ETIQUETA DEL VIAL
4. Posibles efectos adversos
= ⎡⎣( )− − ⎤⎦
∫
( 2 ) sin( )0 1
3 3
2
2 2 1 1
(c)
Thus
x x
i i x
i
i 2
3 3
1
4 1
1 1
− = ⎡⎣− − ⎤⎦
=
∑
∞π ( ) sin(π ) (d)
The series on the right-hand side of Equation d converges to x2 – x with respect to the standard inner product generated norm defined for C[0, 1] in the sense of definition 2.13.
2.10 Linear operators
Mathematical modeling of physical systems leads to the formulation of a mathematical problem whose solution provides required information about the physical system. It is convenient to examine these equations using a con-sistent formulation. Let u represent a vector of dependent variables which is an element of a vector space D, and let f represent a vector, obtained through the modeling process, which is an element of a vector space R. The relation-ship between u and f can be written as
Lu= (2.35)f
where L is an operator determined from the modeling process. A formal defi nition of an operator follows:
Defi nition 2.18 An operator L is a function by which an element of a vector space D, called the domain of L, is mapped into an element of a vector space R, called the range of L.
Equation 2.35 is the general form of an operator equation. Given a vector f and the defi nition of L, it is desired to fi nd the vector or vectors for which Equation 2.35 is satisfi ed. The vectors u which satisfy Equation 2.35 are said to be solutions of the equation. Before considering how the solutions to Equation 2.35 are obtained, two basic questions are considered. (1) Does a solution exist?
(2) If so, how many solutions exist?. The fi rst question can be phrased as, “For a specifi c f in R, is there at least one u in D which solves Equation 2.35?” An alternate form of the second question is, “If a solution exists, is it unique?”. If for each f in R, a unique solution u, a vector in D, exists, then there is a one to one correspondence between the elements of D and R. In this case, an inverse operator L−1 exists whose domain is R and whose range is D, such that
u=L f−1 (2.36)
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Defi nition 2.19 If L is an operator defi ned such that there is a one-to-one correspondence between the elements of its domain D and the elements of its range R, then the inverse of L, denoted by L−1, exists, with domain R and range D, and is defi ned such that if Lu = f, then u = L−1f.
One method of fi nding the solution of Equation 2.35 is to determine the inverse of L and apply Equation 2.36. Note that substitution for f from Equa-tion 2.35 into EquaEqua-tion 2.36 leads to
L−1( )Lu =u (2.37)
If R = D, then u is an element of the domain of L−1, and then Equation 2.35 and Equation 2.36 lead to
L L u
(
−1)
=u (2.38)Defi nition 2.20 An operator L is said to be a linear operator if for each u and v in D and for all scalars α and β,
L u(α +βv)=αLu+βLv (2.39) An equation of the form of Equation 2.35 in which L is linear is said to be a linear equation. If L is not a linear operator, then Equation 2.35 is a nonlinear equation. The focus of this study is on linear equations.
An n × m matrix is a linear operator whose domain is Rm and whose range
The matrix formulation of this set of equations is
a a a a
Equation 2.41 can be summarized as
Ax= (2.42)y
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where A is the matrix operator which represents the coeffi cients of the equa-tions of Equation 2.40 arranged in n rows and m columns. The element in the ith row and jth column is identifi ed as ai,j. The solution x is a vector in Rm, while the input vector y is a vector in Rn.
When n > m, the number of linear equations represented by Equation 2.41 is greater than the dimension of the solution vector. If more than m equations are independent, then a solution does not exist for all y. That is, the range of A is not all of Rn When n < m, the number of equations is less than the dimen-sion of the solution vector. In this case, a solution exists, but is not unique;
the range is all of Rn, but for each y in Rn, there is more than one x in Rm that solves Equation 2.42.
