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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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4 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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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6 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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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8 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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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10 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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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12 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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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14 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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
0f (r )sin nr dr sin nx
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
0f (r )sin nr dr sin nx
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
0sinh( n( y))f n ( )d
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
0sinh( n( y))f n ( )d
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16 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
Exercise 4. The problem is
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 )
0f (r )sin nr dr sin nx
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).
Exercise 4. The problem is
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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18 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
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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20 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
Figure 4. Maple program to solve Exercise 3.
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8. FINITE DIFFERENCE METHODS 21
Figure 5. Solution surface in Exercise 3.
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22 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
Figure 6. Maple program to solve Exercise 4.
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8. FINITE DIFFERENCE METHODS 23
Figure 7. Solution surface in Exercise 4.
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24 1. PARTIAL DIFFERENTIAL EQUATIONS ON BOUNDED DOMAINS
Figure 8. Maple program to solve Exercise 5.
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8. FINITE DIFFERENCE METHODS 25
Figure 9. Solution surface in Exercise 5.
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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