303 WK5TPCSWEB.TEX
ME M.M. Yovanovich
Week 5
Lecture 1
Sturm-Liouville Problem SLP Cartesian Coordinates. u = ux; y or u = ux; t. Partial di erential equations.
uxx + uyy = 0; 0 x L; 0 y H 1D Di usion Equation: uxx = 1 ut; t 0; 0 x L 1D Wave Equation: uxx = c1 utt; t 0; 0 x L where and c are positive constants. Separation of Variables Method SVM is used to obtain two sets of independent ODES. Let ux; y = X xY y and let ux; t = X xT t. One very important ODE appears in all three cases.
2D Laplace Equation:
2 2
X 00 + X = 0;
2
0 x L
Homogenous BCs at x = 0 and x = L are
i X 0 = 0 or ii X 00 = 0 or iii , kX 0 0 = hX 0
where h and k are positive thermophysical parameters. These are homogeneous BCs of the rst kind Dirichlet, of the second kind Neumann, and of the third kind Robin. Similarly
i X L = 0 or ii X 0L = 0 or iii , kX 0 L = hX L
There are nine combinations of these homogeneous BCs.
Solution of ODE.
X x = C cos x + C sin x; 0 x L and its derivative is X 0x = ,C sin x + C cos x
1 2 1 2
Eigenfunctions are
cos x
and
sin x
Three combinations of the homogeneous BCs. a X 00 = 0 and X L = 0 b X 00 = 0 and X 0L = 0 a X 00 = 0 and , kX 0L = hX L Case a requires that
X 00 = ,C sin 0 + C cos 0 = 0
1 2 2
C =0 Therefore C = 0 or = 0. Since = 0 gives a trivial solution, it will be rejected, and we take C = 0. Solution is X x = C cosx and X 0x = ,C sinx
2 2 1 1
Apply second homogeneous Dirichlet condition to get
C cos L = 0 Either C = 0 which gives a trivial solution, and therefore this option is rejected, or cos L = 0. This is satis ed when n L = 2n , 1 ; n = 1; 2; 3; : : : 2
1 1
Eigenvalues are
n = 2n , 1 2 ; n = 1; 2; 3; : : : L
2
Many math texts call 2 the eigenvalues. Most engineering texts call n the n eigenvalues. Eigenfunctions which satisfy the ODE and the two homogeneous Dirichlet conditions at x = 0 and x = L are x ; n = 1; 2; 3: : : : Xn x = Dn cos 2n , 1 2L where Dn are arbitrary constants.
X 00 and X 0L = 0 The rst homogeneous Neumann condition requires C = 0 and X x = C cosx and X 0x = ,C sinx as in Case a. The homogeneous Neumann condition at x = L requires that
2 1 1
Case b
X 0L = ,C sinL = 0
1
Both options C1 = 0 and = 0 give trivial solutions, therefore they are rejected. Eigenfunctions for Case b are sinn L = 0; n = 1; 2; 3; : : : which require
n L = n ; n = 1; 2; 3; : : : n = n ; n = 1; 2; 3; : : : L
Eigenvalues for Case b are
Eigenfunctions which satisfy the ODE and the two homogeneous Neumann conditions at x = 0 and x = L are Xn x = Dn cos nx ; n = 1; 2; 3: : : : L where Dn are arbitrary constants. Case c
X 00
and , k X 0L = hX L 3
The rst homogeneous Neumann condition requires C2 = 0 and X x = C1 cosx and X 0x = ,C1 sinx as in Case a and Caseb. The homogeneous Robin condition at x = L requires that
,kX 0L = hX L or , k ,C sinL = hC cosL
1 1
The option C1 = 0 gives a trivial solution, therefore it is rejected. Cancel the common C1 and rewrite the relation as h cosL = sinL k The relation is dimensional because both and h=k have the same units m,1 . The argument of the cosine and sine functions is L which must be dimensionless. Characteristic Equation Multiple through by L and de ne dimensionless parameters: = L and Bi = hL 0 k Now the characteristic equation becomes: sin = Bi cos and 0 Bi 1 This is a transcendental equation. Numerical methods are required to obtain its roots for a given value of the parameter Bi. The Newton-Raphson iterative method can be used to obtain the roots. For a given value of Bi the in nite set of roots called eigenvalues are:
1 2 3
n
n+1
n+2
Lecture 2
Makeup Lecture Number 2. Discuss Project Number 1. Physical interpretation of equations and the solution. Some Problems from Spiegel's Text: p. 584, A Excercises: 3 4
