Linearization of Second Order Differential Equations

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To linearize second order differential equations for a double pendulum, one should replace the complex function with a linear approximation using the chain rule. This involves evaluating the function and its partial derivatives at a specific point, typically denoted as (x_0, y_0, z_0, w_0). The process is akin to using a Taylor polynomial and retaining only the linear terms, effectively approximating the function by its tangent plane. This method simplifies the original equation, making it more manageable for analysis. Understanding these steps is crucial for successfully linearizing the equations involved in the dynamics of a double pendulum.
Ruby Tyra
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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

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

(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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Ruby Tyra said:
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

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

(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:

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)

where the partials are evaluated at x_0,y_0,z_0,w_0
 
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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