Solution Methodologies
P.4.7 illustrates prescribed polynomials that satisfy the conditions (4.39), (4.41)
4.3.3 Exact and Conditional Polynomial Solutions
4.3.3.1 The Dirichlet/Neumann BVPs
For Dirichlet/Neumann problems, using the contour parametrization functions x(t), y(t) are trigonometric identities, we write l.h.s. of (4.38a) or (4.38b) as
JΛ(t) =
kΛkp
m=1
∑
JΛsmsin(mt) + JΛcmcos(mt) (4.47) where kΛ= kQ+1. Clearly, the functions JΛsm, JΛcmcontain 2kΛfree coefficients Hi jofΛ. There-fore, in order to ensure that JΛ(t) − F3(t) ≡ 0 one should enforce the equations
JΛcm− F3cm= 0, JΛsm− F3sm= 0, m= 1,...,kΛkp, (4.48) namely, a total of 2kΛkp equations. In an elliptical domain where kp= 1 (see Example 3.1), the number of equations and coefficients is 2kΛ, and the above linear system is solvable in an exact manner. When kp> 1, conditional solutions may be obtained by producing a set of
“data relations” between the coefficients of the boundary polynomials, pj,i,s−i, qj,i,s−i, that must be satisfied in order to enable polynomial solutions. P.4.8 presents exact and conditional polynomial solutions for homogeneous Dirichlet/Neumann BVPs, see Examples 4.3, 4.4.
See also application of polynomial solutions involved for the generalized Neumann problem derived in (Kezerashvili, 1986).
Example 4.3 Exact Polynomial Solution of the Neumann BVP in an Ellipse (Low Order).
In order to examine the homogeneous Neumann BVP in an ellipse, see (4.13a), (4.38b), we activate P.4.8 with BVP =−1, iso = 0, k3(= kQ) = 1, k0= 0. In addition, we work with symbolic ai j and set x(t) = acos t, y(t) = bsin t, kp= 1. Accordingly, the given boundary polynomials are P3= p00+ p10x+ p01y, Q3= q00+ q10x+ q01y. Therefore, the r.h.s. of (4.38b) is written as
F3(t) = −1
2ab(q01+ p10) − aq00sin t−bp00cos t
−1
2(b2p01+ a2q10)sin(2t) −1
2ab(p10− q01)cos(2t). (4.49) The condition F3c0= 0 (for the BVP solution existence and uniqueness) is given by the above underlined term as
q01+ p10= 0. (4.50)
Hence, by replacing p10by−q01, we obtain a consistent version of the boundary function F3(t) = −aq00sin t−bp00cos t−1
2(b2p01+ a2q10)sin(2t) + abq01cos(2t). (4.51) P.4.8 shows that the prescribed harmonic polynomial becomes (with kΛ=kQ+1=2)
Λ= H10x+ H01y+ H20x2+ H11xy+a45H11− a44H20
a55 y2. (4.52)
By balancing the trigonometric functions, we arrive at the following system of 2 kΛ= 4 equa-tions:
a45H01− a44H10= −p00, a45H10− a55H01= −q00, a45H11− 2a44H20= − p10, H11(2a452− a44a55− 4a552) − 2H20a45(a44+ 4a55) = −a55(4q10+ p01). (4.53)
The solution of this system is H10= − 1
a0(p00a55+ q00a45), H01= −1
a0(q00a44+ p00a45), H20= 1
2δ1a0[a2a55(a45q10− q01a55) +b2(a45a55p01+ 2q01a452− q01a44a55)], H11= − 1
δ1a0[a2a55(a45q01− q10a44) −b2a44(a55p01+ a45q01)], (4.54) where a0= a44a55− a245and δ1= a55a2+ a44b2.
Example 4.4 Exact Polynomial Solution of the Neumann BVP in an Ellipse (Higher Order).
