help me to solve this second order non-linear differential eq?

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pedram
pedram 2012-12-26
can anyone help me to solve this?
y''+a*sign(y')*|y'|^1.8+by=c*sin(wt)
y=y(t) ;
y(0)=0 ; y'(0)=0
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Teja Muppirala
Teja Muppirala 2012-12-27
Well you're almost there
Rewrite this:
y(2)=y'(1)
y'(2)+a*sign*y'(1)*|y'(1)|^1.8+by(1)-c*sin(wt)=0
Like this:
y'(1) = y(2)
y'(2) = -a*sign(y(2))*abs(y(2))^1.8 - b*y(1) + c*sin(w*t)
And then now it is in a form that you can use with MATLAB's many ODE solvers.
% Just picking some values for a,b,c,w
a = 1.2;
b = 2.4;
c = 0.4;
w = 3;
dYdt = @(t,y) [y(2); -a*sign(y(2))*abs(y(2))^1.8 - b*y(1) + c*sin(w*t)]
ode45(dYdt, [0 30], [0 0])
  2 个评论
Greg Heath
Greg Heath 2012-12-28
As Walter has stated, previously, the answer is y(t)= y'(t)=y''(t)= 0 for all t because at t=0 y, y' and y'' are all zero.
However if x = cos(w*t) there will be a solution that can be obtained numerically.
Walter Roberson
Walter Roberson 2012-12-28
When I used Teja's formulation, then with non-negative "a" I get what appears to be a sine-convolved sine wave. With small negative "a" on the order of -1/10000 then the sine wave still looks more or less sine-convolved but with an expanding function, but once the wave gets to around +/- 1000 then the second derivative heads towards +/- infinity. Also, there are repeated small sections of the primary output which are not especially sine-convolved.
With cos() instead of sin() the output looks much the same.

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更多回答（1 个）

Walter Roberson
Walter Roberson 2012-12-27
The solution is y(t) = 0, x(t) = 0
  2 个评论
Greg Heath
Greg Heath 2012-12-27
Only because this is an ill-posed problem with the silly condition x(0)=0. Typically. the assumption would be that x is an arbitrary known input, not necessarily zero at t=0. For example, your solution is correct for x(t) = sin(t), but not for x(t) = cos(t).
Then
y1' = 0*y1 + 1*y2
y2' = - b*y1 - a*y2*abs(y2)^0.8 + x
Which, as far as I can see, can only be solved numerically.
Walter Roberson
Walter Roberson 2012-12-27
The original problem had x(t) on the right-hand side rather than c*sin(wt)

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