
NAMEequation  data format for the definition of the functional equation for VECFEM.PURPOSEIn the data set equation the user defines the functional equation to be solved by VECFEM. Using the parameters defined in the resource file it is transformed into a FORTRAN source code for the solution of the problem, see vembuild. The equation may be steady or nonsteady. In the case of a steady problem, the NKvalued solution u=(u<i>) is selected from a space H of smooth, NKvalued functions. The equation can be given in the formulation of a steady functional equation: [1] for all v in H with v(i)D<i>=0 for all i=1,..,NK: F{u}(v)=0 and for all i=1,...,NK : u<i>D<i>=b<i>or in the formulation of a steady variational equation [2] F{u}=minimal over all u in H with u<i>D<i>=b<i> for all i=1,..,NK.In both cases the condition 'u<i>D<i>=b<i> for all i=1,..,NK' is called the Dirichlet condition, where D<i> is a subset of the domain of u<i>. b<i> is a smooth function on D<i> which must not depend on the solution u. F is a realvalued functional which depends arbitrarily on the searched solution u in H and in the case [1] linearly on the NKvalued test function v=(v<i>) in H. In the case of a nonsteady problem, the NKvalued solution u=(u<i>) depends on the time t, t0<=t<=tend, and for all t it is selected from a space H of smooth, NKvalued functions. The equation can be given in the formulation of a nonsteady functional equation: [3] for all v in H with v(i)D<i>=0 for all i=1,..,NK: F{t,ut,u}(v)=0 for all t0<t<tend, for all i=1,...,NK : u<i>(t)D<i>=b<i>(t) for all t0<t<tend and for all i=1,...,NK : u<i>(t0)=u0<i>.The condition 'u<i>(t)D<i>=b<i>(t) for all i=1,..,NK' is called the Dirichlet condition, where D<i> is a subset of the domain of u<i>. For all t0<t<tend, b<i>(t) is a smooth function on D<i> which must not depend on the solution u. The condition 'u<i>(t0)=u0<i> for all i=1,..,\INK' is called the initial condition, where u0 is a given function in H. F is a realvalued functional which depends arbitrarily on the searched solution u in H, its derivative with respect to the time t and the time t itself, and linearly on the NKvalued test function v=(v<i>) in H. SYNOPSISThe script for the definition of the functional equation has three sections. In the first section parameters are defined, in the second one the Dirichlet conditions are specified and in the third one the functional F is described. Lines starting with '#' are comment lines. Spaces are insignificant. Upper and lowercase letters are equivalent. Allowed expressions are real constants, reserved symbols, the parameters declared in the parameter section and operator evaluations. The following operators are available:
parameterA Set of parameters can be introduced (e.g. material constants). The statement<parameter>=<expr>assigns the allowed expression <expr> to the parameter <parameter>. The name of the parameter may be a sequence of letters and digits starting with a letter. Reserved symbols and the function names must not be a substring of the declared parameter. The succession of the parameter declarations is free, but a recursive definition is not allowed. So for example the declaration of the first parameter can use the second parameter, but then the second parameter must not use the first parameter. Initial ValuesThe initial condition for component i (i=1,...NK) is prescribed by the statementu0<i>=<expr0>where <expr0> is an allowed expression. After the insertion of the parameters the prescribed value for the initial condition must only depend on real constants, the available operators and the symbols x<i>. If there is no statement declaring a initial condition for component i, the statement 'u0<i>=0' is assumed. The initial solution is modified by VECFEM so that the Dirichlet conditions are fulfilled at the start time t0. In the case of steady problems, 'u0<i>=...' can be used to define an initial guess of the iteration solution algorithm. Dirichlet ConditionsThe Dirichlet condition for component i (i=1,...NK) is prescribed by the statementu<i>=<exprd>where <exprd> is an allowed expression. The support D<i> of the Dirichlet conditions is specified by the FEM mesh. After the insertion of the parameters the prescribed value for the Dirichlet conditions must only depend on real constants, the available operators and the symbols x<i>, t and prevalue, which is the real value assigned to component i by the FEM mesh. If there is no statement declaring a Dirichlet condition for component i, but the D<i> is nonvoid, the