Unstable solution of differential equation

[spaghetti3451]

Homework Statement

Consider the following differential equation

$$\frac{\partial^{2}\phi}{\partial t^{2}}-\nabla^{2}\phi = \phi(a-b\phi^{2}), \qquad a>0.$$

I would like to prove that $\phi=0$ is an unstable solution of this equation.

Homework Equations

The Attempt at a Solution

Do you suggest a numerical approach?

Can this not be done analytically?

 

[hilbert2]

Here we need to do something that's called linear stability analysis... Suppose that at an initial moment $t=t_0$, the function $\phi$ is almost $\phi (x,t_0 ) = 0$, but has a very small-amplitude oscillation in it:

$\phi (x, t_0 ) = \epsilon \sin (k x)$, $\epsilon << 1$.

Now, let's also assume that the initial behavior of this solution (in 1st order with respect to $t$), is such that the amplitude starts growing or decreasing exponentially:

$\phi (x, t) = \epsilon e^{\beta (t-t_0 )} \sin (k x)$

Substitute this trial function $\phi$ in the diff. equation, expand both sides of the resultant equation as Taylor series about $t_0$, discard all terms that are higher than first order in $\epsilon$ and then prove that this gives the parameter $\beta$ a positive value (implying exponential growth).

 

[spaghetti3451]

Here we need to do something that's called linear stability analysis... Suppose that at an initial moment $t=t_0$, the function $\phi$ is almost $\phi (x,t_0 ) = 0$, but has a very small-amplitude oscillation in it:

$\phi (x, t_0 ) = \epsilon \sin (k x)$, $\epsilon << 1$.

Now, let's also assume that the initial behavior of this solution (in 1st order with respect to $t$), is such that the amplitude starts growing or decreasing exponentially:

$\phi (x, t) = \epsilon e^{\beta (t-t_0 )} \sin (k x)$

Substitute this trial function $\phi$ in the diff. equation, expand both sides of the resultant equation as Taylor series about $t_0$, discard all terms that are higher than first order in $\epsilon$ and then prove that this gives the parameter $\beta$ a positive value (implying exponential growth).

Why do we assume, to begin with, that

$\phi (x, t_0 ) = \epsilon \sin (k x)$, $\epsilon << 1$?

In other words, why does $\phi(x)$ have a small-amplitude oscillation? Could the perturbation not be just a constant number in space?

 

[hilbert2]

The constant function is just the special case where $k=0$. Using the sine function trial allows finding out whether the system is unstable for some range of wavenumbers $k$ and stable for others.

EDIT: Sorry, it has to be $\cos (kx)$ for it to contain the constant function case. However, this does not matter if the DE is symmetric in translation $x \rightarrow x + a$

 

[spaghetti3451]

I was wondering why you use the words 'in first order in $t$' when describing the initial behaviour of the system?

 

[hilbert2]

Well, for example, the function $\sqrt{x+1}$ is equal to $1 + \frac{1}{2}x$ in 1st order of the variable $x$. It means that you consider the function only in so small intervals that it's graph is practically linear.

 

[spaghetti3451]

But, what if the system follows some other solution of the differential equation.

I mean,

$\phi (x, t) = \epsilon e^{\beta (t-t_0 )} \sin (k x)$

is just an ansatz and it could be that it does not encapsulate the full set of solutions of the differential equation, right?

 

[hilbert2]

Note that any function $\phi (x, t_0 )$ can be written as a combination of different sine and cosine functions with different wavelengths. In this linear approximation we only consider a time interval or values of $\epsilon$ that are so small that we can assume the different Fourier components evolve independently, like they would in the case of a linear partial differential equation (note that the equation here is nonlinear).

EDIT: also note that a single counterexample like this is enough to prove that a solution is not stable, while proving that a solution is stable requires a lot more effort.

 

[spaghetti3451]

Why do the different Fourier components evolve independently for a small time interval or small value of $\epsilon$?

 

[hilbert2]

Why do the different Fourier components evolve independently for a small time interval or small value of $\epsilon$?

Because if $\epsilon$ is small, then the term $-b\phi^3$ which causes the nonlinearity of the equation is even smaller by several orders of magnitude.

