MATHEMATICS 2 Continuation of Applied Mathematics and Numerical - PowerPoint PPT Presentation

     b   b     2 2           2 min b a L min ( ) x a f x ( ) dx ( ) x a f x ( ) dx 0 2  a a a a a b b                 T T T 2 ( ) x ( ) x a f x ( ) dx 0 ( ) ( ) x x a ( ) ( ) x f x dx 0 a a b b         T T ( ) ( ) x x dx a ( ) ( ) x f x dx a a     1 Aa B a A B   b b b          ( ) x ( ) x dx ( ) x ( ) x dx ... ( ) x ( ) x dx   1 1 1 2 1 m a a a   b b b          ( ) x ( ) x dx ( ) x ( ) x dx ... ( ) x ( ) x dx     b b     T       A ( ) ( ) x x dx ( ) x ( ) x dx 2 1 2 2 2 m a a a      i j   ( m m ) a a  ... ... ... ... i j , 1,... m     b b b          (easy) ( ) x ( ) x dx ( ) x ( ) x dx ... ( x ) ( ) x dx     m 1 m 2 m m a a a   b   ( ) ( ) x f x dx   1 a   b   ( ) ( ) x f x dx     b b        T 2 B ( ) ( ) x f x dx ( ) ( ) x f x dx a       i  a a ( m 1)  ... i 1,..., m   (difficult)   b   ( ) ( ) x f x dx     m a

for monomial basis functions: m      j 1     m 2  m 1 p x ( ) a x a a x ... a x a x  j 1 2 m 1 m  j 1     p q ( ) x x , ( ) x x i j b        p q 1 p q 1 p q 1 x b a b b         p q p q A x x dx x dx i j ,     p q 1 p q 1 a a a   b  f x dx ( )   a   troublesome (due to f ( x ) ), b   x f x dx ( )   b     p B x f x dx ( ) usually numerical integration required a  i a ...     b   m 1 x f x dx ( )     a

numerical integration (division into k subintervals)  b a  t k j k 1        x a k 1 t x b x a   x a t        x a jt x a j 1 t for trapezoidal rule:      x a j 1 t 1 t k b          p  p   p   B x f x dx ( ) x f x x f x , i 1,2,..., m     x a jt i 1 1 2 2 2 a 2  j 1 for Gaussian rule (2 points)     2 a 2 j 1 t t    x 1 2 2 3 k t b               p p p   B x f x dx ( ) x f x x f x , i 1,2,..., m   2 a 2 j 1 t   t i 1 1 2 2 2   x a  j 1 2 2 2 3

Are weights applicable in case of continuous approximation? weight function may control the accuracy in continuous manner (on the whole interval)    y ( ) x 0 smaller approx. error larger approx. error   b b            T A ( ) ( ) ( ) x x x dx ( ) x ( ) ( ) x x dx     i j  ( m m ) a a  i j , 1,... m   b b              T B ( ) ( ) x f x x dx ( ) ( ) x f x x dx    i  a a ( m 1)  i 1,..., m

Are weights applicable in case of continuous approximation?   b b b          ( ) x ( ) x dx ( ) x ( ) x dx ... ( ) x ( ) x dx   1 1 1 2 1 m        a a a   0 0   b b b          ( ) x ( ) x dx ( ) x ( ) x dx ... ( ) x ( ) x dx   2 1 2 2 2 m        a a a    A  0 0   ( m m ) ... ... ... ...     b b b          ( ) x ( ) x dx ( ) x ( ) x dx ... ( x ) ( ) x dx   m 1 m 2 m m       a a a     0 0 orthogonal (with weight) basis functions: orthogonal basis functions:       0 , i j 0 , i j b b            ( ) x ( ) x x dx ( ) x ( ) x dx   i j  i j  0 , i j 0 , i j  a  a orthonormal (with weight) basis functions: orthonormal basis functions:      0 , i j  0 , i j b b            ( ) x ( ) x dx ( ) x ( ) x x dx   i j i j   1 , i j 1 , i j a  a 

