User:Egm6321.f10.team2.oztekin/Newwwwwwwww

=Problem 7.1: Two Dimensional Heat problem with fixed boundary considering static and transient case =

Given: The Following 2D Data Set
Given the data set provided below.

Domain: Bi-Unit Square
1) The following domain information $$ \Omega = \bar \omega  = \square $$ which is for a bi-unit square

Equation: PDE, K, f, and Initial Condition
2) The partial differential equation of the form $$ \displaystyle \operatorname{div} \left ( K  \cdot \nabla{u} \right ) + f = \rho c\frac $$ 2.a) $$  {K}= {I}=I_n = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} $$

Find: 2-D Lagrange Interpolation Basis Functions (LIBF), Linear Lagrange Interpolation Basis Functions(LLEBF) and Solve the PDE for $$ m = n = 2,4,6,8 $$
Use 2-dimensional Lagrange Interpolation Basis Functions (LIBF) and Linear Lagrange Interpolation Basis Functions(LLEBF) with $$ \displaystyle m = n = 2,4,6,8 $$ nodes per element to solve the data set above. Ensure the accuracy of the approximate solution $$ \displaystyle u^h(x = 0, y = 0) = 10^{-6} $$ at the center of the element $$ \displaystyle (x,y) = (0,0) $$

1-) Static case (steady state) $$f\left(\boldsymbol{x} \right)=1$$ in $$\Omega =\square $$

1(a) using 2D LIBF

'''1(b). 2D LLEBF'''

1(b).1 uniform mesh

1(b).2 non-uniform mesh

2-) Dynamic Case(Transient)

$$\rho c=3$$

initial condition: $$u\left( x,t=0 \right)=xy$$, $$\forall x\in \Omega $$

2(a). using 2D LIBF.

$$f\left(\boldsymbol{x},t \right)=0$$, $$\forall \boldsymbol{x}\in \Omega $$, $$\forall t>0$$

$$g\left( \boldsymbol{x},t \right)=2$$ on $${{\Gamma }_{g}}=\partial \Omega $$

2(b). With new conditions:

$$f\left( \boldsymbol{x},t \right)=1$$, $$\forall \boldsymbol{x}\in \Omega $$, $$\forall t>0$$

$$g\left( \boldsymbol{x},t \right)=2$$ on $${{\Gamma }_{g}}=\partial \Omega $$

2(b).1 2D LIBF

2(b).2 2D LLEBF

2(b).2.1 uniform mesh

2(b).2.1 non uniform mesh

Solution
The 2-D LIBF can be defined as in (6.5.1)

Where $$ \displaystyle L_{i,m}(x) $$ and  $$ \displaystyle L_{j,n}(y) $$ represent Lagrange Interpolation Basis Functions

For LLEBF we will have only four basis function for each element, which can be constructed from the coordinates of that particular element in the same way as LIBF are constructed.

Initial condition is given for transient case as


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$$  \displaystyle \begin{matrix} u(x,t=0)=u_{o}(x)=xy & \forall x\in \Omega \end{matrix} $$     (7.1.3) Now in order to find values at nodes at time zero we have to use projection method to get matrices.
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$$  \displaystyle = $$     (7.1.4) Our goal is
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Find $$\displaystyle \begin{matrix} d_{j}(0) & st \forall x_{j} \in \Omega \end{matrix}$$


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$$  \displaystyle \sum_{i} c_{i}\left [ \sum_{j}- \right ]=0 $$     (7.1.5)
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$$  \displaystyle \begin{matrix} \forall c_{i} & \forall x_{i}\in \Omega =c_{F} \end{matrix} $$     (7.1.6)
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$$  \displaystyle G_{i}= $$     (7.1.7)
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$$  \displaystyle c_{F}[\Gamma _{FF}d_{F}-G]=0 $$     (7.1.8)
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$$  \displaystyle d(0)=d_{F}(0)=\Gamma _{FF}^{-1}G $$     (7.1.9)
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$$  \displaystyle \Gamma _{FF}=[,i,j\in \eta _{F}] $$     (7.1.10)
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Where $$\displaystyle\eta _{F}$$ is a set of global node numbers not on essential boundary.

DWF form was given in Meeting 23


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$$  \displaystyle c_{E}\left [ (M_{EE}g^{(s)}+K_{EE}g)+M_{EF}d_{F}^{(s)}+K_{EF}d_{F}-F_{E} \right ]=0 $$     (7.1.11)
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$$  \displaystyle c_{F}\left [ (M_{FE}g^{(s)}+K_{FE}g)+M_{FF}d_{F}^{(s)}+K_{FF}d_{F}-F_{F} \right ]=0 $$     (7.1.12)
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$$  \displaystyle M_{FF}d_{F}^{(s)}+K_{FF}d_{F}=F_{F}-(M_{FE}g^{(s)}+K_{FE}g) $$     (7.1.13)
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$$  \displaystyle Md^{(s)}+Kd=F $$     (7.1.14) For static case
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$$  \displaystyle \begin{matrix} Kd=F\\ where\\ F=F_{F}-K_{FE}g \end{matrix} $$     (7.1.15) For transient heat transfer case
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$$  \displaystyle \displaystyle \dot{d}=M^{-1}[F-Kd] $$     (7.1.16)
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1(a) using 2D LIBF





Temperature (u) at the center when m=n=5. This value is taken as accurate and error is calculated with reference to this value


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$$  \displaystyle
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{T_{center}} = {\text{2}}{\text{.294921821297876}} $$     (N)
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1(b) using 2D LLEBF





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The plot of $$u-u_h$$ is zig-zag in this case because in this plot both odd number of global node(3*3,5*5,7*7) and even number of global nodes (4*4,6*6) are considered together. If we consider only odd numbers and even numbers separately, curve will smoothly approach to accurate solution (Tcenter = 2.294921821297876).

2(a) using 2D LIBF

first five plot represent the state of temperature with respect to time. Last plot represent system approaching steady state with time. In this case as heat source is zero system should approach to (u=2) at center with time













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First five plot represent the state of temperature with respect to time. Last plot represent system approaching steady state with time. In this case as heat source is zero system should approach to (u=2) at center with time.

2(b)1 using 2D LIBF









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First three plot represent the state of temperature with respect to time. Last plot represent system approaching steady state with time. In this case as heat source is one system should approach to (u=2.294921821297876) at center with time.

2(b)2 using 2D LLEBF









comment
First three plot represent the state of temperature with respect to time. Last plot represent system approaching steady state with time. In this case as heat source is one system should approach to (u=2.294921821297876) at center with time