Linearise the euqations near each equilibrium point?

[pokerfan91]

linearise the euqations near each equilibrium point?

Write each of the following differential equations as two first order equations, find all the equlibrium points, and linearise the equations near each equlibrium point

(a) X.. +x -1/4X³=0

first part done getting X.=y and Y.=-x+1/4X³

second part done getting (0,0) (2,0) and (-2,0)

however i don't even know what the third part means let alone how to do it, assistance required please

 

[Defennder]

I don't understand your notation. What is X.. ? Which is the term representing the derivative? Maybe it's something I have not learned before.

 

[HallsofIvy]

Write each of the following differential equations as two first order equations, find all the equlibrium points, and linearise the equations near each equlibrium point

(a) X.. +x -1/4X³=0

first part done getting X.=y and Y.=-x+1/4X³

second part done getting (0,0) (2,0) and (-2,0)

however i don't even know what the third part means let alone how to do it, assistance required please

I assume the two dots are supposed to be superscripts denoting derivatives. In ASCII it is better to use ' and " for derivatives: x"+ x- (1/4)x³. Letting y= x', y'= x" so that equation becomes y'+ x- (1/4)x³ or x'= y, y'= -x+ (1/4)x³ as you say.

The equilibrium points are points where both derivatives are equal to 0: were x'= y= 0 and y'= -x+ (1/4)x³= 0. We can factor that as x((1/4)x²-1)= 0 and, yes, the equilibrium points are (0, 0), (2, 0), and (-2, 0).

Now, "linearize" means simply to replace any nonlinear function with a linear approximation. The best linear approximation to x³ for x close to 0 is just 0 itself (because the slope of x³ at 0 is 0). At (0,0) the linear approximation is just
x'= y, y'= -x.

The derivative of x³ at x= 2 is 3(2²)= 12 and 2³= 8 so the tangent line to y= x³ at x= 2 is y= 12(x-2)+ 8. At (2, 0) the linearization is x'= y, y'= -x+ 12(x- 2)+ 8= 11x- 16.

You could also do this by writing the "Jacobian" for the system and evaluating it at (0,0), (2,0) and (-2,0). Does your textbook say anything about the "Jacobian"?

 

[pokerfan91]

.. no not X'' like X with 2 dots above it but i can't write that on here
X