# # # #控制方程构成方程不反应的示踪剂测试是由流体由于平流(粘性流)和流体由于扩散。管理这一过程的偏微分方程给出:$ $ \压裂{\部分C}{\部分t} + \微分算符。\离开(v C - D \微分算符{C} \右)= 0美元美元,美元加元是示踪剂的浓度空间(x, y, z)美元和时间t美元;v是美元注入示踪流体的速度,和D是美元色散系数。# # # #单维的例子让我们考虑一个单维的核心中示踪剂注入后左手边的脸(B点)作为heavyside单位阶跃函数。输出的示踪剂浓度剖面观察到右手边的脸(A): [![在这里输入图像描述][1]][1]这个通用场景中给出的边界条件:$ $ \眉题C_B = 0, x > 0, t = 0 $ $ $ $ \眉题C_B = 1, x = 0, t > = 0 $ $ $ $ \眉题C_B = 0, x \ rightarrow \ infty, t > = 0 $ $现在,存在一个解析解单维的例子可以给你列示踪剂浓度在任何地方,说x美元,在任何时候,说台币美元。我们可以估计C (x, t)使用美元以下解析解:$ $ \眉题C_B = \压裂{1}{2}\离开[1 -{小块土地}\离开(\压裂{x-vt} {2 \√{Dt}} \) \右]$ $,$ v $是已知的流体速度和D是美元液弥散系数未知。# #如何估算弥散系数# #之一,我们从示踪实验获得的数据列输出的示踪剂浓度随时间(点),也称为污水浓度。这个浓度剖面看起来像:[![在这里输入图像描述][2]][2]让我们称之为数据作为美元C_{测量}$。 Similarly, using the above analytical function, we would estimate the tracer effluent concentration (at $x=L$) by guessing some value of dispersion coefficient $D$. However, since this $D$ is just a guess, so the estimated effluent concentration using the formula is not correct. The correct effluent concentration is the one we have measured i.e .$C_{measured}$. You would have guessed by now that we use the measured concentration in the analytical expression to estimate $D$, and the appropriate way of doing this is by devising an objective function and minimizing its square as follow: $$f_{obj}=min\left[\sum (C_{measured}-C_{calculated})^2\right]$$ So you find $C_{calculated}$ using multiple guesses of $D$ and stop the iteration for that particular $D$ which gives you minimum $f_{obj}$ [1]: http://i.stack.imgur.com/LEL3O.png [2]: http://i.stack.imgur.com/0eU70.png