Example          f x sin x x 0, 2 Given is . Find its in    - discrete LS approximation (apply three nodes regularly spaced), - weighted LS discrete approximation (apply  = 10 for the first node) - continuous LS approximation - continuous weighted LS approximation (apply  ( x ) = x )     2 p x a a x Apply 1 2 Each time, evaluate the approximation error (as a function and a norm).      f x sin x     2 p x 0.2110 0.3483 x     2 p x 0.0306 0.4343 x     2 p x 0.3279 0.3754 x     2 p x 0.4782 0.2694 x

      f x sin x , n 8     x 10,10 2nd approach to overcome Runge effect: optimal distribution of nodes n    ( ) n   estimation of the interpolation error ( ) x f ( x x ) max i  i 1 n   min max ( x x ) Chebyshev problem: find min of max error i    x 1 x 1 i  i 1 Solution: the solution to the above problem are the roots of Chebyshew polynomials

Chebyshev polynomials        T ( ) x cos m 1 arc cos x iterative definition m   T x ( ) 1 recursive definition 1   T x ( ) x  2          T ( ) x 2 x T ( ) x T ( ) x , m 3 1 x 1    m m 1 m 2 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

Formula for Chebyshev polynomials’ roots      2 i 1     x cos , i 1,2,..., m 1      1 x 1 i   m 1 2     z a b , , x 1,1 Transformation between intervals: 1       x z : z [( b a ) x ( b a )] 2   2 z ( b a )   z x : x  b a Orthogonality of Chebyshev polynomials   0, i j   1 T x T x ( ) ( )     i j    I dx , i j 1  ij 2  2 1 x   1     , i j 1 

Example          z 0, 2 f z sin z Given is in . Find:    - approximation data with nodes as roots of fourth Chebyshev polynomial - SLAE for interpolation method using Chebyshev polynomials            p z a T z a T z a T z Apply 1 1 2 2 3 3 solution:           0.8660 1.4656,0.9945 , 0 0.7854,0.7071 , 0.8660 0.1052,0.1050    a      1 0.8660 0.5 0.9945 0.6022 1             1 0 1 a 0.7071 a 0.5135         2           1 0.8660 0.5 a 0.1050 0.1049         3   2  4   4                       p z a T z a T z a T z 0.6022 1 0.5135 z 1 0.1049  2 z 1 1        1 1 2 2 3 3             2 0.3401 z 1.1881 z 0.0162

y = sin( x ) Lagrange Chebyshev 3 points – 2nd order error = 0.0162 error = 0.0187 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5

y = sin( x ) Lagrange Chebyshev 8 points – 7th order error = 1.0825 error = 5.3767 5 4 3 2 1 0 -1 -2 -3 -4 -5 -10 -8 -6 -4 -2 0 2 4 6 8 10

y = sin( x ) Lagrange Chebyshev 12 points – 11th order error = 1.3704 error = 36.0440 30 20 10 0 -10 -20 -30 -15 -10 -5 0 5 10 15

Orthogonality (with weight) of Chebyshev polynomials SLAE for Chebyshev interpolation:    1 0.8660 0.5  a  0.9945  1           1 0 1 a 0.7071 cond 1.4143       2        1 0.8660 0.5 a 0.1050       3 SLAE for monomial interpolation:    1 1.4656 2.1480  a  0.9945  1          1 0.7854 0.6169 a 0.7071 cond 15.8567       2       1 0.1052 0.011 a 0.1050       3   0, i j   1 T x T x ( ) ( )     i j    I dx , i j 1  ij 2  2 1 x   1     , i j 1 

Generation of orthogonal (orthonormal) (with weight) basis functions - Gramm – Schmidt method orthogonal (with weight) basis functions: orthogonal basis functions:       0 , i j 0 , i j b b            ( ) x ( ) x dx ( ) x ( ) x x dx   i j  i j  0 , i j 0 , i j   a a orthonormal (with weight) basis functions: orthonormal basis functions:      0 , i j  0 , i j b b            ( ) x ( ) x dx ( ) x ( ) x x dx   i j i j   1 , i j 1 , i j a  a  scalar product     b b                    , ( ) x ( ) x dx , x ( ) x ( ) x dx i j i j i j i j a a

