考虑:
\documentclass[12pt,a4paper,oneside]{book}
\usepackage{amsfonts, graphicx, verbatim, mathtools,amssymb, amsthm, mathrsfs,amsmath}
\begin{document}
\begin{align*}
K_1&=\max \left \lbrace c_1(k)\int_M [D_1(k,y)-log(1+D_1(k,y))]\nu(dy), c_2(k)\int_M [D_2(k,y)-log(1+D_2(k,y))]\nu(dy)\right \rbrace.
\end{align*}
\begin{align*}
d V&= LV dt+((S-S_k^*+I-I_k^*)[\sigma_1(k)SdB_1 + \sigma_2(k) I dB_2]+a(k)\left(1- \frac{I_k^*}{I} \right)\sigma_2(k)dB_2\\
&+\int_M \left \lbrace \frac{1}{2}[D_1(k,y)S+D_2(k,y)I]^2+(S-S_k^*+I-I_k^*)[D_1(k,y)S+D_2(k,y)I]+a(k)[D_2(k,y)I-I_k^*\log(1+D_2(k,y))]\right \rbrace\tilde{N}(dt,dy)
\end{align*}
\end{document}
答案1
如果我编辑这篇文章我可能会这么做
\documentclass[12pt,a4paper,oneside]{book}
\usepackage{mathtools,amssymb}
\begin{document}
\begin{align*}
K_1={} & \begin{aligned}[t]
\max\Bigl \lbrace & c_1(k)\int_M
\bigl[D_1(k,y)-\log(1+D_1(k,y))\bigr]\nu(dy),
\\
& c_2(k)\int_M
[D_2(k,y)-\log(1+D_2(k,y))]\nu(dy)\Bigr\rbrace.
\end{aligned}
\\
d V={} & LV dt+((S-S_k^*+I-I_k^*)[\sigma_1(k)SdB_1 + \sigma_2(k) I dB_2]
\\
&+a(k)\left(1- \frac{I_k^*}{I} \right)\sigma_2(k)dB_2\\
&+\int_M
\begin{aligned}[t]
\smash[b]{\Bigl \lbrace}
&
\tfrac{1}{2}[D_1(k,y)S+D_2(k,y)I]^2
\\
&+(S-S_k^*+I-I_k^*)[D_1(k,y)S+D_2(k,y)I]
\\
& +a(k)[D_2(k,y)I-I_k^*\log(1+D_2(k,y))]\Bigr\rbrace\tilde{N}(dt,dy)
\end{aligned}
\end{align*}
\end{document}
请注意,这里使用的技巧aligned
只有在没有方程编号的情况下才有效。
正如其他人已经提到的,使用一个align*
,当它们跨越行分隔符时使用手动缩放的栅栏。用于aligned
进行子对齐。