Limiting cases II: repeated roots of a differential equation

The solution to a polynomial-ish differential equation (the formal name being "linear homogenous time-invariant differential equation") with repeated roots is not completely unintuitive. While it's not immediately obvious where the solution to $(D-rI)^2y(t)=0$

comes from, it is pretty clear in the case $r=0$, where $D^2y(t)=0$ is solved by

... so it seems that the linear function comes from integrating twice, or more correctly, inverting the same differential operator twice.

Let's try to derive our desired equation $y=(c_1+c_2t)e^{rt}$ via a limit. It doesn't seem like this would arise in the limit of an equation like $y=c_1e^{r_1t}+c_2e^{r_2t}$, but once again -- this is an arbitrary-constant-problem. Much like how we switched to definite integrals (i.e. fixed the limits/boundary conditions of the integral) before taking the limit in Part 1, we must fix the initial conditions here too.

For those new to this series, here's the reason we switch to an initial conditions approach/co-ordinate system:
Most people have the right idea, that you need to take the solution for non-repeated roots, and take the limit as the roots approach each other. This is correct, but it's a mistake to take the limit of the general solution $c_1e^{r_1t}+c_2e^{r_2t}$, which is what most people try to do when they see this problem, and are then puzzled since it gives you a solution space of the wrong dimension.

This is wrong, because $c_1$ and $c_2$ are arbitrary mathematical labels, and have no reason to stay the same as the roots approach each other. You can, however, take the limit while representing the solution in terms of your initial conditions, because these can stay the same as you change the system.

You can think of this as a physical system where you change the damping and other parameters to create a repeated-roots system as the initial conditions remain the same -- this is a simple process, but if you instead try to ensure $c_1$ and $c_2$ remain the same, you'll run into infinities and undefined stuff.

This is exactly what happens here, there simply isn't a repeated-roots solution with the same $c_1$ and $c_2$ values, but you obviously do have a system/solution with the same initial conditions.
Taken from my answer on Math Stack Exchange.

We consider the differential equation

And tend $r\to1$. The solution to the equation in general is

$$y(t) = {c_1}{e^t} + {c_2}{e^{rt}}$$
 If we let $y(0) = a,\,\,y'(0) = b$, then it shouldn't be hard to show that the solution we're looking for is

This is where we must tend $r\to1$. Doing so is simply algebraic manipulation and a bit of limits:

$$\begin{array}{c}y(t) = \frac{{\left( {ra - b} \right){e^t} - \left( {a - b} \right){e^{rt}}}}{{r - 1}} = \frac{{\left( {ra - b} \right) - \left( {a - b} \right){e^{(r - 1)t}}}}{{r - 1}}{e^t}\\ = \frac{{(r - 1)a + \left( {a - b} \right) - \left( {a - b} \right){e^{(r - 1)t}}}}{{r - 1}}{e^t}\\ = \left[ {a + \left( {a - b} \right)\frac{{1 - {e^{(r - 1)t}}}}{{r - 1}}} \right]{e^t}\\ = \left[ {a - \left( {a - b} \right)\frac{{{e^{(r - 1)t}} - {e^{0t}}}}{{r - 1}}} \right]{e^t}\\ = \left[ {a - \left( {a - b} \right){{\left. {\frac{d}{{dx}}\left[ {{e^{xt}}} \right]} \right|}_{x = 0}}} \right]{e^t}\\ = \left[ {a - \left( {a - b} \right)t} \right]{e^t}\end{array}$$
Which indeed takes the form

$$y(t) = \left( {{c_1} + {c_2}t} \right){e^t}$$
With $c_1,\,\,c_2$ such that $y(0)=a,\,\,y'(0)=b$.

Here's a visualisation of the limit, with varying values of $r$:

And here's an interactive version with a slider for r.

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