Generation of orthogonal (orthonormal) (with weight) basis functions - Gramm – Schmidt method       f x , j 1,2,..., m , x a b , j Algorithm t  for orthogonal 1 1       g x t f x 1.     12       1 1 1 g , g 1 t f , f for orthonormal 1 1 1 1 1    f , g             g x f x g x      2 1 2. g , g 0   2 2 1,2 1 2 1 1,2  g g , 1 1         12          g x t g x   for orthonormal g , g 1 t g , g 2 2 2 2 2 2 2 2                g x f x g x g x 3. 3 3 1,3 1 2,3 2      f , g  f , g          3 2 g , g 0      3 1 g , g 0     3 2 2,3  3 1 1,3 g , g  g g , 2 2 1 1         12          g x t g x   for orthonormal g , g 1 t g , g 3 3 3 3 3 3 3 3

Generation of orthogonal (orthonormal) (with weight) basis functions - Gramm – Schmidt method       f x , j 1,2,..., m , x a b , j General formula for all cases  p 1                  g x f x g x / x g x p p i p , i j  i 1  p 1               g , g f , g g , g p j p j i p , i j       i 1 0      g , g j p , j j    f , g p j      , p 1,2,..., m , j 1,2,..., p 1   j p ,  g , g j j for orthonormal (with weight) case     12             g  x t g x , g  , g  1 t g , g p p p p p p p p

Example Given is          i 1 f x x , i 1,2,3 , x 0, 2  i   - orthogonal series     - orthonormal series with weight x x - use orthogonal series as basis functions for SLAE from previous example Solution:    g x 1    f x 1 1 1          f x x g x x 2 2 4    2 f x x      3 2   2          2 2 g x x x x x   3 12 2  4  2 24    1 0.6802 0.2571  a  0.9945  1           1 0 0.2056 a 0.7071 cond 4.6358       2        1 0.6802 0.2571 a 0.1050       3

Interpolation using spline functions 12 10 8 6 4 2 0 -2 -4 -4 -2 0 2 4 6 8 10

Construction of quadratic spline   x , y n n 12 10         2 2      s x p x b x x b x x 2 2 3 3   8       2 6    s x p x b x x 2 2  4   x , y 3 3 2   x , y 0 2 2  n 1       2 s x ( ) p x b x ( x )   -2 x y ,  i i 1 1  i 2 -4 -4 -2 0 2 4 6 8 10         2 s x p x a x a x a 1 2 3

Construction of the k- th order spline  ( , x y ), i 1,2,..., n Basic data: i i  x x x , spline on the 1st interval 1 2 .    k 1             k 1 k 2 k 1 i p x ( ) a x a x ... x a a a x  k k 1 2 1 i  i 1    n 1 k 1 n 1            k k 1 i k s x ( ) p x ( ) b x ( x ) a x b x ( x )   i i i i i spline on    i 2 i 1 i 2 other intervals   k ( x x ) , for x > x    k i i , ( x x )  i  0, for x x  i  n 1      s x ( ) a x a b x ( x ) linear spline ( 2 + n - 2 = n unknows)       1 2 i i  i 2   p x  n 1   2     2 s x ( ) a x a x a b x ( x ) quadratic spline ( 3 + n - 2 = n+ 1 unknows)      1 2 3 i i  i 2 p x ( )  n 1        3 2 3 s x ( ) a x a x a x a b x ( x ) cubic spline     1 2 3 4 i i  i 2 ( 4 + n - 2 = n+ 2 unknows)   p x

Additional data for quadratic spline 12 10 8 6 4 2   x , y 0 2 2     f ' x 1   x y , -2 1 1 -4 -4 -2 0 2 4 6 8 10         2 s x p x a x a x a 1 2 3

Additional data for cubic spline   x , y n n 12 10     f ' x n n 8 6   x , y   n 1 n 1 4 2   x , y 0 2 2     f ' x 1 1   x y , -2 1 1 -4 -4 -2 0 2 4 6 8 10          3 2 s x p x a x a x a x a 1 2 3 4

Additional data for cubic spline   x , y n n 12 10 8 6   x , y   n 1 n 1 4   x , y 2 2 2      0 f ' x  1        f '' x  x y ,  -2 1 1 1 -4 -4 -2 0 2 4 6 8 10          3 2 s x p x a x a x a x a 1 2 3 4

