Linearization of Second Order Differential Equations


by Ruby Tyra
Tags: linearization, pendulum
Ruby Tyra
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#1
Jan2-13, 02:48 PM
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I'm having some difficulties figuring out how to linearize second order differential equations for a double pendulum.

I have an equation that is in the form of

[itex]\theta_{1}''[/itex][itex]\normalsize = function[/itex] [[itex]\theta_{1}[/itex],[itex]\theta_{2}[/itex],[itex]\theta_{1}'[/itex],[itex]\theta_{2}'[/itex]]

(The original equation is found at http://www.myphysicslab.com/dbl_pendulum.html, the equations inside the orange rectangle.)

I was told to replace that function by a linear function of all four variables but I don't know where to start with that since the original equation is much more complex than the simple pendulum example we were given.

Thank you!
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Chestermiller
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Jan2-13, 08:33 PM
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Quote Quote by Ruby Tyra View Post
I'm having some difficulties figuring out how to linearize second order differential equations for a double pendulum.

I have an equation that is in the form of

[itex]\theta_{1}''[/itex][itex]\normalsize = function[/itex] [[itex]\theta_{1}[/itex],[itex]\theta_{2}[/itex],[itex]\theta_{1}'[/itex],[itex]\theta_{2}'[/itex]]

(The original equation is found at http://www.myphysicslab.com/dbl_pendulum.html, the equations inside the orange rectangle.)

I was told to replace that function by a linear function of all four variables but I don't know where to start with that since the original equation is much more complex than the simple pendulum example we were given.

Thank you!
If f = f(x,y,z,w) and you want to linearize, use the chain rule:

[tex]f(x,y,z,w) = f(x_0,y_0,z_0,w_0)+\frac{\partial{f}}{\partial{x}}(x-x_0)+\frac{\partial{f}}{\partial{y}}(y-y_0)+\frac{\partial{f}}{\partial{z}}(z-z_0)+\frac{\partial{f}}{\partial{w}}(w-w_0)[/tex]

where the partials are evaluated at [tex]x_0,y_0,z_0,w_0[/tex]
HallsofIvy
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Jan3-13, 05:54 AM
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Note that what Chestermiller is saying is essentially the same as replacing the function by a Taylor polynomial in all variables, then dropping all but the linear terms. And that, in turn, is the same as replacing the "surface" by its "tangent plane".


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