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=Problem 6.5 Solve G2DM1.0/D1 using 2D LIBF(Lagrange Interpolation Basis Function)=

Given
The G2Dm1.0/D1 PDE

where: $$\left[ \right]$$ is an identity matrix The essential boundary condition is $$g(x,t) = 2\ \ \ \ on\ \ \ \ \partial \Omega $$.

Find
1a) Solve G2DM1.0/D1 using 2D LIBF until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for steady state and $$f(x) = 1\ \ \ \ in\ \ \ \ \Omega = \square $$.

1b1) Solve G2DM1.0/D1 using 2D LLEBF with uniform mesh until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for steady state and $$f(x) = 1\ \ \ \ in\ \ \ \ \Omega = \square $$.

1b2) Solve G2DM1.0/D1 using 2D LLEBF with non-uniform mesh until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for steady state and $$f(x) = 1\ \ \ \ in\ \ \ \ \Omega = \square $$.

2a) Solve G2DM1.0/D1 using 2D LIBF until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for transient state and $$f(x) = 0\ \ \ \ in\ \ \ \ \Omega = \square $$.

2b1) Solve G2DM1.0/D1 using 2D LIBF until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for transient state and $$f(x) = 1\ \ \ \ in\ \ \ \ \Omega = \square $$.

2b2a) Solve G2DM1.0/D1 using 2D LLEBF with uniform mesh until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for transient state and $$f(x) = 1\ \ \ \ in\ \ \ \ \Omega = \square $$.

2b2a) Solve G2DM1.0/D1 using 2D LLEBF with non-uniform mesh until accuracy $${10^{ - 6}}$$ at center (x,y)=(0,0) for transient state and $$f(x) = 1\ \ \ \ in\ \ \ \ \Omega = \square $$.

Solution
For Eq.6.5.1, we can write the weak form

1a) 2D LIBF
For 2D LIBF, we choose Lagrange Interpolation Basis Function for each element. For example, we choose n=m=2


 * $$N_{I}(x,y)=L_{i,m}(x)\cdot L_{j,n}(y),\;\; where \;\;I=i+(j-1)m$$

After substituting Eq.7.1.3 into Eq.7.1.2 in which Lagrange Interpolation Basis Function are both trial function and weight function, we integrate over the whole domain to obtain the capacitance, conductance, and flux matrices. These matrices are used to achieve final results for the temperatures at each node using the following Matlab codes.

1b) 2D LLEBF
For the 2D LLEBF case, we choose Linear Lagrange elemental basis functions for each element, i.e.,

After substituting Eq.7.1.4 into Eq.7.1.2 in which Linear Lagrange elemental basis function are both trial function and weight function, we integrate over the whole domain to obtain the capacitance, conductance, and flux matrices. These matrices are used to achieve final results for the temperatures at each node using the following Matlab codes.

1b1) Uniform Mesh
The first solution is calculated using a uniform mesh (a mesh where the nodes are equidistant from each other).



One sign that the solution was correctly coded is that the essential boundary conditions are met in these figures. Notice that along the edges of the temperature distribution graph the temperature is equal to 2. Additionally, because there is a positive heat source within the boundaries of our model we would expect the temperature to rise from the boundaries of the model, which is also apparent. These are good indicators that the solution provided is correct, or at least we are heading in the correct direction.

1b2) Non-Uniform Mesh
The second solution is calculated using a non-uniform mesh (an irregularly shaped mesh that displays the abilities of the robust code adopted from the textbook by Fish and Belytschko).



Similarly, the same boundary conditions are evident, and the rise in temperature in the center of the object is obvious.