    1 Aa B a A B Determination of p ( x ) ( parameters a ) linear spline              s x ( ) y a x a y x 1 a y a ...  1 1  1 1 2 1  1 1  1  1              s x ( ) y a x a y x 1 a ... a y           2 2 1 2 2 2 2 2 2 2 quadratic spline      2    2      s x ( ) y a x a x a y x x 1 a y  a ... 1 1 1 1 1 1 2 1 3 1 1 1 1                   2 2 s x ( ) y a x a x a y x x 1 a y a ...          2 2 1 2 2 2 3 2 2 2 2 2 2                I  s ( ) x 2 a x a 2 x 1 0 a a ...          1 1 1 2 1 3 3 cubic spline         3  2    3 2 a  y   s x ( ) y a ... a x a x a x a y x x x 1 1 1 1 1 1  1 1 2 1 3 1 4 1 1 1 1               s x ( ) y 3 2 3 2 a y a ...   a x a x a x a y x x x 1        2 2   2  2  2 1 2 2 2 3 2 4 2 2 2 2             I      s ( x ) 2 2 a a ... 3 a x 2 a x a 3 x 2 x 1 0    1 3 3 1 1 2 1 3 1 1          II        s ( x ) 6 a x 2 a  6 x 2 0 0  a a ...          1 4 1 1 2 1 4

Determination of parameters b   for x , y 3 3  y p x ( )       k s x ( ) p x ( ) b x ( x ) y b 3 3 3 3 2 3 2 3 2  k ( x x ) 3 2   for x , y 4 4    k y p x ( ) b x ( x )         k k s x ( ) p x ( ) b x ( x ) b x ( x ) y b 4 4 2 4 2 4 4 2 4 2 3 4 3 4 3  k ( x x ) 4 3 …    for x x 1 , j 2,3,..., n 1  j j     k   s x ( ) p x ( ) b x ( x ) y     j 1 j 1 i j 1 i j 1  i 2  j 1        k k p x ( ) b x ( x ) b x ( x ) y     j 1 i j 1 i j j 1 j j 1  i 2  j 1     k y p x ( ) b x ( x )    j 1 j 1 i j 1 i     b i 2 , j 2,3,..., n 1 j  k ( x x )  j 1 j

Example Given is the set of points: x 0 1 2 3 0 1 3 8 y Find quadratic spline, assume y’(0) = 0    2  x , x 0 1         2  2      s x x x 1 2 x 1 , x 1 2         2 2         2 2 x x 1 3 x 2 3 x 10 x 11 , x 2 3  

2nd ORDER PARTIAL DIFFERENTIAL EQUATIONS • Classification • Elliptic equations: local and global (variational) formulations • Fundamentals of FDM (finite difference method) • Laplace and Fourier elliptic equations • Parabolic equations: explicit and implicit time formulas • Hyperbolic equations

Classification      2 2 2 u u u u u             a b c d e g u f , a 0 b 0 c 0       2 2 x x y y x y    u u x y , unknown function       x y , problem domain a b c d e f g , , , , , , known coefficients: functions or numbers    2 b 4 ac   0 elliptic type   0 parabolic type   0 hyperbolic type Equation type may change within Ω

Elliptic equations Examples: static analysis in elastic range (plane stress, plane strain, bar torsion) stationary heat flow analysis filtration through the porous system distribution of the magnetic and electric charges linear elasticity heat flow analysis (0) natural (Neumann b.c.) t  q  n f  b     q u   T (0) u essential (Dirichlet b.c.)

Heat flow analysis (a good start for local / standard (M)FDM) local formulation  2 T C     T T    2 2 T T   T         k k f in           t D T f in x y x  y    2 2 x y      k 0       T T on T T on ,   x   T T D      0 k  T       t   n D T q on y      k q on  q n q   n     n n n   x y variational formulation    1 1 find such T H H T 0     v T v T                  1 k k d v x y f x y d , , v x y qd , , v H x y 0     x x y y    q                         t 1 v x y , D T x y , d v x y f x y d , , v x y qd , , v H 0    q a good start for most of the computational approaches (FEM, variational (M)FDM)

Example Given is the differential equation 1          u 1 a u u u f x y , xx xy yy x 2 Problem domain constitutes the figure limited by: - the bottom right quarter of the circle   2 2 x y 1    - the skew line y x 1 - the horizontal straight line y = 0 . Along the semicircle and skew line, natural b.c. are ascribed, along the horizontal line – essential b.c. - determine such values of parameter a that the above equation remains elliptic. - determine the right hand side function f ( x , y )   knowing the solution of this equation is  2  2 u x y , x y - write the b.c.: essential and natural ones for any point ( x , y )

