本文介绍了Scilab中ODE的奇数行为:方程dy/dx = A未正确求解的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我仍在学习Scilab(5.5.2),因此我正在编写和运行测试代码以熟悉该软件.

I am still learning Scilab (5.5.2), so I am writing and running test codes to familiarize myself with the software.

要测试数值微分方程求解器,我从方程 dy/dx = A 轻松开始,该方程具有 y = Ax + c (线性方程)

To test the numerical differential equation solver, I started easy from the equation dy/dx = A, which has as solution y = Ax+c (line equation).

这是我写的代码:

// Function y = A*x+1
function ydot=fn(x, A)
    ydot=A
endfunction
A=2;
//Initial conditions
x0=0;
y0=A*x0+1;
//Numerical Solution
x=[0:5];
y= ode(y0,x0,x,fn);
//Analytical solution
y2 = A*x+1;
clf(); plot(x, y); plot(x, y2, '-k');
//End

这些是意想不到的结果:

And these are the unexpected results:

y2 = 1. 3. 5. 7. 9. 11.

y2 = 1. 3. 5. 7. 9. 11.

看来 y = e ^ x .有人可以解释出什么问题了或我做错了什么吗?

It appears that y = e^x. Can someone explain what is going wrong, or what I did wrong?

推荐答案

仅重命名变量并不会更改ODE求解器在内部使用变量的方式.由于该求解程序需要一个带有参数time,state的函数,因此它将以这种方式解释所提供的函数.

Just renaming the variables does not change how they are used internally by the ODE solver. Since that solver expects a function with arguments time,state it will interpret the provided function that way.

重命名变量,您所编程的内容等同于

Renaming the variables back, what you programmed is equivalent to

function ydot=fn(t,y)
    ydot = y
endfunction

它确实具有指数函数作为解决方案.

which indeed has the exponential function as solution.

手册中,您可以看到包含参数的方法是将功能作为列表传递,

From the manual you can see that the way to include parameters is to pass the function as a list,

function ydot=fn(t,y,A)
    ydot = A
endfunction

y= ode(y0,x0,x,list(fn,A));

这篇关于Scilab中ODE的奇数行为:方程dy/dx = A未正确求解的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

07-22 15:14
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