# 多变量方法 Multivariate Methods MATH5745M01

0

$$0=g_{x}^{0}\left(x-x_{0}\right)+g_{y}^{0}\left(\phi(x)-\phi\left(x_{0}\right)\right)+E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)$$
Now divide through by $x-x_{0}$ to get
$$0=g_{x}^{0}+g_{y}^{0}\left(\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}\right)+\frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}$$
Thus,
$$g_{y}^{0}\left(\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}\right)=-g_{x}^{0}-\frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}$$
and assuming $g_{x}^{0} \neq 0$ and $g_{y}^{0} \neq 0$, we can solve to find
$$\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}=-\frac{g_{x}^{0}}{g_{y}^{0}}-\frac{1}{g_{y}^{0}} \frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}$$

Define $h(t)=u\left(x_{0}, t\right)$. The second order Taylor expansion of $h$ is then
$$h(t)=h\left(t_{0}\right)+h^{\prime}\left(t_{0}\right)\left(t-t_{0}\right)+\frac{1}{2} h^{\prime \prime}\left(c_{t}\right)\left(t-t_{0}\right)^{2}$$
where $c_{t}$ is some point between $t_{0}$ and $t$. Using the chain rule for functions of two variables, it is easy to see $h\left(x_{0}\right)=u\left(x_{0}, y_{0}\right), h^{\prime}(t)=u_{t}\left(x_{0}, t\right)$ and $h^{\prime \prime}(t)=u_{t t}\left(x_{0}, t\right)$. Hence, we can rewrite the expansion as
$$u(x, t)=u\left(x_{0}, t_{0}\right)+u_{t}\left(x_{0}, t_{0}\right)\left(t-t_{0}\right)+\frac{1}{2} u_{t t}\left(x_{0}, c_{t}\right)\left(t-t_{0}\right)^{2} .$$
We can then write these another way by letting $\Delta t=t-t_{0}$. This gives
$$u\left(x_{0}, t_{0}+\Delta t\right)=u\left(x_{0}, t_{0}\right)+u_{t}\left(x_{0}, t_{0}\right) \Delta t+\frac{1}{2} u_{t t}\left(x_{0}, c_{t}\right)(\Delta t)^{2} .$$