Finite difference method (FDM) recap difference operators: replacement of the analytical derivatives with numerical ones y      i 1 y f ' x , f '' x 1D: i i i y  i 1    y f x FD star x h h x  x x  x i 1 i i 1 i FD formulas, FD schemes FD operators: 2D: formulas composition   T T xx yy 1   y y  i 1 i  h  h   y y 1    h h y ' i i 1  1 -4 1 i h  2 h   y y   i 1 i 1 h x   2 h   y 2 y y 1    y '' i 1 i i 1 i 2 h

Example prismatic bar torsion 5a  2  2 F F      2 G in     2 2  x y   3a C    F 0 on  F – Prandtl function 0 0 0   F F           2 2 , , zx  zy  z zx zy y x 0 1 2     v F v Fd         1 C v d , F v , H  h a     0 x x y y   0 Generate FD system of equations, by means of the: - local formulation - variational formulation Use matrix form of SLAE. Dealing with natural b.c. in FDM, evaluating stresses -> Computational methods in civil engineering

Parabolic equations Examples: diffusion non-stationary heat flow analysis           u u f , x t , , a x b , t t xx t 0    u u x t , unknown temperature function    f f x t , intensity of heat generation inside the domain boundary conditions initial condition       u a t , u t       u x t , u x a  0 0      u b t , u t   b

Parabolic equations t     u u f xx t      u t u t a b  t t 0   a a b b x u x 0  x a) problem formulation b) FD discretization

  i k    i 1, k 1 , 1 i 1, k 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k k t , known level k   x x the simplest approach : three nodes at known level, one node at unknown level i k  , 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k known level k t , k   x x

Numerical approximation of the 2nd space derivative:   u 2 u u      i 1, k i k , i 1, k u xx  2 i k , x Numerical approximation of the 1st time derivative:   u u    i k , 1 i k , u t  i k , t     u u f After substitution into differential equation : xx t    u 2 u u u u       where: u u x t , , ...  i 1, k i k , i 1, k  i k , 1 i k ,  f i k , i k i k ,   2 x t    f f x t , i k , i k One constant instead of three ones:   t    2 x Final form:             u u 1 2 u u t f    i k , 1 i 1, k i k , i 1, k i k ,

Explicit, open scheme             u u 1 2 u u t f    i k , 1 i 1, k i k , i 1, k i k , 1   t f  t i k , conditionally stable   x x λ λ 1-2λ  2 1 1 x       t t crit  2 2 leads to one value at time (no SLAE) i k  , 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k known level k t , k   x x

  i k    i 1, k 1 , 1 i 1, k 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k k t , known level k   x x more complex approach : three nodes at unknown level, one node at known level     i 1, k 1 i 1, k 1 i k  , 1  k 1, t  unknown level k 1  t i k , known level k t , k   x x

Numerical approximation of the 2nd space derivative:   u 2 u u         i 1, k 1 i k , 1 i 1, k 1 u xx   2 i k , 1 x Numerical approximation of the 1st time derivative:  u u     i k , 1 i k , u t   i k , 1 t     u u f After substitution into differential equation : xx t    u 2 u u u u          where: u u x t , , ...  i 1, k 1 i k , 1 i 1, k 1  i k , 1 i k ,  f   i k , 1 i k 1  i k , 1   2 x t    f f x t ,   i k , 1 i k 1 One constant instead of three ones:   t    2 x Final form:              u 1 2 u u t f u       i 1, k 1 i k , 1 i 1, k 1 i k , 1 i k ,

Simple implicit, closed scheme              u 1 2 u u u t f       i 1, k 1 i k , 1 i 1, k 1 i k , i k , 1 unconditionally stable   x x -λ -λ 1+2λ     t  t   t f  i k , 1 1 leads to SLAE at each time level     i 1, k 1 i 1, k 1 i k  , 1  k 1, t  unknown level k 1  t i k , known level k t , k   x x

Crank – Nicolson scheme     i 1, k 1 i 1, k 1 i k  , 1  k 1, t  unknown level k 1  t i k , known level k t , k   x x 1 1 1            u u 2 u u u u        i 1, k 1 i 1, k i k , 1 i k , i 1, k 1 i 1, k 2 2 2    u xx   i k , 1 2 x                      u 1 u u u 1 u u t f         i 1, k i k , i 1, k i 1, k 1 i k , 1 i 1, k 1 i k , 1 2 2 2 2 Complex implicit, closed scheme, unconditionally stable

Example Solve the equation:        u u , x t , , 0 x 4 , t 0 xx t     u 0, t 10   u (4, ) t 10          u x ,0 x x 4 10  Assume 5 nodes at each time level, as well as the time step, being the half of the critical time step. Perform calculations for two time steps. Apply explicit and implicit time schemes. Use symmetry conditions, if available.