In view of the superposition principle, we shall now restrict our attention to the second-order homogeneous polynomials P3= p20x2+ p11xy+ p02y2, Q3= q20x2+ q11xy+ q02y2. Hence, this example is identical to Example 4.3 except for the fact that we now activate P.4.8 with k3(= kQ) = 2 and k0= 2. P.4.8 shows that (4.39) is satisfied whenΛis expressed as
Λ= H10x+ H01y+ H30x3+ H21x2y+ H12xy2+ H03y3, (4.55) where we have chosen to eliminate the coefficients
H12=2a45H21− 3a44H30
a55 , H03= −(a44a55− 4a245)H21+ 6a45a44H30
3a255 . (4.56)
The resulting expressions for the remaining four coefficients are quite lengthy and require some “careful organization”. P.4.9 shows that the expressions documented below are identical to those obtained by P.4.8:
H01δ2= a2a45a55(a55a2+ a44b2)(b2p02+ a2q11)
+b2a44[(3a55a44− 2a452)a2+ a442b2](b2p11+ a2q20)
+ [2a552a4+ (9a55a44− 4a452)b2a2+ 3a442b4]a2a45p20 (4.57a) + [3a552a4+ (7a55a44− 2a452)b2a2+ 2a442b4]b2a44q02,
H10δ2= a2a55[a552a2+ (3a55a44− 2a452)b2](b2p02+ a2q11) +b2a44a45(a55a2+b2a44)(b2p11+ a2q20)
+ [2a552a4+ (7a55a44− 2a452)b2a2+ 3a442b4]a2a55p20 (4.57b) + [3a552a4+ (9a55a44− 4a452)b2a2+ 2a442b4]b2a45q02
H30δ2= a55[(2a452− a55a44)b2−1
3a255a2](b2p02− a2p20+ a2q11) + a45[a552a2+ (a55a44−4
3a452)b2](b2p11−b2q02+ a2q20), (4.57c) H21δ2= a45a55(3a44b2− a55a2)(b2p02− a2p20+ a2q11)
+ a44[3a552a2+ (a55a44− 2a452)b2](b2p11−b2q02+ a2q20), (4.57d) where a0= a44a55− a245, δ1= a55a2+ a44b2and δ2= a0(3δ21+ 4a0a2b2).
4.3.3.2 The Biharmonic BVP
Employing the parametrization functions x(t), y(t) and employing trigonometric identities, en-able us to write the l.h.s. of (4.40) as
JΦx(t) =
∑
(km=1Φ−1)kpJΦxsmsin(mt) + JΦxcmcos(mt),JΦy(t) =
∑
(km=1Φ−1)kpJΦysmsin(mt) + JΦycmcos(mt). (4.58) Similar to S.4.3.3.1, in order to ensure that JΦx(t) + F1(t) ≡ 0 and JΦy(t) − F2(t) ≡ 0 one should enforce the following equations for m= 1,...,(kΦ− 1)kp:JΦxcm+ F1cm= 0, JΦycm− F2cm= 0, JΦxsm+ F1sm= 0, JΦysm− F2sm= 0, (4.59) which yields a total of 4(kΦ− 1)kpequations. For kp= 1 there are 4kΦ− 4 equations. Yet, in view of the single-value condition for Φ, the rank of this linear system is 4kΦ− 5, see Remark 4.2. Hence, for kp= 1 the number of equations and coefficients Bi jis the same (see discussion of S.4.3.2) and the above system is solvable in an exact manner. Again, when kp>
1, conditional polynomial solutions may be obtained under assumed relations between the boundary polynomials coefficients pj,i,s−i, qj,i,s−i. P.4.10 presents such exact and conditional polynomial solutions for homogeneous biharmonic BVPs.
Remark 4.2 The rank of the linear system (4.59) is lower by one than the number of equations due to the single-value conditions discussed in S.4.3.3. The conditions F1c0 = F2c0= 0 are identically fulfilled since we earlier assumed JΦxc0= 0 and JΦyc0= 0 by taking the summation in (4.58) from m= 1. Hence only one condition should be imposed, i.e., Rc0= 0, see (4.44). This additional condition lowers the system rank from 4 kΦ− 4 to 4kΦ− 5. For higher kpvalues (depending on the geometry, and mainly on the amount of “symmetry ” it possesses), larger reduction in the system’s rank is possible.
Example 4.5 Exact Polynomial Solution of the Biharmonic BVP in an Ellipse (Low Order).
For solving the homogeneous biharmonic problem for an ellipse, we activate P.4.10 with case= caseb, iso = 0, k1(= kQ) = 1, k0= 0. In addition, we work with symbolic bi j, set x(t) = acos t, y(t) = bsin t and kp= 1, and close “Ex=rot” command. We therefore assume the following third-degree polynomial solution (with kΦ= kQ+ 2 = 3)
Φ= B20x2+ B11xy+ B02y2+ B30x3+ B21x2y+ B12xy2+ B03y3, (4.60) while Pj= pj 00+ pj10x+ pj 01y, Qj= qj 00+ qj10x+ qj 01y . The single-valued conditions in this case become
p100= q200, p110= −q101, p210= −q201. (4.61) We therefore arrive at 4kΦ− 4 = 8 equations, but the system rank is only 4kΦ− 5 = 7, which is the number of unknowns in (4.60). The system solution yields
B20=1
2q100, B02=1
2p200, B11= −q200, B12= −1
2q201, B21=1 2q101, B30= 1
6a2(b2p101+ a2q110−b2q201), B03= 1
6b2(a2q101+b2p201+ a2q210). (4.62) As shown, the above result is not material-dependent since it deals with a low-order polynomial that satisfies the biharmonic equation identically.
