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1、1.8 一阶微分方程的应用应用微分方程去解决一些实际问题应用介绍：应用大意应用一：曲线族的等角轨线应用二：雨滴的下落应用三：人口增长模型应用四：静脉注射给药应用五：水流问题,1：应用大意,适应范围 与变化率有关的各种实际问题应用三步曲(1)建立模型 Modelling(2)模型求解 Solving(3)模型应用 Application建议:模型要详略得当,应用一：曲线族的等角轨线,设给定一个平面上以C为参数的曲线族,（*）,我们设法求出另一个以k为参数的曲线族,（*）,使得曲线族（*）中的任一条曲线与曲线族,样的曲线族（*）是已知曲线族（*）的,等角轨线族。,正交轨线族。,的正交轨线族。,设
2、y=y(x)为(C)中任一条曲线，于是存在相应的C，使得 因为要求x,y,y 的关系，将上式对x求导数，得(1.84)这样，将上两式联立，即由,上述关系式成为曲线族满足的微分方程,解：对方程两边关于x求导得,正交，故满足方程,曲线族为,这是一个椭圆，如右图,放大此图,图2.16,应用二：雨滴的下落,考虑雨滴在高空形成后下落的过程中速度的变化 三种不同的假设(1)自由落体运动(2)小阻力的情况(3)大阻力的情况,(1)自由落体运动,下落过程中没有任何阻力,小阻力的情况,下落过程中阻力与速度和半径的乘积成比例,(3)大阻力的情况,下落过程中阻力与速度和半径的乘积平方成比例,三、药物设计,医生给病人
3、开处方是必须注意两点：服药的剂量和服药的时间间隔。超剂量的药物会对患者产生严重不良后果，甚至死亡；剂量不足，则不能达到治疗的效果。,一次给药的药时曲线,治疗窗口,药物消除类型,1 一级动力学消除（恒比消除）：单位时间内按血药浓度的恒比进行消除。消除速度与血药浓度成正比。若以血药浓度（C）的对数与时间（t）作图，为一直线。,零级动力学消除（恒量消除）：单位时间内始终以一个恒定的数量进行消除。消除速度与血药浓度无关。,是指包括零级和一级动力学消除在内的混合型消除方式。如当药物剂量急剧增加或患者有某些疾病，血浓达饱和时，消除方式则可从一级动力学消除转变为零级动力学消除。如乙醇血浓0.05 mg/ml
4、时，则可转成按零级动力学消除。,3米氏消除动力学（混合型消除）：,模型及其数值实现,阅读材料:服药问题,医生给病人开处方时必须注明两点:服药的剂量和服药的时间间隔.超剂量的药品会对身体产生严重不良后果,甚至死亡,而剂量不足,则不能达到治病的目的.已知患者服药后,随时间推移,药品在体内逐渐被吸收,发生生化反应,也就是体内药品的浓度逐渐降低.药品浓度降低的速率与体内当时药品的浓度成正比.当服药量为A、服药间隔为T,试分析体内药的浓度随时间的变化规律.,体内药的浓度随时间的变化规律,Model 3:Population dynamics,In this section we examine equa
5、tions of the form y=f(y),called autonomous equations,where the independentvariable t does not appear explicitly.,The main purpose of this section is to learn how geometricmethods can be used to obtain qualitative informationdirectly from differential equation without solving it.,Simplest model:popul
6、ation growth rate is proportional tocurrent size of the population:,Solution:exponential growth):,Model 3:Population dynamicsLogistic Growth,An exponential model y=ry,with solution y=ert,predictsunlimited growth,with rate r 0 independent of population.,Assuming instead that growth rate depends on po
7、pulationsize,replace r by a function h(y)to obtain y=h(y)y.,We want to choose growth rate h(y)so that h(y)r when y is small,h(y)decreases as y grows larger,and h(y)0 when y is sufficiently large.,The simplest such function is h(y)=r ay,where a 0.,Our differential equation then becomes,This equation
8、is known as the Verhulst,or logistic,equation.,The logistic equation from the previous slide is,This equation is often rewritten in the equivalent form,where K=r/a.The constant r is called the intrinsic growth rate,and as we will see,K represents the carrying capacity of the population.,A direction
9、field for the logisticequation with r=1 and K=10is given here.,Equilibrium solutions of the logistic equation,Our logistic equation is,Two equilibrium solutions are clearly present:,In direction field below,with r=1,K=10,note behavior ofsolutions near equilibrium solutions:y=0 is unstable,y=K=10 is
