln p1 ln p2)By the mean value theorem

| ln a ln b||a b|

= d ln x

dx |x = c = 1c

for some c between a and b. Thus

|r 1 r 2| = 1t| ln p1 ln p2|

1Mt| p1 p2|

Exercise 2. We have (x)u tt = uxx . Putting u = Y (x)g(t) gives, upon separatingvariables,

y = (x)y, y(0) = y(1) = 0We have

yf = (x) f yf , yf (0) = yf (1) = 0

Integrating from x = 0 to x = s gives

yf (s) + yf (0) = f s

0(x)yf (x)dx


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7. PARAMETER IDENTIFICATION PROBLEMS 17

Now integrate from s = 0 to s = 1 to get

yf (0) = f 1

0 s

0(x)yf (x)dxds

= f 1

0(1 x)(x)yf (x)dx

The last step follows by interchanging the order of integration. If (x) = r ho0 is aconstant, then

0 =yf (0)

f 1

0 (1 x)yf (x)dx

Exercise 3. From Exercise 2 in Section 4.6 we have the solution

u(x, t ) = 2

n =1

1kn 2

(1 en2 kt )


Therefore

U (t) = u(/ 2, t ) =

0

2

n =1sin nr sin(n/ 2)

1kn 2

(1 en2 kt ) f (r )dr

We want to recover f (x) if we know U (t). This problem is not stable, as thefollowing example shows. Let

u(x, t ) = m3/ 2(1 em2 t )sin mx, f (x) = m sin mx

This pair satises the model. If m is sufficiently large, then u(x, t ) is uniformlysmall; yet f (0) is large. So a small error in measuring U (t) will result in a largechange in f (x).


u t = uxx , x, t > 0u(x, 0) = 0 , x > 0

u(0, t ) = f (t), t > 0

Taking Laplace transforms and solving gives

U (x, s ) = F (s)ex s

Here we have discarded the unbounded part of the solution. So, by convolution,

u(x, t ) = t

0f ( )

x

4(t )3ex

2 / 4( t ) d

Hence, evaluating at x = 1,

U (t) = t

0f ( )

1

4(t )3e1/ 4( t ) d

which is an integral equation for f (t). Suppose f (t) = f 0 is constant and U (5) = 10.

Then 20 f 0

= 5

0

e1/ 4(5 )

4(5 )3d = 2 erfc (1/ 5)

where erfc = 1 erf . Thus f 0 = 59 .9 degrees.


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Exercise 5. The problem

u t = Du xx vux , xR, t > 0; u(x, 0) = eax2

, xR

can be solved by Fourier transforms to get

u(x, t ) = a

a + 4 Dt e(xvt )

2 / (( a +4 Dt )

Thus, choosing a = v = 1 we getU (t) = u(1, t ) =

1 1 + 4 Dt e

(1 t )2 / ((1+4 Dt )

8. Finite Difference Methods

Exercise 1. The Cauchy-Euler algorithm for this problem isY n +1 = Y n + h(2nhY n + Y 2n ), n = 0 , 1, 2, . . . ; Y 0 = 1

where h is the step size. With h = 0 .1 the values of Y n at the points tn = nh, n =0, . . . , 10 are:

1, 1.1, 1.199, 1.295, 1.385, 1.466, 1.534, 1.585, 1.615, 1.617, 1.587

Exercise 2. A time snapshot of the solution surface at t = 1 is shown in theaccompanying gure. The Maple program in Figure 4.8 of the text was run withh = 0 .1 and k = 0 .1, which gives r = k/h 2 = 10. This violates the stabilitycondition, and one can observe the highly oscillatory behavior of the numericalscheme.

Exercise 3. The Maple worksheet and surface plot is given in the two gures.

Exercise 4. The Maple worksheet and surface plot is given in the two gures.

Exercise 5. The Maple worksheet and surface plot for the rst problem in Exercise5 is given in the accompanying gures.


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8. FINITE DIFFERENCE METHODS 19

Figure 3. Time t = 1 prole of the numerical solution in Exercise2 when h / k = 10.