Example 4.6 Conditional Polynomial Solution of the Biharmonic BVP in a Triangle.
To examine the conditional solution of the homogeneous biharmonic problem in a triangle, we activate P.4.10 with case= caseb, iso = 0, k1(= kQ) = 1, k0= 1. In addition, we work with symbolic bi j and set x(t) = −sin t +13sin(2t), y(t) = cos t +13cos(2t) and kp= 2 (which puts the system origin at the center of the triangle). The solution is therefore limited
to the loading Pi= pi10x+ pi01y, Qi= qi10x+ qi01y (i = 1,2). P.4.10 shows that the three single-valued conditions of (4.41) impose the data relations
p110= −q101, q201= −p210 (4.63)
and that this solution is a conditional one. In other words, an exact polynomial solution is possible only for specific loading distributions. In the present case, in order to form a valid loading distribution, the following additional relations are required:
q210= −q101, p210= −p101. (4.64)
The stress function therefore becomes Φ=16p201y3−12p101xy2+12q101x2y+16q110x3. As a special case, suppose that q110= 1 and all three other coefficients vanish. Thus, we obtain Φ=
1
6x3. This yields F2= 0 and F1(t) = −16sin t−12sin(2t) +12sin(3t) −19sin(4t). According to (3.65), this case may be generated by zero body-force and a surface loading of Xs(t) = 0, Ys(t) = F1(t), as schematically shown in Fig. 4.1.
Figure 4.1:A conditional solution for the triangular domain of Example 4.6.
Example 4.7 Exact Polynomial Solution of the Biharmonic BVP in an Ellipse (Higher Order).
We shall present here a symbolic solution of the homogeneous biharmonic problem in an ellipse, when the boundary functions, Fj= Pjcos(¯n,x) + Qjcos(¯n,y), j = 1,2, are described by (4.42) with kQ= 3. Such a case encompasses most of the applications encountered in beam analysis with low-order loading, see Chapter 7. Activating P.4.10 as in Example 4.5 but with k1(= kQ) = 3 yields the fifth-degree prescribed biharmonic polynomial
Φ= B50x5+ B41x4y+ B32x3y2+ B23x2y3+ B14xy4+ B05y5+ B40x4+ B31x3y+ B22x2y2 + B13xy3+ B04y4+ B30x3+ B21x2y+ B12xy2+ B03y3+ B20x2+ B11xy+ B02y2, (4.65) while the three single-value conditions are
4p210= −4q201−a2(q221+ 3p230) −b2(p212+ 3q203), 4p110= −4q101−a2(q121+ 3p130) −b2(p112+ 3q103),
4p100= 4q200+ a2(q220− 3p120− q111) +b2(p211+ 3q202− p102). (4.66) The underlined terms in the above and the following expressions should be ignored if one uses k1(= kQ) = 2 only. Since the detailed formulas for the involved Bi jterms are quite lengthy, we
have organized them as shown below by suitable algebraic simplification. P.4.11 proves that the following presentations and those obtained directly from P.4.10 are identical:
B50= − 1
80a4[(−40B32+4q112−4p121+q221−p230)a2b2+(4p103+p212−q203)b4−4q130a4], B41= − 1
16a2δ3
[2b2(b16a2+ b26b2)δ4+ a2(5b22b4+ a2b2(2b12+ b66) + b11a4)δ5],
B32= − 1 8a2δ3
[2a4(b16a2+ b26b2)δ5+ (5b11a4+ a2b2(2b12+ b66) + b22b4)δ4],
B23= − 1
8b2[(q121− p130− 16B41)a2+ (p112− q103)b2], B14= − 1
16b2[(p230− q221− 8B32)a2+ (q203− p212)b2], B05= − 1
40b4[(−40B41+3q121+2q230−3p130)a4+(3p112+2p221−3q103−2q212)a2b2−2p203b4], B40= − 1
12a2[(−6B22+ q102− p111)b2− q120a2], B31= − 1
24a2[(3p120− 3q111− q220)a2+ (q202− p211− 3p102)b2],
B22= − 1
2(3b11a4+ a2b2(2b12+ b66) + 3b22b4)[b11(q211− p220)a4 + b22(p111− q102)b4− (6b26B31+ 6b16B13− b22q120− b11p202)a2b2], B13= − 1
