# Unsolvable differential equations?

1. Sep 26, 2009

### RedX

Suppose you have the differential equation:

$$dg/dt=g^3$$

for a function g(t).

I got that the solution works out to be:

$$g(t)=\pm \left( \frac{1}{-2t}\right)^{1/2}$$

Does this mean that the original differential equation has no solution for t>0, since you can't have a negative in a square root?

If so, how did this happen?

2. Sep 26, 2009

### lurflurf

Re: unsolveable differential equations?

There is an arbitrary constant
$$g(t)=\pm \left( \frac{1}{C-2t}\right)^{1/2}$$

3. Sep 26, 2009

### Ben Niehoff

Re: unsolveable differential equations?

First, you forgot the arbitrary constant:

$$\int \frac{dg}{g^3} = \int dt$$

$$-\frac{1}{2g^2} = t + C$$

$$g(t) = \pm \frac{1}{(-2t + C)^{1/2}}$$

Also, take a look at the slope field for this differential equation. You will see that all trajectories sweep up (or down) away from the t axis, and they diverge in finite time. So, your function simply grows too fast.

In fact, for the general equation

$$\frac{dg}{dt} = g^r$$

you will find that for r > 1, solutions will blow up in finite time. Note that r = 1 corresponds to exponential growth. To get faster than exponential growth while still not blowing up in finite time, you need something that is not quite a power...for example

$$\frac{dg}{dt} = g \ln g$$

has the solution

$$g = e^{e^t}$$

4. Sep 26, 2009

### RedX

Re: unsolveable differential equations?

oops. In general, is the number of constants equal to the highest order derivative?

But can you really see vertical asymptotes from a graph? If you have y=x^2, then at x=1000 all the slopes look vertical.

Anyways, the differential equation I chose is of the form of the beta function in QED calculated to 1-loop. I was just wondering if you could integrate it to get the coupling as a function of energy. So would t=C/2 in your formula correspond to what is called a Landau pole?

5. Sep 26, 2009

### lurflurf

Re: unsolveable differential equations?

Not in general no. Though there are theorems that establish that fact with restrictions. One problem is constants can arise due the form of an equation rather than the order of differentiation.

(y')^2-3yy'+2y^2=0
has only first derivatives but the solution has two constants
y=C1*exp(x)+C2*exp(2x)

6. Sep 26, 2009

### RedX

Re: unsolveable differential equations?

Would it be safe to say that a differential equation has at least a number of constants equal to the order? So for your example you have a first order equation having two constants, and two is greater than one. So for example a 2nd order differential equation would have at least two constants, but perhaps more?

7. Sep 27, 2009

### lurflurf

Re: unsolveable differential equations?

No, consider
(y')^2+y^2=0
which has one solution y=0 and no constants
or
(y')^2+y^2=-1
which has no solutions

Each order differentiation gives the potantial for a constant, but the relation of the derivatives (the form of the differential equation) can add or subtract from that. This is why your hypothesis hold when the differntial equation is simple.