When n = m and the number of equations is equal to the dimension of the solution vector, then the domain of A is Rn and the range of A is a subspace of Rn. It can be shown that the range is all of Rn when the matrix A is nonsin-gular. That is, its determinant is not equal to zero. When the matrix is non-singular, then A−1 exists and has the property
A−1(Ax)=x (2.43)
An associative property can be used on Equation 2.43, giving (A−1A)x = x.
The operator in parentheses defi nes the n × n identity matrix (a matrix with ones along the diagonal and zeros for all off-diagonal elements). A formal procedure exists to determine the inverse of a nonsingular square matrix.
When the determinant of a square matrix is zero, the matrix is said to be singular. In this case, the range of A is only a subset of Rn. That is, a solution does not exist for all y in Rn. When a solution does exist for a system with a singular matrix, the solution is not unique.
Example 2.24 The equation of a plane in three-dimensional space is of the form
ax+ + = (a)by cz d
where a, b, c, and d are constants. If n=nxi+nyj+nzk is a unit vector normal to the plane, Equation a can be rewritten as
n xx +n yy +n zz = ˆ (b)d where ˆd=d/ a2+ +b2 c2. Defi ning r = xi + yj + zk as the position vector from the origin to the point (x,y,z), Equation b can be written in the form
r n⋅ = ˆd (c)
Equation c can be interpreted as a statement that the length of the projection of a position vector from the origin to any point on the plane is the constant ˆd.
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The equations of three separate planes may be written as
a x1 +b y1 +c z1 = (d)d1
a x2 +b y2 +c z2 = (e)d2
a x3 +b y3 +c z3 =d3 (f) Values of x, y, and z which satisfy Equation d, Equation e, and Equation f are the coordinates of a point that lies on all three planes, a point on the intersec-tion of the planes. Equaintersec-tion d, Equaintersec-tion e, and Equaintersec-tion f may be summarized in matrix form as
a b c
Use the matrix formulation given in Equation g to discuss the existence and uniqueness of a point of intersection of three planes.
Solution Three planes may (a) intersect at a single point, (b) intersect along a line, or (c) have no common point of intersection. If the planes intersect at a single point, then Equation g has a unique solution. In this case, the determi-nant of the matrix on the left hand side of Equation g is nonzero. Noting the geometric interpretation of Equation c, this implies that the normal vectors to the planes are linearly independent.
If the vectors are linearly dependent, then without loss of generality it can be assumed that there exist constants α and β such that
n1=αn2+β (h)n3
where n1, n2 and n3 are unit vectors normal to the planes, and that a position vector, if it exists, satisfi es equations of the form of Equation c for each plane.
These equations are used in the following derivation:
r n
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Equation i must be satisfi ed for a solution of Equation g to exist when the normal vectors to the planes are linearly dependent. A condition of the form of Equation i is called a solvability condition. When the solvability condition is satisfi ed, the three planes intersect in a line, and the solution is not unique.
If the solvability condition of Equation i is not satisfi ed, then there is no intersection between the three planes.
Example 2.25 The state of stress at a point in a material is uniquely defi ned by the stress tensor at the point,
σ = unusual phenomena such as body moments or intrinsic angular momentum.
Let n = nxi + nyj + nzk be a unit vector normal to a plane through the point for which the stress tensor is known. The stress vector acting on the plane is calculated by
The stress tensor is a matrix operator. In general, its domain and range are R3. However, in application, its domain is restricted to the set of all unit vec-tors in R3. The resulting range is thus also restricted. In operator notation,
An= σ (c)
The stress tensor at a point in a solid is determined to be
σ =
Determine (a) the stress vector, (b) the normal stress, and (c) the shear stress acting on a plane whose normal is n=2 3/ i+1 3/ j−2 3/ k.