24b2[(p120− q111− 3q220)a2+ (3q202− 3p211− p102)b2], B04= − 1
12b2[(−6B22− q211+ p220)a2− p202b2], B30= 1
6a2[(−6B32+ q112)a2b2+ (p103− q203)b4+ (p101− q201)b2+ q110a2], B21= −1
2[(p130+ 4B41)a2+ p110], B12=1
2[(−2B32+ p230)a2+ p210], B03= 1
6b2[(−12B41+ q230− p130+ q121)a4+ (p112+ p221)a2b2+ (q210+ q101)a2+ p201b2], B11= −1
8[(5p120+ 3q111+ q220)a2+ (3p102+ p211− q202)b2] − p100, B20=1
2[(−2B22+ q102)b2+ q100], B02=1
2[(−2B22+ p220)a2+ p200], (4.67) where
δ3= [5b22b4+ a2b2(2b12+ b66) + b11a4][5b11a4+ a2b2(2b12+ b66) + b22b4]
− 4a2b2(b16a2+ b26b2)2,
δ4= [2(q121− p130)b16+ (q221− p230)b11]a6+ (p230− q221+ 4p121− 4q112)b22a2b4 + [2b16(p112− q103) + (p212− q203)b11+ 4b22q130]a4b2+ (q203− p212− 4p103)b22b6,
δ5= 2b11(3p130− 2q230− 3q121)a4+ [(2b12+ b66)(p130− q121) + 2b16(p230− q221) + 2b11(3q103+ 2q212− 2p221− 3p112)]a2b2
+ [(2b12+ b66)(q103− p112) + 2(q203− p212)b16+ 4b11p203]b4. (4.68)
Example 4.8 Exact Polynomial Solution of Non-Homogeneous Biharmonic BVP in an Ellipse.
Suppose now that the biharmonic problems discussed in Examples 4.5, 4.7 are non-homoge-neous, and we wish to transform them into their homogeneous form by expressing the poly-nomial Φ(x,y) as the sum Φ=Φp+Φ0 of a particular and a homogeneous solution.
By keeping the kQ= 3 and kΦ= 5 selections (and again ignoring the underlined terms for kQ= 2 and kΦ = 4), the maximal degree level of F0 is kΦ− 4 = 1 (see S.4.2.1). Hence,
∇(4)1 Φp= F00+ F10x+ F01y, and we reach a particular solution of the form Φp= F10
120b22
x5+ F01
120b11
y5+ F00
24b11
y4. (4.69)
According to (4.14c), we may use the expressions derived in Example 4.7 to determineΦ0, by modifying the coefficients of Pj, Qjas
p203⇒ p203− F01
6b11, p202⇒ p202−F00
2b11, q130⇒ q130− F10
6b22. (4.70) Substituting the above pi jk, qi jkcoefficients in the Bi jexpressions derived in Example 4.7, we reach the following fifth-order polynomial Φ(x,y):
Φ= (B50+ F10
120b22)x5+ B41x4y+ B32x3y2+ B23x2y3+ B14xy4+ (B05+ F01
120b11)y5+ B40x4 + B31x3y+ B22x2y2+ B13xy3+ (B04+ F00
24b11)y4+ B30x3+ B21x2y+ B12xy2+ B03y3
+ B20x2+ B11xy+ B02y2. (4.71)
4.3.3.3 The Coupled-Plane BVP
Combining the definitions of S.4.3.3.1, S.4.3.3.2, we write the boundary conditions of the homogeneous Coupled-Plane BVP as the system (4.48), (4.59), where kΛ= kΦ− 1.
For kp= 1 the total number of equations and coefficients is the same, 6kΦ−7, and the above linear system is solvable in an exact manner. Conditional solutions may be obtained for kp> 1 in this case as well. P.4.12 presents exact and conditional polynomial solutions for the above Coupled-Plane BVP.
Example 4.9 Exact Polynomial Solution of Coupled-Plane BVP in an Ellipse (Low Order).
To examine the homogeneous problem presented by (4.16a), (4.38a), (4.40), we activate P.4.12 with k1(= kQ) = 1, k0= 0 and hence with kΦ= 3, kΛ= 2. The program yields
Φ= B20x2+ B11xy+ B02y2+ B30x3+ B21x2y+ B12xy2+ B03y3,
Λ= H10x+ H01y+ H20x2+ H11xy+ H02y2, (4.72) where H02=b155(b45H11−b44H20−3b15B03+B12b14+B12b56−B21b25−B21b46+3b24B30).
Here and below, the underlined terms are not used for the case of k1(= kQ) = 0 (i.e., kΦ= 2, kΛ= 1). According to the above selection,
Pj= pj 00+ pj10x+ pj 01y, Qj= qj 00+ qj10x+ qj 01y, j= 1,2,
P3= p300, Q3= q300. (4.73)
P.4.12 also shows that the four conditions of (4.39), (4.41) impose (only) the following three