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Figure 4. Maple program to solve Exercise 3.


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Figure 5. Solution surface in Exercise 3.


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CHAPTER 2

Appendix

1. The equation y + 2y = ex is rst order, linear. Multiply by the integratingfactor e2x and the equation becomes ( ye2x ) = ex Integrate both sides and multiplyby e2x to get

y(x) = Ce2x + ex

2. Here, y = 3y. Separate variables and integrate to obtainy(x) = Ce3x

3. The equation y +8 y = 0 is second-order, linear, with constant coefficients. Thecharacteristic equation is m 2 + 8 = 0 which has roots m = 8. Then

y(x) = A cos 8x + B sin 8x

4. The equation y xy = x2y2 is a Bernoulli equation. Make the substitutionw = 1 /y and the equation turns into a linear equation w + xw = x2 . Theintegrating factor is exp( x2 / 2). Multiplying by the integrating factor gives(wex

2 / 2) = x2ex2 / 2

Integrating both sides from 0 to x gives

wex 2 / 2

w(0) = x

0 r2

er 2 / 2

drThen

w(x) = 1 /y (x) = ex2 / 2w(0) ex

2 / 2 x

0r 2er

2 / 2dr

5. The equation x2y3xy +4 y = 0 is a Cauchy-Euler equation. The characteristicequation is m(m 1) 3m + 4 = 0 which has roots m = 2 , 2. Thusy(x) = ax 2 + bx2 ln x

6. The equation y + x(y)2 = 0 does not have y appearing explicitly. So let v = yto obtain

v + xv2 = 0We separate variables to get

1v

= 12

x2 + A

27


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28 2. APPENDIX

Then

y(x) = dxx2 / 2 + A + B = 2 2A arctan x 2A + B7. The equation y + y + y = 0 is linear, second-order, with constant coefficients.The characteristic equation is m2 + m +1 = 0, which has roots m = 1/ 2 3i/ 2.Thus

y(x) = ex/ 2 a cos 3x

2 + bsin

3x2

8. In the equation yy (y)3 = 0 the independent variable does not appearexplicitly. So let v = y which gives y = v dvdy . Then we get

yvdvdy

= v3

Separating variables and solving gives v = 1/ (C + ln y). Then(C + ln y)dy = dx

ThenCy + y ln y y = x + B

9. The equation 2 x2y+3 xy y = 0 is a Cauchy-Euler equation with characteristicequation 2 m(m 1) + 3 m 1 = 0. The roots are m = 1, 1/ 2. Theny(x) = a x + b

x

10. The equation y3y4y = 2sin x is a linear, nonhomogeneous equation. Thehomogeneous solution is yh (x) = ae4t + bet . Guess a particular solution of the formy p

(x) = A sin x + B cos x. Substitute into the equation to nd A = 5 / 8, B =

3/ 8.Then

y(x) = ae4t + bet + 58

sin x 38

cos x

11. The homogeneous solution of y + 4 y = x sin x is yh (x) = a cos2x + bsin2x. Aparticular solution can be found by undetermined coefficients or using the variationof parameters formula from this appendix. We choose the latter. We have

y p(x) = x

0

cos2s sin2x cos2x sin2s2

s sin2s ds

The calculation is left as an exercise. The solution is the sum of yh and y p.

12. The equation y 2xy = 1 is linear with integrating factor exp(x2). Multi-plying by the integrating factor leads to(yex

2

) = ex2


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2. APPENDIX 29

Integrating from 0 to x then gives

y(x) = y(0)ex2

+ x

0ex

2

r2

dr

13. The second order linear equation y + 5 y + 6 y = 0 has characteristic equationm2 + 5 m + 6 = 0 with roots

3,

2. Hence

y(x) = ae3x + be2x

14. Separate variables to obtain

(1 + 3 y3)dy = x2dx

Integrating gives the implicit solution

y + 34

y4 = 13

+ C

Documents

Ecuaciones diferenciales parciales de Logan - Chapter 4 Solutions