Hyperbolic equations Examples: forced bar vibrations problems of theory of plasticity           u u f , x t , , a x b , t t    0 xx tt 0 E    u u x t , unknown deflection function    f f x t , mass forces boundary conditions initial conditions           u x t , u x    u a t , u t  0 0 a            u ' x t , v x    u b t , u t   t 0 0 b

Hyperbolic equations t    u u f xx tt      u t u t a b  t t 0 a a     b b x u x , v x 0 0  x b) FD discretization a) problem formulation

    i k  i 1, k 1 , 1 i 1, k 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k k t , known level k  t     i k  i 1, k 1 , 1 i 1, k 1  known level k 1, t  k 1   x x i k  , 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k known level k t , k  t i k  , 1  known level k 1, t  k 1   x x

Numerical approximation of the 2nd space derivative:   u 2 u u      i 1, k i k , i 1, k u xx  2 i k , x Numerical approximation of the 2nd time derivative:   u 2 u u      i k , 1 i k , i k , 1 u tt  2 i k , t    u u f After substitution into differential equation : xx tt     u 2 u u u 2 u u        where: u u x t , , ... i 1, k i k , i 1, k   i k , 1 i k , i k , 1  f i k , i k i k ,   2 2 x t    f f x t , i k , i k One constant instead of three ones:  2 t     2 x Final form:  2 t            u u u 2 2 u u f     i k , 1 i 1, k i k , i 1, k i k , 1 i k , 

Explicit, open scheme  2 t            u u u 2 2 u u f     i k , 1 i 1, k i k , i 1, k i k , 1 i k ,  1  2 t  f  t i k ,  conditionally stable   x x λ λ 2-2λ      x    1 t t  t kryt -1 leads to one value at time (no SLAE) i k  , 1  k 1, t  unknown level k 1  t   i 1, k i k , i 1, k known level k t , k  t i k  , 1  known level k 1, t  k 1   x x

Example Solve the equation (forced bar vibrations):        u 4 u , x t , , 0 x 4 , t 0 xx tt     u 0, t 0   u (4, ) t 0         u x ,0 x x 4      v x ,0 0  Assume 5 nodes at each time level, as well as the time step, being the half of the critical time step. Perform calculations for two time steps. Apply explicit time scheme. Use symmetry conditions, if available.

Fourier series analysis • General formulation for periodic function • General formulation for function in arbitrary interval • Beam deflection problem • Bending plate problem

Motivation Fourier analysis: - representation of the wave-like functions as the simple sin or / and cos functions - decomposition of any periodic function / signal into a finite sum of simple oscillating functions Applications: - function approximation method, - harmonic analysis (frequencies, periods, amplitudes), - noise filtering, signal processing - „exact” analytical solution of boundary value problems .

General formulation (periodic function) For any periodic function with period 2 π  a                   x f x 0 a cos kx b sin kx k k 2  k 1 Trigonometric base fulfils the orthogonality properties             cos kx dx sin kx dx 0 k               sin kx cos mx dx 0 k m ,       0 , k m        sin kx sin mx dx   , k m      0 , k m            cos kx cos mx dx , k m 0        2 , k m 0 

General formulation In search of coefficients a k and b k                f x a cos kx b sin kx / cos mx k k  k 0        f x cos mx dx 0    0 , k m            cos kx cos mx dx , k m 0        2 , k m 0    1 1               a f x cos 0 x dx , a f x cos kx dx , k 1,2,... 0 k        1        a f x cos kx dx , k 0,1,2,... k   

General formulation In search of coefficients a k and b k                f x a cos kx b sin kx / sin mx k k  k 0        f x sin mx dx 0    0 , k m         sin kx sin mx dx   , k m     1        b f x sin kx dx , k 1,2,... k   

