Differential Equations

Introduction

Even though I’m only a senior in high school, I’m taking a course in differential equations. We came to the section where we learn to solve high order linear differential equations with constant coefficients. We went over the characteristic polynomial, and the case where it has a repeated root. We learned that if r is a repeated root then, e^{rt} and te^{rt} are both solutions to the differential equation. There was a proof of this using reduction of order. Our teacher also told us that if r is repeated more times, then we can keep multiplying our previous solutions by t for every time r was repeated. This, he said without proof. So, I decided to prove it myself.

Trial 1

So, let’s say you have a differential equation with constant coefficients, second order to make it easy, as shown below:
y'' + ay' + by = 0

This is completely general, as, if there was a coefficient in front of the y'' term, I could merely divide it out. I haven’t worked with differential operators very much, but I did notice that this could be written as the product of differential operators like this:
\left(\frac{d}{dt} - c_1\right)\left(\frac{d}{dt} - c_2\right)y = 0

because, if you multiply that out you get
\left(\frac{d^2}{dt^2} - (c_1 + c_2)\frac{d}{dt} + c_1c_2\right) y = 0

which obviously simplifies to the above equation of -(c_1 + c_2) = a and c_1c_2 = b.

The advantage of putting it in this form, is that it’s obvious that the solutions to the resulting characteristic equation are c_1 and c_2. It’s also obvious that higher order equations can be made easily by multiplying by \left(\frac{d}{dt} - c\right) where c will be another solution to the characteristic equation. It’s also worth noting that the order that they’re multiplied in doesn’t matter. You get the same thing.

Now, I considered what happened when \left(\frac{d}{dt} - c\right) is applied to t^ne^{ct}. Doing this you get
\left(\frac{d}{dt} -c\right) t^ne^{ct} = \left(nt^{n-1}e^{ct} + ct^ne^{ct} - ct^ne^{ct}\right) = nt^{n-1}e^{ct}

Notice that this is very similar to what the normal differentiation operator does to polynomials. It’s exactly the same except that t^n is replaced by t^ne^{ct}. Now, it’s obvious why t^ne^{ct} are also solutions. If we take the nth derivative of t^k, we get 0 if n >= k. In the same way, if we take \left(\frac{d}{dt} - c\right)^n t^ke^{ct}, we get 0 if n>= k.

Generaliziminization
I was satisfied with this result at the time, and didn’t think to take this any further. But recently we started going over Euler Equations. That is,  differential equations of the form:
x^2y'' + \alpha xy' + \beta y = 0

Basic solutions are of the form x^r, and by substituting that for y you can find the characteristic polynomial and solve it like you would the differential equations with constant coefficients. Repeated roots are treated the same way, except you multiply by \ln x instead.

This diff eq. can also be written as a product of differential operators, but this time of the form
(x\frac{d}{dx} - c)

What if one could derive that the natural log was the appropriate choice using  this fact? That is find a function g(x) that causes the above differentiation operator to act on g(x)^nx^c like the regular differentiation operator does on t^n. Actually, this is pretty easy to do. Merely solve the equation (x\frac{d}{dx} - c)g(x)^nx^c = ng(x)^{n-1}x^c for g(x).

(x\frac{d}{dx} - c)g(x)^nx^c = ng(x)^{n-1}x^c
\left[x\left(ng(x)^{n-1}g'(x)x^c + cg(x)^nx^{c-1}\right) - cg(x)^nx^c\right] = ng(x)^{n-1}x^c
ng(x)^{n-1}g'(x)x^{c+1} = ng(x)^{n-1}x^c
g'(x) = 1/x

Obviously g(x) = \ln x is an appropriate function. To generalize even further if you have any differential equation that can be produced by products of

(h(x)\frac{d}{dx} - c),

and the solutions are of the form f(x,c), solutions to equations with repeated roots can be found by repeatedly multiplying the base solution, f(x,c) by any g(x) that satisfies:
f(x,c) = e^{c\cdot g(x)}

For example, with linear differential equations with constant coefficients f(x,c) = e^{cx} and g(x) = x. In the case of Euler equations f(x,c) = x^c and g(x) = \ln x.

Knowing this probably isn’t very useful but finding connections in mathematics is always interesting. I do wonder, though, if there is a name for this. But whether or not there is, I think this is a fun way to see things.

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About rioghasarig

Just a high school student with an interest in math, science, programming, philosophy, religion, and...no that's pretty much it.
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