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# 数学代写|有限元方法代写finite differences method代考|The Galerkin method

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## 数学代写|有限元代写Finite Element Method代考|The Galerkin method

For the choice of weight function $\psi_i$ equal to the approximation function $\phi_i$, the weighted-residual method is known as the Galerkin method. When $A$ is a linear operator, the algebraic equations of the Galerkin approximation are
$$\begin{gathered} \sum_{j=1}^N A_{i j} c_j=F_i \ A_{i j}=\int_{\Omega} \phi_i A\left(\phi_j\right) d x d y, \quad F_i=\int_{\Omega} \phi_i\left[f-A\left(\phi_0\right)\right] d x d y \end{gathered}$$
We note that $A_{i j}$ are not symmetric (i.e., $A_{i j} \neq A_{j i}$ ).
In general, the Galerkin method is not the same as the Ritz method. This should be clear from the fact that the former uses the weighted-integral form whereas the latter uses the weak form to determine the coefficients $c_j$.
Consequently, the approximation functions used in the Galerkin method are required to be of higher order than those in the Ritz method. The method that uses the weak form in which the weight function is the same as the approximation function is sometimes called the weak-form Galerkin method, but it is the same as the Ritz method. The Ritz and Galerkin methods yield the same solutions in two cases: (i) when the specified boundary conditions of the problem are all of the essential type, and therefore the requirements on $\phi_i$ in the two methods become the same and the weighted-integral form reduces to the weak form; and (ii) when the approximation functions of the Galerkin method are used in the Ritz method.

## 数学代写|有限元代写Finite Element Method代考|The least-squares method

In least-squares method, we determine the parameters $c_j$ by minimizing the integral of the square of the residual $R$ in Eq. (2.5.55):
$$\frac{\partial}{\partial c_i} \int_{\Omega} R^2\left(x, y, c_1, c_2, \cdots, c_n\right) d x d y=0$$
Or
$$\int_{\Omega} \frac{\partial R}{\partial c_i} R d x d y=0, i=1,2, \ldots, n$$
Comparison of Eq. (2.5.59) with Eq. (2.5.56) shows that $\psi_i=\partial R / \partial c_i$. If $A$ is a linear operator, we have $\psi_i=\partial R / \partial c_i=A\left(\phi_i\right)$, and Eq. (2.5.59) becomes
$$\begin{gathered} \sum_{j=1}^N\left[\int_{\Omega} A\left(\phi_i\right) A\left(\phi_j\right) d x d y\right] c_j=\int_{\Omega} A\left(\phi_i\right)\left[f-A\left(\phi_0\right)\right] d x d y \ \sum_{j=1}^N A_{i j} c_j=F_i \ A_{i j}=\int_{\Omega} A\left(\phi_i\right) A\left(\phi_j\right) d x, \quad F_i=\int_{\Omega} A\left(\phi_i\right)\left[f-A\left(\phi_0\right)\right] d x \end{gathered}$$
Note that the coefficient matrix $A_{i j}$ is symmetric whenever $A$ is a linear operator, but it involves the same order of differentiation as in the governing differential equation $A(u)-f=0$.

## 数学代写|有限元代写Finite Element Method代考|The Galerkin method

$$\begin{gathered} \sum_{j=1}^N A_{i j} c_j=F_i \ A_{i j}=\int_{\Omega} \phi_i A\left(\phi_j\right) d x d y, \quad F_i=\int_{\Omega} \phi_i\left[f-A\left(\phi_0\right)\right] d x d y \end{gathered}$$

## 数学代写|有限元代写Finite Element Method代考|The least-squares method

$$\frac{\partial}{\partial c_i} \int_{\Omega} R^2\left(x, y, c_1, c_2, \cdots, c_n\right) d x d y=0$$

$$\int_{\Omega} \frac{\partial R}{\partial c_i} R d x d y=0, i=1,2, \ldots, n$$

$$\begin{gathered} \sum_{j=1}^N\left[\int_{\Omega} A\left(\phi_i\right) A\left(\phi_j\right) d x d y\right] c_j=\int_{\Omega} A\left(\phi_i\right)\left[f-A\left(\phi_0\right)\right] d x d y \ \sum_{j=1}^N A_{i j} c_j=F_i \ A_{i j}=\int_{\Omega} A\left(\phi_i\right) A\left(\phi_j\right) d x, \quad F_i=\int_{\Omega} A\left(\phi_i\right)\left[f-A\left(\phi_0\right)\right] d x \end{gathered}$$

## MATLAB代写

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