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Solution The stress vector acting on this plane can be calculated using
The normal stress is the projection of the stress vector onto the unit vector normal to the plane, or
σn= ⋅ n (f)σ
Substitution of Equation c into Equation f leads to
σn=(An n, ) (g)
where the inner product of Equation g is the standard inner product for R3. Thus
(c) The shear stress is the component of the stress vector perpendicular to the unit normal. Thus if τ is the shear stress vector, then τ ⋅ n = 0. Also τ = σ − σn
n. Hence, for this problem,
τ =
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The magnitude of the shear stress is calculated as
τ= ⋅τ τ1 2/ =1 07 10. × 3 Pa (j)
Example 2.26 Consider an n-degree-of-freedom linear mechanical system as illustrated in Figure 2.6. The displacements of the particles as functions of time are defi ned as x1, x2,…, xn and are referred to as generalized coordi-nates. The potential energy of the system, V, at any instant is a function of the generalized coordinates, V = V(x1, x2,…, xn). The stiffness matrix for a linear system is the matrix K whose element in the ith row and jth column, ki,j, is calculated by
k V
i j x x
i j
, = ∂
∂ ∂
2
(a)
Since the order of differentiation is interchangeable, ki,j = kj,i , and therefore the stiffness matrix is symmetric.
The potential-energy functional for a linear system can be written in a quadratic form as
V k x xij i j j
n
i n
=
=
=
∑
∑
1
2 1 1 (b)
(a) Determine the quadratic form of potential energy for the system shown in Figure 2.7 and use it to determine the stiffness matrix for the system.
x1 x2
k1 k2
xn
k3 kn
m1 m2 mn
Figure 2.6 The stiffness matrix for the system of Example 2.26 is obtained using the potential-energy functional written at an arbitrary instant.
x1 x2 x3
k k 3k
m m m
Figure 2.7 The stiffness matrix for this three-degree-of-freedom mechanical system is a 3 × 3 matrix.
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(b) Show that, for this system, if x is the vector of generalized coordinates at an arbitrary instant,
Kx x,
( )= 2V (c)
where the inner product is the standard inner product for Rn.
Solution (a) Note that the potential energy developed in a spring of stiff-ness k when the spring is subject to a force which leads to a change in length of x, measured from the spring’s unstretched length, is V = 1/2(kx2). The potential energy for the system of Figure 2.7 at an arbitrary instant is
V kx k x x k x x
The stiffness matrix is obtained through comparison of Equation d with the general quadratic form of Equation c as
K=
To obtain Equation d, the symmetry of the stiffness matrix is used, so that two terms within the double summation in Equation b are combined to yield
k x xi j i, j+k x xj i, j i= 2k x xi j i, j. Equation e is used to calculate
Kx=
The required inner product is evaluated as Kx x,
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There are many examples of linear operators defi ned for infi nite-dimen-sional vector spaces. The proof of existence and uniqueness of solutions of equations of the form of Equation 2.35 when the domain of L is an infi nite-dimensional vector space is beyond the scope of this study. However, the problems considered are formulated from the viewpoint of the mathematical modeling of a physical system. The physics of the system often dictate that a solution must exist and it must be unique. For example, the temperature distribution in a solid must be continuous and single-valued, requiring a unique solution. If the mathematical problem correctly models the physics, then unique solutions should exist.
Example 2.27 The nondimensional differential equation for the steady-state temperature distribution in a thin rod subject to an internal heat generation is
d
dx Bi x
2 2
Θ− Θ= +α βsin(π ) (a)
where α and β are nondimensional constants and Bi = hp/kA is the Biot num-ber. The left end of the rod is maintained at a constant temperature, while the right end is insulated. The temperature distribution satisfi es the boundary conditions
Θ( )0 = (b)0
d dx
Θ( )1 = (c)0
Write this problem in the operator form of Equation 2.35 and defi ne D and R.
Show, by obtaining the solution of the differential equation, the existence and uniqueness of the solution.
Solution (a) Equation a can be written in the form LΘ = f, where LΘ = d 2Θ/ and dx2 − BiΘ and f (x) = α + βsin(πx) The domain D is the subspace of C2[0,1] of all functions g(x) such that g(0) = 0 and dg/dx(1) = 0. The range R is the set of all elements of PC[0,1], the space of all piecewise continuous func-tions defi ned on [0,1].
