Take a look at this nonlinear differential equation and linearizing it about $(x,\dot{x})=(0.5,1)$
$$
\ddot{x} + 2x^2\dot{x} + 3\dot{x}^2 + x = 0
$$
In the book I'm reading, the author solved it as letting $y=\dot{x}$ and $z=2x^2\dot{x}+3\dot{x}^2$
$$
\ddot{x} + 2x^2y + 3y^2 + x = 0
$$
And he computed
$$
\frac{dz}{dx}\Big\vert_{0.5,1} = 4xy = 2, \qquad \frac{dz}{dy}\Big\vert_{0.5,1} = 2x^2+6y = 6.5 \\
(z-z_0) = 2(x-x_0) + 6.5(y-y_0) \\
z = 2x + 6.5y + 4
$$
Finally, he got the linearized ODE as
$$
\ddot{x} + 6.5\dot{x} + 3x – 4 = 0
$$
Now, I'm trying to reach same result using Jacobain matrix. If I let $x_1=x, x_2=\dot{x}$, I get
$$
\begin{align}
\dot{x}_1 &= x_2 \\
\dot{x}_2 &= -2x_1^2x_2 – 3x_2^2 – x_1
\end{align}
$$
The jacobain matrix about (0.5, 1) is
$$
J=
\begin{bmatrix}
0&1\\
-3&-6.5
\end{bmatrix}
$$
I got this linearized ODE as
$$
\ddot{x} +6.5\dot{x} + 3x = 0
$$
It is missing the constant -4. What is the problem?
Best Answer
The linearization of $F(u)$ is $$ F(u_0)+J_F(u_0)(u-u_0)=J_F(u_0)u+F(u_0)-J_F(u_0)u_0. $$ In your second method you are missing the last two terms. ($u=[x,\dot x]$)