 

[spaghetti3451]

Ok, so following your prescription,

Substituting the trial function $\phi(x,t)$ in the differential equation, I get

$\partial^{\mu}\partial_{\mu}\phi = a\phi - b\phi^{3}$

$(\partial_{t}^{2}-\partial_{i}^{2})\phi = a\phi - b\phi^{3}$

$(\beta^{2}+\vec{k}^{2})\phi = a\phi - b\phi^{3}$

Then,

$(\beta^{2}+\vec{k}^{2})\phi = a\phi - b\phi^{3}$

$(\beta^{2}+\vec{k}^{2})\left(\epsilon e^{\beta (t-t_0 )} \sin (\vec{k}\cdot{\vec{x}})\right) = a\left(\epsilon e^{\beta (t-t_0 )} \sin (\vec{k}\cdot{\vec{x}})\right) - \underbrace{b\left(\epsilon e^{\beta (t-t_0 )} \sin (\vec{k}\cdot{\vec{x}})\right)^{3}}_{\text{ignore}}$

I ignored the second term as it is cubic in $\epsilon$. Also, I have decided to use wavevector $\vec{k}$ instead of wavenumber $k$.

I don't why I need to expand in taylor series about $t_0$. After all, we already have an equation for $\beta$ in terms of $a$ and $\vec{k}$, right?

 

[hilbert2]

Yeah, you don't really need a Taylor series when the nonlinearity is a power of $\phi$ ($\phi^3$ in this case) and not some more complicated function. Just divide the equation with $\epsilon e^{\beta (t-t_0)} \sin kx$ and find the conditions for $\beta$ to be a positive real number.

EDIT: oh, looks like the equation only gives information about $\beta^2$... Maybe we'd need an ansatz that's proportional to $e^{\sqrt{\beta}(t- t_0)}$. I've only done this for equations that have a first order time derivative, before...

 

[spaghetti3451]

Why can we divide by $\sin (kx)$? After all, the sine function can be $0$ and then we'd be dividing by $0$.

Why can't we just divide by the exponential and $\epsilon$ on both sides of the equation?

 

[hilbert2]

I corrected the last post. I'll try to find some article where the linear stability analysis is done for an equation that is second order in time derivative.

EDIT: You probably need to use a perturbation term that is a plane wave (proportional to $e^{ikx}$) instead of a sine or cosine function, to get rid of the division by zero problem. I've seen that kind of an approach being used in some articles.

 

[hilbert2]

If you use the ansatz $\phi (x,t) = \epsilon e^{\sqrt{\beta}t}e^{ikx}$, you get the equation $\beta = a - k^2$, which means that there's some maximum value the wavenumber $k$ can have for the solution to be unstable. This kind of relations between the "Lyapunov exponent" $\beta$ and the perturbation wavelength are called dispersion relations.

 

[spaghetti3451]

If you use the ansatz $\phi (x,t) = \epsilon e^{\sqrt{\beta}t}e^{ikx}$, you get the equation $\beta = a - k^2$, which means that there's some maximum value the wavenumber $k$ can have for the solution to be unstable. This kind of relations between the "Lyapunov exponent" $\beta$ and the perturbation wavelength are called dispersion relations.

So, solution is unstable for $a> \sqrt{k}$?

 

[hilbert2]

So, solution is unstable for $a> \sqrt{k}$?

It appears to be if $a>k^2$ , and particularly the constant solution ($k=0$) is always unstable if $a$ is a positive real number.

 

[spaghetti3451]

If $\beta < 0$, then the solution decreases exponentially. Why is this not an unstable solution also?

 

[hilbert2]

If $\beta < 0$, then the solution decreases exponentially. Why is this not an unstable solution also?

Because then any deviation from the constant solution disappears really quickly. The situation is similar to comparing these images:

and assuming that there's friction between the sphere and the convex/concave surface. The constant solution is similar to the equilibrium situation where the sphere is positioned exactly on the top/bottom of the hill.

 

[spaghetti3451]

Because then any deviation from the constant solution disappears really quickly.

I don't quite understand. How does the deviation disappear? It's still there, right?

After all, an exponential solution with negative exponent takes the initial value of $\phi$ exponentially (i.e. takes forever) to zero, right?

 

[hilbert2]

It's the same thing as the difference between positive and negative feedback in systems theory: http://serc.carleton.edu/introgeo/models/loops.html

In the positive feedback case, a small deviation from equilibrium blows up uncontrollably, like the number of grains in the "wheat and chessboard problem".

 

[spaghetti3451]

Ok, on another note, why is this analysis called the linear stability analysis?