The homogeneous solution of Equation a is of the form
Θh( )x =C1cosh
(
Bix)
+C2sinh(
Bix)
(d)where C1 and C2 are arbitrary constants of integration. The particular solu-tion of Equasolu-tion a is
Θp x
Bi Bi x
( )= − − sin
+ ( )
α β
π2 π (e)
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The general solution of Equation a is Application of the boundary condition of Equation b to Equation f leads to
0=C1− ⇒ 1= Bi C
Bi
α α (g)
Application of the boundary condition of Equation c to Equation f used Equation g leads to
0 2 2
Substitution of Equation g and Equation h into Equation f leads to
Θ( ) cosh
Equation i satisfi es the differential equation, Equation a, as well as the bound-ary conditions of Equation b and Equation c. Thus a solution exists. If a sec-ond solution ˆ ( )Θ x exists such that ˆ ( )Θx ≠Θ( )x for some values of x, then the temperature is multi-valued at those values, a physical impossibility. There-fore, if Equation a, Equation b, and Equation c are a true mathematical model of the physical system, the solution must be unique.
Example 2.28 The nondimensional differential equations for the trans-verse displacements w x1( )andw x2( ) of two elastically coupled, statically loaded beams with an axial load P are
d w
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where ε=PL E I2 1 1 η=kL E I4 μ=E I E I
(a) Write this problem in the operator form of Equation 2.33 and defi ne D and R.
(b) Show, by obtaining the solution of the differential equation, the exis-tence and uniqueness of the solution. Use ε = 2, η = 1, μ = 1 and Λ = 1 with f(x) = sin(πx).
Solution (a) The problem defi ned by Equation a, Equation b, Equation c, and Equation d can be written in the form of Equation 2.35 as Lw = f with two-dimensional vector of functions with each element a member of C4 [0,1].
That is, if g is in Q, then g =⎡
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(b) Substitution of the given parameters into Equation a and Equation b leads to
d w dx
d w
dx w w
4 1 4
2 1
2 1 2
2 0
− +( − )= (g)
d w dx
d w
dx w w x
4 2 4
2 2
2 2 1
−2 +( − )=sin(π ) (h)
The solutions of Equation g and Equation h are of the form
w=wh+wp (i)
where wh is the homogeneous solution, the solution obtained if f = 0, and wp
is the particular solution, the solution particular to the specifi c form of f.
The homogeneous solution is assumed to be of the form
wh a x b e
=⎡
⎣⎢
⎢
⎤
⎦⎥
⎥ α (j)
Substitution of Equation j into Equation g and Equation h with f = 0 leads to α4−2α2+1 0
( )
a b− = (k)− +a
(
α4−2α2+1)
b=0 (l)A nontrivial solution to Equation k and Equation i exists only if
α α
α α
4 2
4 2
2 1 1
1 2 1 0
− + −
− − + = (m)
Evaluation of the determinant in Equation k leads to the following eighth-order polynomial equation:
α8−4α6+6α4−4α2= (n)0 The solutions of Equation l are
α =0 0 1 414, , . ,−1 414 1 0987. , . ±0 4551. i,−1 0987. ±0..4551i (o) Equation k implies that for an appropriate value of ∝, α, b =(α4−2α2+1 . The ) homogeneous solution is a linear combination of all possible solutions of the form of Equation j, where the appropriate values of α are given in Equation o.
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Arbitrarily choosing a = 1 in each case and using trigonometric functions to replace complex exponentials leads to a homogeneous solution of
w t
The particular solution is obtained using the method of undetermined coef-fi cients, discussed in Chapter 3. A particular solution is assumed as
w x
Substitution of Equation q into Equation g and Equation h, using each to develop a relation between W1 and W2and solving simultaneously leads to
w x
The general solution is the sum of the homogeneous solution, Equation q and the particular solution Equation r. Application of the boundary conditions, Equation c and Equation d leads to a set of eight simultaneous equations to solve for the constants of integration for which a unique solution is attained.