B Excercises: 2, 3 ******************** p. 592, A Excercises: 1, 3 ******************** P. 593, B Excercises: 2, 3 **************** p. 615, A Excercises: 1, 2, 3, 4, 5 *********************** p. 616-617, B Excercises: 1, 2, 6, 7 ********************** p. 617, C Excercises: 8 ************** Limiting values of parameter Bi. For Bi = 0, the characteristic equation becomes sin = 0. Since cannot be set to zero it gives a trivial solution, we set sin = 0. The roots eigenvalues are n = n ; n = 1; 2; 3; : : : For Bi = 1 write the equation as cos = Bi sin and now set Bi = 1 to give cos = 0. The roots eigenvalues are n = 2n , 1 ; n = 1; 2; 3; : : : 2 Location of Roots Eigenvalues. Location of the roots are easily found graphically. They lie in the intervals: 0 Bi 1 0 =2 1 3=2 2 2 5=2 3 n , 1 2n , 1=2 n 5
Maple, Mathcad and Matlab can nd these roots quickly and accurately. Di erence between two consecutive eigenvalues. as n ! 1 n+1 , n ! Eigenfunctions which satisfy the ODE and the homogeneous Neumann and Robin conditions at x = 0 and x = L respectively are ! n Xn x = Dn cos Lx ; n = 1; 2; 3: : : : where Dn are arbitrary constants. Approximation of the First Eigenvalue, 1. p As Bi ! 0; Bi 1 ! and As Bi ! 1; 1 ! 2 These limits are used to develop the following approximation proposed by M.M. Yovanovich: =2 and m = 2:15 1 = " ! =2 m 1=m 1+ p Bi This correlation is equation valid of all values of Bi, i.e., 0 Bi 1, and it provides acceptable accuracy for the calculation of 1 for most engineering applications. Newton-Raphson Iterative Method. The the nth root n of the arbtrary function f is obtained by means of the relation: k k +1 = nk , f0 nk ; k = 1; 2; 3; : : : n fn where k represents the kth iteration. For the characteristic equation, sin , Bi cos = 0, we have the relation: k k k k +1 = nk , k n sink n , Bi cos n k ; k = 1; 2; 3 : : : n n cos n + 1 + Bi sin n 6
1 The rst guess of the rst root 1 this characteristic equation can be based on the approximation given above. Generally three to four iterations will provide very accurate values for the rst root. The rst guess for the second root should be based on 1 = 1converged value + 2 This process can be followed to calculate all required roots which may be as many as several hundred for a particular problem. Maple and Mathcad can calculate these roots very quickly and accurately. See ME 303 Web site: Maple WS called CECART.MWS.
Lecture 4
See Spiegel's Text: Sections 2.1 and 2.2 on pages 585-591. Separation of Variables Method SVM applied to 1D wave equation. uxx = c12 utt; t 0; 0 x L and the system constant is c2 = T= where T is the tension in the elastic string and is the linear density of the string. The units of c2 m2=s2 and c m=s . See the development of PDE in the text. Boundary and Initial Conditions. The ends are xed, therefore the homogeneous boundary conditions of the rst kind Dirichlet are: t 0; u0; t = 0; uL; t = 0 The two initial conditions are based on the initial displacement and the initial velocity: @ux; 0 = gx t = 0; 0 x L; ux; 0 = f x; @t Separated ODES. Let ux; t = X xT t; substitute into the PDE to get the separated relationship: X 00 = 1 T 00 X c2 T The identity must hold for all time t 0 and any value of x in interval 0; L . There are three options: lhs = rhs = i 0; ii , 2 ; iii 2. 7
Option i gives the separated ODEs:
X 00 = 0;
and their solutions are:
T 00 = 0
and
X x = C x + C ;
1 2
T t = C t + C
3
4
Both solutions are linear in time and space. Option ii gives the separated ODEs:
X 00 + X
2
and and
T 00 + c T = 0
2 2
and their solutions are :
X x = C cosx + C sinx
1 2
T t = C cosct + C sinct
3 4
Both solutions are periodic functions of space and time respectively. Option iii gives the separated ODEs:
X 00 , X;
2
and and
T 00 , c T = 0
2 2
and their solutions are :
X x = C coshx+C sinhx
1 2
T t = C coshct+C sinhct
3 4
Both solutions are non-periodic functions of space and time respectively. The solutions for option iii can also be written in terms of exponentials:
X x = C expx+C exp,x
1 2
and
T t = C expct+C exp,ct
3 4
The xed end conditions require:
u0; t = X 0T t = 0 = X 0 = 0 and uL; t = X LT t = 0 = X L = 0
These homogeneous BCs require C1 = 0 and C2 = 0 in options i and iii. These solutions are not applicable to the xed ends string. These conditions when applied to option ii solution require:
C =0
1
and 8
C sinL = 0
2
Since C2 = 0 leads to a trivial solution, it is rejected...