       Re: Differential Eq.

• To: mathgroup at smc.vnet.net
• Subject: [mg106534] Re: Differential Eq.
• From: dh <dh at metrohm.com>
• Date: Fri, 15 Jan 2010 03:21:44 -0500 (EST)
• References: <himsn6\$j93\$1@smc.vnet.net>

```
Hi Jamil,

g only depends on x'. We therefore define it as a function with one

parameter (== x'). It is convenient using a Piecewise function for this.

Here is the code:

===================================

b = 5;

v = 0.2;

u = 0.1;

tmax = 1;

g[der_?NumericQ] = Piecewise[{{b*v, der > v},

{b*(der - u), Abs[der - u] < v},

{-b*v, True}}];

eq = {

x''[t] - x[t] + g[x'[t]] == 0,

x == 0, x' == 0

};

sol = x /. NDSolve[eq, {x}, {t, 0, tmax}][];

Plot[sol[t], {t, 0, tmax}, AxesLabel -> {"t", "x[t]"}]

ParametricPlot[{sol[t], sol'[t]}, {t, 0, tmax},

AxesLabel -> {"x[t]", "x'[t]"}]

=============================================

Daniel

Jamil Ariai wrote:

> Hi All,

>

>

>

> Can anybody kindly tell me how I can solve the following differential equation, with (x, x') = (0, 0):

>

>

>

> x''[t] -x[t] + g[t] = 0,

>

>

>

> where

>

>

>

> g[t] = b*v, for x'[t] > v,

>

> g[t] = b*(x'[t]-u), for Abs[x'[t]-u] < v, and

>

> g[t] = -b*v.

>

>

>

> Take b = 5, v = 0.2, and u = 0.1. Draw x[t] vs t, and x'[t] vs x[t].

>

>

>

> Thanks.

>

>

>

> J. Ariai

>

>

```

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