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# 数学代写|有限元方法代写finite differences method代考|CIVE602

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## 数学代写|有限元代写Finite Element Method代考|Time approximations

The time approximation is discussed with the help of a single first-order differential equation for $u_i$, and then we generalize for a vector of unknowns, u. Suppose that we wish to determine $u_i(t)$ for $t>0$ such that $u_i(t)$ satisfies
$$a \frac{d u_i}{d t}+b u_i=f_i(t), 0<t<T \quad \text { and } u_i(0)=u_i^0$$
where $a \neq 0, b$, and $u_i^0$ are constants and $f_i$ is a function of time $t$. The exact solution of the problem consists of two parts: the homogeneous and particular solutions. The homogeneous solution is
$$u_i^h(t)=A e^{-k t}, k=\frac{b}{a}$$
where $A$ is a constant of integration. The particular solution is given by
$$u_i^p(t)=\frac{1}{a} e^{-k t}\left(\int_0^t e^{k \tau} f_i(\tau) d \tau\right)$$
The complete solution is given by
$$u_i(t)=e^{-k t}\left(A+\frac{1}{a} \int_0^t e^{k \tau} f_i(\tau) d \tau\right)$$
Finite difference approximations. The finite difference methods are based on truncated (using the desired degree of accuracy) Taylor’s series expansions. For example, Taylor’s series expansions of function $F(t)$ about $t=t_S$ and $t=$ $t_{s+1}$ are given by
$$F(t)=F\left(t_s\right)+\left(t-t_s\right) \dot{F}\left(t_s\right)+\frac{1}{2 !}\left(t-t_s\right)^2 \ddot{F}\left(t_s\right)+\cdots$$
$$F(t)=F\left(t_{s+1}\right)+\left(t-t_{s+1}\right) \dot{F}\left(t_{s+1}\right)+\frac{1}{2 !}\left(t-t_{s+1}\right)^2 \ddot{F}\left(t_{s+1}\right)+\cdots$$
In particular, for $t=t_{\mathrm{s}+1}$ in Eq. (7.4.12a), we have
$$F\left(t_{s+1}\right)=F\left(t_s\right)+\left(t_{s+1}-t_s\right) \dot{F}\left(t_s\right)+\frac{1}{2 !}\left(t_{s+1}-t_s\right)^2 \ddot{F}\left(t_s\right)+\frac{1}{3 !}\left(t_{s+1}-t_s\right)^3 \ddot{F}\left(t_s\right)+\cdots$$
If we truncate the series after the second term and solve for $\dot{F}\left(t_s\right)$, we obtain
$$\dot{F}\left(t_s\right)=\frac{F\left(t_{s+1}\right)-F\left(t_s\right)}{t_{s+1}-t_s}+\mathrm{O}\left(\Delta t_{s+1}\right)$$

## 数学代写|有限元代写Finite Element Method代考|Numerical stability

Since the time-marching scheme in Eq. (7.4.17b) uses the previous time step solution $u_i^s$, which itself is an approximate solution, there is a possibility that the error introduced in $u_i^s$ may be amplified when $u_i^{s+1}$ is computed and may grow unboundedly with time. When the error grows without bounds, the underlying scheme is said to be unstable. Here we wish to determine the conditions under which the error remains bounded.
Consider Eq.(7.4.17b) in operator form
$$u_i^{s+1}=B u_i^s+\bar{F}^{s, s+1}$$
where
$$B=\frac{a-(1-\alpha) \Delta t b}{a+\alpha \Delta t b}, \bar{F}{s, s+1}=\Delta t \frac{\alpha f{s+1}+(1-\alpha) f_s}{a+\alpha \Delta t b}$$
If the magnitude of the operator $B$, known as the amplification operator, is greater than $1,|B|>1$, the error will be amplified during each time step. On the other hand, if the magnitude is equal to or less than unity, the error will not grow with time. Therefore, in order for the scheme to be stable it is necessary that $|B| \leq 1$ :
$$|B|=\left|\frac{a-(1-\alpha) \Delta t b}{a+\alpha \Delta t b}\right| \leq 1$$
The above equation places a restriction on the magnitude of the time step for certain values of $\alpha$. When the error remains bounded for any time step [i.e., condition (7.4.24) is trivially satisfied for any value of $\Delta t$ ], it is known as a stable scheme. If the error remains bounded only when the time step remains below certain value [in order to satisfy (7.4.24)], it is said to be conditionally stable scheme. In the finite element method, the coefficients $a$ and $b$ appearing in Eq. (7.4.24) depend on problem material parameters (which cannot be changed) and the element length (which can be selected). Hence, for a given mesh there is a value of $\Delta t$ that makes the scheme conditionally stable (more details on this topic can be found in the book by Surana and Reddy [4]).

## 数学代写|有限元代写Finite Element Method代考|Time approximations

$$a \frac{d u_i}{d t}+b u_i=f_i(t), 0<t<T \quad \text { and } u_i(0)=u_i^0$$

$$u_i^h(t)=A e^{-k t}, k=\frac{b}{a}$$

$$u_i^p(t)=\frac{1}{a} e^{-k t}\left(\int_0^t e^{k \tau} f_i(\tau) d \tau\right)$$

$$u_i(t)=e^{-k t}\left(A+\frac{1}{a} \int_0^t e^{k \tau} f_i(\tau) d \tau\right)$$

## MATLAB代写

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