Locate the bifurcation values of a for the one-parameter family and describe the bifurcation that takes place at each such value.
dy/dt=e^(-y^2)+a
I got that the equilibrium points will be at -sqrt(ln(a)). I don't know if that is fine. I took the partial derivative with respect to y and got -2ye^(-y^2) and that is zero when a is 0 if I plug in -sqrt(ln(a)) as y.
To locate the bifurcation values of a for the given one-parameter family, we need to find the values of a that result in a change in the stability or behavior of the equilibrium points.
First, let's find the equilibrium points by setting dy/dt to zero:
e^(-y^2) + a = 0
Solving for y, we have:
e^(-y^2) = -a
Taking natural logarithm on both sides, we get:
-y^2 = ln(-a)
Since the logarithm of a negative number is undefined, we need to consider only positive values of a for valid solutions. Thus, we will consider a > 0.
Taking the square root of both sides, we have:
y = ±sqrt(-ln(a))
Now, to analyze the bifurcation values, we will examine the nature of equilibrium points for different ranges of a.
1. For a = 0:
Plugging in a = 0 into the equation, we have:
e^(-y^2) + 0 = 0
e^(-y^2) = 0
Since the exponential function is always positive, there are no equilibrium points at a = 0.
2. For a > 0 (excluding a = 0):
Plugging in a > 0 into the equation, we have:
e^(-y^2) + a = 0
Since a is positive, the first term (e^(-y^2)) will also be positive. Therefore, there are no real solutions for y in this case. However, we can consider the complex solutions.
Taking the square root of -ln(a), we get imaginary solutions:
y = ±i * sqrt(ln(a))
These complex solutions represent the equilibrium points for a > 0.
In summary, the bifurcation values for a occur at a = 0 and a > 0.
1. At a = 0, there are no equilibrium points.
2. For a > 0, the equilibrium points are complex numbers, specifically y = ±i * sqrt(ln(a)).
The bifurcation that takes place at each value depends on the change in the nature of the equilibrium points as a varies. In this case, at a = 0, there are no equilibrium points, so there is a change from no equilibrium points to having complex equilibrium points as a > 0.