statement 'u<i>=0' is assumed. Functional EquationsThe functional equations [1] and [3] are described by the statement: volume{<exprv>}+area{<expra>}+line{<exprl>}+point{<exprp>}=0or for the variational problem [2] volume{<exprv>}+area{<expra>}+line{<exprl>}+point{<exprp>}=minimizewhere <exprv> denotes the kernel of the volume integration, <expra> the kernel of the area integration, <exprl> the kernel of the line integration and <exprp> the terms of the sum over specified points. For the 3dimensional case DIM=3, the integration kernels <exprv>, <expra>, <exprl> and <exprp> are allowed expressions. After the insertion of the parameters, the kernels must only depend on real constants, the available operators and the following symbols:
For the 2dimensional case DIM=2, the integration kernels <expra>, <exprl> and <exprp> are allowed expressions. The volume integration is not allowed. After the insertion of the parameters, the kernels must only depend on real constants, the available operators and the following symbols:
For the 1dimensional case DIM=1, the integration kernels <exprl> and <exprp> are allowed expressions. The volume and area integrations are not allowed. After the insertion of the parameters, the kernels must only depend on real constants, the available operators and the following symbols:
DISCRETIZATIONThe domains of integration are specified by the elements in the finite element mesh, where only order two elements are allowed. The portions D<i> of the domain where the Dirichlet condition of component i has to be satisfied are specified by the list of the mesh nodes in D<i>. The hexahedrons, prisms and tetrahedron elements describe the domain of the volume integration, the triangle and quadrilateral elements the domain of the area integration and the line elements the domain of the line integration. The point elements (mostly these are the nodes where node forces are specified) give the points which are considered in sums over points. The mesh has to be continuous, i.e. if the elements used for the subdivision of the volume are reduced to the area, they are the elements for the subdivision of the area itself (analogously for the line elements and the point elements). If you use tangential derivatives you have to consider the following rules: The vector tau<.>1 points from the first node of the element to the second one and the vector tau<.>2 points from the first node of the element to the third one in the case of a triangle element and to the fourth one in the case of a quadrilateral element. The vector tau<.>, which is the tangential direction at the line, points from the first node of the element to the second one:line quadrilateral triangle 473 3    \    \  ^   ^ \ 8  6 6  5  tau<.>2   tau<.>2 \      \ tau<.>>  tau<.>1>   tau<.>1> \ 132 152 142 EXAMPLESSee also vemexamples. For the computation of a temperature distribution via the 2dimensional Poisson equation the equation file has the following form, see vembuild. Here the variational form of the Poisson equation is used. The thermal conductivity will depend on the location:# # chr=0 for x1<=1 and chr=1 x1>1 # chr=1./(1.+exp(100*(x11.)) # # / 900. x1<=1. # k= for = thermal conductivity # \ 1. x1>1. # k= chr * 900. + (1chr) * 1. # # no heat generation: # qb=0 # # convection boundary condition: # qs= (u1  20) ^ 2 # # the prescribed temperature is given by the input file: # u1=prevalue # # and here is the variational problem: # volume { k/2 * ( u1x1^2 + u1x2^2 )  qb*u1 } + area { qs/2 } = minimalThe second example is the timedependent 3dimensional NavierStokes equation, see vembuild. The components 1, 2 and 3 are the velocity and the fourth component is the pressure: # # the Reynolds number: # Re=10. # # the input and output profile is a parabolic profile : # u1=(1.x1)*x1 u2=0. u3=0. u4=1. # # the initial solution : # u01=0. u02=0. u03=0. u04=1. # volume{ # # the first momentum equation # v1*u1t+v1x1*u1x1+v1x2*u1x2+v1x3*u1x3 + Re*v1*(u1*u1x1+u2*u1x2+u3*u1x3)+ v1x1*u4 + # # the second momentum equation # v2*u2t+v2x1*u2x1+ v2x2*u2x2+ v2x3*u2x3 + Re*v2*(u1*u2x1+u2*u2x2+u3*u2x3)+ v2x2*u4 + # # the third momentum equation # v3*u3t+v3x1*u3x1+ v3x2*u3x2+ v3x3*u3x3 + Re*v3*(u1*u3x1+u2*u3x2+u3*u3x3)+ v3x3*u4 + # # the equation of continuity # v4*(u1x1+u2x2+u3x3) }=0 REFERENCES[FAQ].SEEVECFEM, vemexamples, xvem.COPYRIGHTSProgram by L. Grosz and S. Scheffrahn , 19941996. Copyrights by Universitaet Karlsruhe 19891996. All rights reserved. More details see VECFEM.by L. Grosz, Auckland , 6. June, 2000. 