General formulation (arbitrary function) For function in arbitrary interval:          a k x k x           x L L f x 0 a cos b sin       k k 2  L   L     k 1 L    1 k x     a f x cos dx   k L  L   L    x a b L    1 k x     b f x sin dx   k L  L   L  b b a 1 b    0   For even functions:  L , dx k 2 L a a  For odd functions: 0 k

For which functions their Fourier series are convergent to them? Dirichlet theory If y = f ( x ) , bounded on interval [ -L, L ], is: 1. Monotonic in subintervals inside [ -L, L ], 2. Continuous in [ -L, L ], expect for finite number of discontinuity points of the first order         lim f x , lim f x     x x x x 0 0 3. Function f ( x ) has finite number of local minima and maxima in [ -L, L ].       f L f L 4. Values at both interval ends should be the same Additionally if y = f ( x ) is periodic and its period is 2L , then its development holds for all x

a 0 =-2.9155e-16 a = -0.0000 b = 1.0000 0.0000 -0.0000 Examples -2.0000 -0.0000 n = 13 0.0000 -0.0000 -0.0000 0.0000 -0.0000 -0.0000 -0.0000 5.0000 -0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 0.0000         -0.0000 -0.0000    f x sin x 2cos 3 x 5sin 7 x 0.0000 0.0000

a 0 = 0.0016 a = -0.0016 b = 0.9997 0.0016 0.0005 Examples -2.0016 -0.0008 n = 13 0.0016 0.0010 -0.0016 -0.0013 0.0017 0.0016 -0.0017 4.9981 0.0017 0.0023 -0.0017 -0.0026 0.0018 0.0031             f x sin x 2cos 3 x 5sin 7 x -0.0018 -0.0035 0.0019 0.0040        2 (cos 47 x sin 32 x ) x -0.0020 -0.0046

  Examples  2 f x x n = 10 n = 5 n = 20 n = 15

Examples n = 10 n = 5 n = 20 n = 15

Simply supported beam deflection problem   q x L EJ ,   y x static relations        y 0 y L 0   4 q x d y  distributed load,   2 2 d y d y L differential equation     4 dx EJ   0 0   2 2 dx dx   3 T x d y  shear force 3 dx EJ   M x 2 d y   bending moment 2 dx EJ

Simply supported beam deflection problem Main idea  n  k x     a-priori fulfillment of the boundary   y x y sin   k conditions  L   k 1 Procedure 1. Formulate distributed load as the Fourier series L   n  k x  2  k x          q x q sin , q q x sin dx     k k  L  L  L   k 1 0   4 q x d y  2. Substitute the above formulas into differential equation 4 dx EJ    4 4     n k k x 1 n k x    y sin q sin     k k 4 L  L  EJ  L     k 1 k 1  k

3. Evaluate coefficients y k  4 4 q k        , EJ , y k , k 1,2,..., n k k k k  4 L k 4. Compose solution and its derivatives    n k x     y x y sin deflection   k  L   k 1   2   n k x     2 M x EJ k y sin bending moment   k 2 L  L   k 1   3   n k x      3 T x EJ k y cos   shear force k 3 L  L   k 1

Numerical integration for arbitrary type of load  L      n k x 2 k x         q x q sin , q q x sin dx     k k  L  L  L   k 1 0 numerical integration (division into m subintervals) L  h m j m 1   L EJ ,    x k 1 h x h  x  x L 0      x j h x j 1 h h         x j 1 h , x x , x x h for Simpson rule: 1 2 1 3 1 2    2 k x L    q q x ( )sin dx   k L  L  0            1 h m k x k x k x         3 q x ( ) sin 1 4 ( q x ) sin 2 q x ( ) sin , k 1,2,..., m         1 2 3 3 L  L   L   L     j 1

Simply supported beam with constant load q 0 4 d y q  0 4 dx EJ L EJ ,   y x  n  k x      q x q sin   k  L   k 1 L L       2 q k x 2 q L k x      q 0 sin dx 0 cos     k  L  L  L k  L  0 0    4 q   0  , odd k 2 q       0 cos k cos0  k        k     1 0 , even k  1 , odd k    1 , even k

