这是一份uwa西澳大学MATH1722的成功案例

$$
\hat{\mathbf{f}}(\mathbf{u})=\mathbf{f}(\mathbf{u})+h \mathbf{f}{1}(\mathbf{u})+h^{2} \mathbf{f}{2}(\mathbf{u})+\cdots
$$
Assume that at some initial time $\mathbf{u}=\hat{\mathbf{u}}$. Using a simple Taylor series we can expand the solution of the modified ordinary differential equation so that
$$
\hat{\mathbf{u}}(t+h)=\mathbf{u}+h\left(\mathbf{f}+h \mathbf{f}{2}+h^{2} \mathbf{f}{3}+\cdots\right)+\frac{h^{2}}{2}\left(\mathbf{f}^{\prime}+h \mathbf{f}^{\prime}+\cdots\right)(\mathbf{f}+\cdots)+\cdots
$$
We want to compare this solution with the action of the discrete operator $\Psi_{h}$. In general, for a consistent scheme, there will be functions $\mathbf{d}{k}(h)$ so that $$ \Psi{h} \mathbf{u}=\mathbf{u}+h \mathbf{f}(\mathbf{u})+h^{2} \mathbf{d}{2}(\mathbf{u})+h^{3} \mathbf{d}{3}(\mathbf{u})+\cdots
$$
We can now equate the above two expressions. Taking terms at orders $h^{2}$ and $h^{3}$ gives
$$
\mathbf{f}{2}=\mathbf{d}{2}-\frac{1}{2} \mathbf{f}^{\prime} \mathbf{f} \quad \text { and } \quad \mathbf{f}{3}=\mathbf{d}{3}-\frac{1}{6}\left(\mathbf{f}^{\prime}(\mathbf{f}, \mathbf{f})(\mathbf{u})+\mathbf{f}^{\prime} \mathbf{f}^{\prime} \mathbf{f}(\mathbf{u})\right)-\frac{1}{2}\left(\mathbf{f}^{\prime} \mathbf{f}{2}(\mathbf{u})+\mathbf{f}{2}^{\prime} \mathbf{f}(\mathbf{u})\right)
$$

MATH1722 COURSE NOTES ：

$$
\frac{\mathrm{d}}{\mathrm{d} t} F\left(\mathbf{u}{i}(t)\right)=\left(D{i} F\right)\left(\mathbf{u}{i}(t)\right)=F^{\prime}\left(\mathbf{u}{i}\right) f_{i}\left(\mathbf{u}{i}\right), $$ where $\mathbf{u}{i}(t)$ is the solution of the associated differential equation. Then if $\mathrm{d} \mathbf{u}{i} / \mathrm{d} t=$ $\mathbf{f}{i}\left(\mathbf{u}{i}\right)$ we have $\mathrm{d} \mathbf{u}{i} / \mathrm{d} t=D_{i} \mathbf{u}{i}(t), \mathrm{d}^{2} \mathbf{u}{i} / \mathrm{d} t^{2}=D_{i}^{2} \mathbf{u}{i}(t)$, etc. As a consequence we may express the flow $\varphi{t}^{i}$ as an exponential of the form
$$
\varphi_{i}^{i} \mathbf{u}{0}=\exp \left(t D{i}\right) \mathbf{u}{0}=\sum{n} \frac{t^{n} D_{i}^{n}}{n !} \mathbf{u}{0} . $$ and compose two flows in the manner $$ \left(\varphi{t}^{1} \circ \varphi_{s}^{2}\right) u_{0}=\exp \left(t D_{1}\right) \exp \left(s D_{2}\right) \mathbf{u}_{0} .
$$
Thus we see how a flow can be thought of in terms of the exponential of an operator. Now suppose that $X$ and $Y$ are general operators with associated exponentials
$$
\exp (h X)=I+h X+\frac{h^{2}}{2} X^{2}+\cdots, \quad \exp (h Y)=I+h Y+\frac{h^{2}}{2} Y^{2}+\cdots,
$$