Discretization- Finite difference, Finite element methods

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1 Discretization- Finite difference, Finite element methods Q. Identify the natural and essential boundary conditions of the following differential equation: d d y a( ) b( ) + =, for < < ; subject to the following boundary conditions: y = and dy = at = ; a( ) d y = = A, d d y a( ) =. = d d y a( ) b( ) v + = d d y a( ) v + b( ) v = d d y dv d d y v a( ) a( ) + b( ) v = d d y dv d y d v d y v a( ) a( ) a( ) b( ) v + + = It is clear from the boundary terms that the primary variables are y and dy and d d y secondary variables are a( ) and a( ) d y respectively. Hence, essential boundary conditions are y = and dy = at = and natural boundary conditions are a( ) d y = = A, d d y a( ) =. Q. Consider the following differential equation on the domain d d du a b cu f + = a, b, c for all. ( ) ( ) ( ) =, where a, b, c are known functions of, (a) Develop the variational formulation for this differential equation in the form: Fin a u, v = v for all v such that ( ) ( ) (b) Identify the function spaces that u and v should lie in, and the appropriate essential and natural boundary conditions.

2 (c) Show that a (.,.) is symmetric and positive definite. (d) Formulate the minimization problem corresponding to the above variational formulation. (a) d d du a v b v cuv fv + = + + = d d dv du du dv v a a vb b cuv fv = d dv d v du du dv v a a a vb b cuv fv (, ) = ( v) a u v d v du dv a ( u, v) = a + b cuv + d dv du ( v) = fv v a + a + vb (b) du < H < H dv d v < H < H (c) Possible boundary conditions are as follows Essential boundary conditions Natural boundary conditions (i) (ii) (iii) u specified u specified u specified (c) a ( u, v) = a( v, u) It implies that a is symmetric. d v dv, = + + ( ) a v v a b cv Therefore, a is positive. (d) π = (, ) ( ) a u u u d a specified a specified du b specified

3 Q3. = + + du a b cu d du du fu u a + a + ub Consider the following heat conduction problem ( ) : d dt k + S = dt π (i) Is T = cos( π ) + sec( ) a valid trial function? Eplain. T = sin( π ) + cosec π a valid weighting function? Eplain. The boundary conditions specified are as follows: T ( ) =, (ii) Is ( ) (i) π T = cos( π ) + sec( ) π T ( ) = cos( π ) + sec( ) = + = It violates the given essential boundary condition ( ) = ( ) π T = cos( π ) + sec( ) is not a valid trial function. (ii) d dt k + S = d dt k + S v = where v is the variation in T which is taken as weighting function ( w ). w = δt Since T is given at =, δ T = at =. w = sin( π ) + cos ec π Now, ( ) w( ) sin( π ) cosec( π ) Hence T sin( π ) cosec( π ) = + =undefined = + is not a valid weighting function. T =. Hence Q. One-dimensional steady fully developed fluid flow takes place between two parallel plates (this effectively implies that the transient and advection terms in the momentum equation turn out to be identically zero) with zero pressure gradient. However, the flow is

4 subjected to other body forces dependent on the position and velocity, so that the corresponding governing equation for velocity in a non-dimensional form becomes + u + =, with () () u = u =. Considering a trial function of u = asinπ, determine the value of the parameter a by following the least square method, the point collocation method ( considering a single collocation point as the mid point of the domain), Galerkin s Method and Rayleigh Ritz Method + u + = u = u = ( ) ( ) Trial function is given as u = asin π appro Differentiating the trial function with respect to twice, we get du appro = aπ cosπ appro = aπ sin π The residual becomes appro R = + u appro + = aπ sinπ + a sinπ + R = + a sinπ π or, ( ) i) east Square Method. R = a R R = a { ( )} ( ) { } + asinπ π sinπ = { π ( π ) π } + asin sin = ( ) sinπ π sin π + a = cosπ sinπ + + ( π ) a = π π

5 a + ( π ) = π a = u = ( ) ( ) sinπ ii) Point collocation Method : R = =.5 π R = + asin ( π ) = a = π u = sinπ π iii) Galerkin s Method According to Galerkins Method, weighting function is equal to trial function. Therefore, Weighting function w = sinπ { + ( )} asinπ π sinπ ( ) a π sin π + sinπ = a = = ( π ) sinπ sin ( ) u = ( ) sinπ π iv) Rayleigh Ritz Method u + u + v =

6 u v + uv + v = du dv du v uv v + = du dv uv = (, ) = l ( v) a u v v π = (, ) ( ) a u u l u du = u u π = ( a π cos π a sin π ) asinπ d a π a a π = π π = a aπ a a = π a = u = ( ) ( ) sinπ