L=10m Simply supported beam with constant load q=10 kN/m 4 4 5 q L q L   0 0 y 0.0130 EJ=10^6 kNm 2 true 384 EJ EJ 4 4 4 q L q L   y 0 0.0131 0 n = 1, k = 1 Fourier  5 EJ EJ (error = 0.48%) 1   M q L 2 0.125 q L 2 true 0 0 8 (error = 3.2%) 4  2  2 M q L 0.129 q L Fourier  0 0 3  T 0.5 q L true 0 (18.94%) 4   T q L 0.4053 q L Fourier  0 0 2

L=10m Simply supported beam with constant load q=10 kN/m EJ=10^6 kNm 2 n = 15, k = 15 max. error = 0.0005% 125; 124.97 (0.02%) 50 48.91 (2.18%)

Simply supported beam with constant piecewise load q 0  L  q 1 , x  4  q 0 d y 2  0  4 dx EJ L   L EJ , 1 , x   2   y x    n k x     q x q sin   k  L   k 1 L L L  L    2         2 q k x 2 q k x 2 q L k x 2 q L k x 2         q 0 sin dx 0 sin dx 0 cos 0 cos         k   L  L  L  L  L k  L  L k  L  L L 0 0 2 2    2 q   2 q   2 q   k k k           0 cos 1 0 cos k cos 0 cos k 2cos 1          k  2  k  2  k  2 

Simply supported beam with constant piecewise load L=10m q=20 kN/m EJ=10^6 kNm 2 n = 2, k = 2

Simply supported beam with constant piecewise load L=10m q=20 kN/m EJ=10^6 kNm 2 n = 15, k = 1

Simply supported beam with concentrated force P 4 EJ    d y P L   x   4 dx  2  L EJ , Dirac delta properties: d H x           x x x y x 0 0 dx            f x x x dx f x selective property 0 0     n k x     q x q sin   k  L   k 1   L  k   L         2 P L k x 2 P 2 P k 2       q x sin dx sin sin         k L  2   L  L L L  2    0  

n = 1 n = 10 L=10m, P=10 kN, EJ=10^6 kNm 2 n = 50 n = 20 n = 2, k = 2

(all formulas taken from the handouts Bending rectangular plate written by dr M.Jakubek) load as the Fourier sum deflection as the Fourier sum differential equation

After substitution of deflection function and load function into differential equation Deflection coefficients Post processing

Bending rectangular plate – uniform load m or n is even m and n are odd

Bending rectangular plate – concentrated force

plate subjected to uniform load a = 2 m n = m = 50 v = 0.3 E = 20*10^6 kN/m 2 q = -1000 kN/m h = 0.2 m plate subjected to concentrated force a = 2 m n = m = 50 v = 0.3 E = 20*10^6 kN/m 2 P = -1000 kN h = 0.2 m x0 = y0 = a/2 = 1 m

maximum deflection convergence with respect to the number of Fourier series – uniform load maximum deflection convergence with respect to the number of Fourier series – concentrated force

Elements of „applied” probability • Deterministic vs. probabilistic approaches • Fuzzy logic – data and results randomness • Monte Carlo simulations for integration • Monte Carlo + random walk for differential equations • Genetic algorithms for optimization and inverse problems

Deterministic vs. probabilistic approaches Deterministic approach one set of data parameters (geometry, load, material parameters, …) - usually average values one unique solution methods : FEM, FDM, meshless methods, … Probabilistic approach variety of sets of data parameters, each with different prescribed stochastic (probability) family of solutions methods: soft computing (Monte Carlo simulations, genetic algorithms, artificial neural networks) + FEM, FDM, meshless methods

Application of probabilistic approaches In mathematics: • integration (in many dimensions) • analysis of differential and integral equations In civil engineering: • influence of data parameters variability on the solution sensibility • determination of reliability of a structure (probability that the failure state does not appear) • optimization problems (the optimal load / supports position that minimizes e.g. the total potential energy) • inverse problems (known is the answer, unknown are the parameters)

FUZZY SETS x  (2.5,3.5) 2 Boundary value problem (beam deflection) with x  (1.5,2.5) fuzzy data 1 (variable concentrated force and moment positions )   u ''( ) x f x ( ), x (0,4) x  x  (1.5,2.5) (2.5,3.5) 1 2   u (0) u (4) 0 x  (1.5,2.5) 1 x  (2.5,3.5) 2 Membership functions