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How would I solve the following problem?

Linearize around the fixed points

$$\left\{\begin{align}\frac{\text{d}x}{\text{d}t}&=y-x^2\\\frac{\text{d}y}{\text{d}t}&=y-x\end{align}\right.$$

I tried taking the derivative of each, and then I am stuck and Taylor expansion.

user2789433
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  • We need fixed points $x_0=x(t_0)$, $y_0=y(t_0)$. Since they are fixed the velocity is zero $\frac{dx}{dt}(t_0)=\frac{dy}{dt}(t_0)=0$. We get to solve the system $0=ax_0-bx_0y_0-ex_0^2=-fy_0+cx_0y_0.$ Then either $y_0=0$, or $x_0=f/c$ and $y_0=(ax_0-ex_0^2)/bx_0=(af-ef^2/c)/bf$. – OR. Mar 28 '14 at 00:32
  • Now we only need the first order terms of the Taylor of the righ-hand sides at these points. Their coefficients are going to be the derivatives of the right-hand sides evaluated at these points. – OR. Mar 28 '14 at 00:34
  • Since you have the fixed points from your other question [http://math.stackexchange.com/questions/726754/nullclines-with-variables/], toss out the non-linear terms (those with $ \ xy \ $ and $ \ x^2 \ $ and replace the $ \ x \ $ and $ \ y \ $ with $ \ (x - X) \ $ and $ \ (y - Y) \ $ , where $ \ (X,Y) \ $ are the coordinates of each of your fixed points (so, yes, you're being asked to do up to five linearizations). – colormegone Mar 28 '14 at 00:46
  • Rather, there are three fixed points. – colormegone Mar 28 '14 at 01:22
  • And the question I linked wasn't your question, but it did involve the same system of DEs. – colormegone Mar 28 '14 at 02:08

2 Answers2

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Question Updated

You need simultaneous zeros for $x' = y' = 0$.

From the second equation, we have $y = x$, sub back into first and we have:

$$x(1 - x) = 0 \implies x = 0, 1$$

Now find the two $y$ values, which are $y = 0, 1$.

So, we have two critical points as:

$$(x, y) = (0, 0), (1, 1)$$

Try those in each equation and verify that you get zero for each one.

Next, find the eigenvalues of the Jacobian matrix for each critical (fixed) point. We have:

$$J(x,y) = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -2x & 1 \\ -1 & 1 \end{bmatrix}$$

Do you see what the linearized system is now?

Now, evaluate the eigenvalues of $J(0,0)$ and $J(1,1)$.

For $J(0,0)$, we get $\lambda_1 = (-1)^{1/3},~ \lambda_2 = -(-1)^{2/3}$.

For $J(1,1)$, we get $\lambda_1 = \dfrac{1}{2}(-1 - \sqrt{5}),~ \lambda_2 = \dfrac{1}{2}(-1 + \sqrt{5})$.

Update 2

To find the linear approximation to the function $f(x)$ near a point $x^*$, we make use of Taylor’s theorem:

$$f(x) ≈ f(x^∗) + f′(x^∗)(x − x^∗).$$

There is a similar version of this theorem for functions depending on two variables. Indeed,

$$F(x, y) ≈ F(x^∗, y^∗) + \dfrac{\partial F}{\partial x}(x^∗,y^∗)(x − x^∗) + \dfrac{\partial F}{\partial y} (x^∗, y^∗)(y − y^∗).$$

At a critical point, $F(x^∗, y^∗) = 0 = G(x^∗, y^∗)$. So the linear approximation is:

$$ \begin{bmatrix} x \\ y \end{bmatrix}' = \begin{bmatrix} \dfrac{\partial F}{\partial x} (x^∗, y^∗) & \dfrac{\partial F}{\partial y} (x^∗, y^∗) \\ \dfrac{\partial G}{\partial x} (x^∗, y^∗) & \dfrac{\partial G}{\partial y} (x^∗, y^∗) \end{bmatrix} \begin{bmatrix} x-x^* \\ y-y^* \end{bmatrix}$$

We can clean this up a little if we set $u = x − x^∗$ and $v = y − y^∗$. Then the linearization near $(x^∗, y^∗)$ becomes:

$$ \begin{bmatrix} u \\ v \end{bmatrix}' = \begin{bmatrix} \dfrac{\partial F}{\partial x} (x^∗, y^∗) & \dfrac{\partial F}{\partial y} (x^∗, y^∗) \\ \dfrac{\partial G}{\partial x} (x^∗, y^∗) & \dfrac{\partial G}{\partial y} (x^∗, y^∗) \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = J\begin{bmatrix} u \\ v \end{bmatrix}$$

Notes: this is exactly what was derived above!

Notes: Do the calculations and make sure you get the same result, here are the intermediate calculations.

  • $\dfrac{\partial F}{\partial x} = -2x$
  • $\dfrac{\partial F}{\partial y} = 1$
  • $\dfrac{\partial G}{\partial x} = -1$
  • $\dfrac{\partial G}{\partial y} = -1$
  • $(x^*, y^*) = (0, 0), (1, 1)$

Here are some Nonlinear Differential Equations, Section 3.2 on the Taylor approach for another perspective and for you to follow along with.

Amzoti
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You need to find the Jacobian matrix, which is the first term of the Taylor expansion.

$$A(x,y) = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} a - by - 2e & -bx \\ cy & -f + cx \end{bmatrix}$$

Now, you need to put the fixed points in $A(x,y)$ to obtain a linear dynamical system around each fixed point as $\dot{z}_i = A(x_i^*, y_i^*) z_i$ where $(x_i^*, y_i^*)$ is $i$th fixed point and $z = \begin{bmatrix} x & y \end{bmatrix}^T$.

obareey
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