# Unstable solution of differential equation

1. Dec 7, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

Do you suggest a numerical approach?

Can this not be done analytically?

2. Dec 7, 2016

### 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).

3. Dec 7, 2016

### spaghetti3451

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?

4. Dec 7, 2016

### 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 $x \rightarrow x + a$

5. Dec 7, 2016

### spaghetti3451

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

6. Dec 7, 2016

### 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.

7. Dec 7, 2016

### 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?

8. Dec 7, 2016

### 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.

Last edited: Dec 7, 2016
9. Dec 7, 2016

### spaghetti3451

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

10. Dec 7, 2016

### hilbert2

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.

11. Dec 7, 2016

### 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?

12. Dec 7, 2016

### 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...

Last edited: Dec 7, 2016
13. Dec 7, 2016

### 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?

14. Dec 7, 2016

### 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.

Last edited: Dec 7, 2016
15. Dec 7, 2016

### 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.

16. Dec 7, 2016

### spaghetti3451

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

17. Dec 7, 2016

### hilbert2

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.

18. Dec 7, 2016

### spaghetti3451

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

19. Dec 7, 2016

### hilbert2

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.

20. Dec 7, 2016

### spaghetti3451

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?