10、asymptotically stable.,Qualitative analysis of the logistic equation,To better understand the nature of solutions to autonomousequations y=f(y),we start by graphing f(y)vs.y.,In the case of logistic growth,that means graphing thefollowing function and analyzing its graph using calculus.,Qualitative
11、analysis,critical points,The intercepts of f occur at y=0 and y=K,correspondingto the critical points of logistic equation.,The vertex of the parabola is(K/2,rK/4),as shown below.,Qualitative analysis,increasing/decreasing,Note dy/dt 0 for 0 y K,so y is an increasing function oft there(indicate with
12、 right arrows along y-axis on 0 y K).,Similarly,y is a decreasing function of t for y K(indicatewith left arrows along y-axis on y K).,In this context the y-axis is often called the phase line.,Qualitative analysis,concavity,Next,to examine concavity of y(t),we find y:,Thus the graph of y is concave
13、 up when f and f have samesign,which occurs when 0 K.,The graph of y is concave down when f and f have oppositesigns,which occurs when K/2 y K.,Inflection point occurs at intersection of y and line y=K/2.,Qualitative analysis,curve sketching,Combining the information on the previous slides,we have:,
14、Graph of y increasing when 0 K.Slope of y approximately zero when y 0 or y K.Graph of y concave up when 0 K.Graph of y concave down when K/2 y K.Inflection point when y=K/2.,Using this information,we cansketch solution curves y fordifferent initial conditions.,Qualitative analysis,curve sketching,Us
15、ing only the information present in the differential equationand without solving it,we obtained qualitative informationabout the solution y.,For example,we know where the graph of y is the steepest,and hence where y changes most rapidly.Also,y tendsasymptotically to the line y=K,for large t.,The val
16、ue of K is known as the carrying capacity,orsaturation level,for the species.,Note how solution behavior differsfrom that of exponential equation,and thus the decisive effect ofnonlinear term in logistic equation.,Model 3:Population dynamics Exact solution:separating variables,Provided y 0 and y K,w
17、e can rewrite the logistic ODE:,Expanding the left side using partial fractions,Thus the logistic equation can be rewritten as,Integrating the above result,we obtain,Exact solution:resolving for explicit solution,We have:,If 0 y0 K,then 0 y K and hence,Rewriting,using properties of logs:,Exact solut
18、ion:resolving for explicit solution,We have:,for 0 y0 K.,It can be shown that solution is also valid for y0 K.Also,this solution contains equilibrium solutions y=0 and y=K.,Hence solution to logistic equation is,模型预测的动态行为与大量的实验和观测数据吻合,水的流出问题,一横截面积为 A，高为 H 的水池内盛满了水，有池底一横截面积为 B 的小孔放水。设水从小孔流出的速度为，求在任意时刻的水面高度和将水放空所需的时间,问题分析,从水面1降到水面2所失去的水量等于从小孔流出的水量 容器内水的体积为零 即为容器内水的高度为零,建模与求解,1）从水面1将到水面2所失去的体积为,-在时间内，实际损失的体积是,2）在同样时间内，水从小孔流出的体积为,-是水在 时间内从小孔流出保持水平前进时所经过的距离,则,两端同除以，并令 取极限得,由于,可得一阶方程：,解为,下面求将水放空的时间 t*,令h=0 代入上式得,例如,设 A=0.54 m2,B=0.001 m2,H=4.9m,g=9.8m/s2.则将水放空的时间为,作业,P61 2，3，6,
