Unsolvable differential equations?

In summary, the given differential equation has a solution of g(t)=\pm \left( \frac{1}{C-2t}\right)^{1/2} but it does not have a solution for t>0 due to the presence of a negative value within the square root. This can be seen in the slope field for the equation, which shows that all trajectories diverge in finite time. In general, the number of constants in a solution does not necessarily equal the order of the differential equation, as it also depends on the form of the equation.
  • #1
RedX
970
3
Suppose you have the differential equation:

[tex]dg/dt=g^3 [/tex]

for a function g(t).

I got that the solution works out to be:

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

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?
 
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  • #2


There is an arbitrary constant
[tex]g(t)=\pm \left( \frac{1}{C-2t}\right)^{1/2} [/tex]
 
  • #3


First, you forgot the arbitrary constant:

[tex]\int \frac{dg}{g^3} = \int dt[/tex]

[tex]-\frac{1}{2g^2} = t + C[/tex]

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

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

[tex]\frac{dg}{dt} = g^r[/tex]

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

[tex]\frac{dg}{dt} = g \ln g[/tex]

has the solution

[tex]g = e^{e^t}[/tex]
 
  • #4


lurflurf said:
There is an arbitrary constant
[tex]g(t)=\pm \left( \frac{1}{C-2t}\right)^{1/2} [/tex]

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

Ben Niehoff said:
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.

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


RedX said:
In general, is the number of constants equal to the highest order derivative?

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


lurflurf said:
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)

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


RedX said:
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?

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.
 

FAQ: Unsolvable differential equations?

What makes a differential equation unsolvable?

There are several factors that can make a differential equation unsolvable. One common reason is that the equation is nonlinear, meaning it cannot be written in the form of y' = f(x,y). Another reason could be that the equation involves functions that are not elementary, such as trigonometric or exponential functions. Additionally, the initial or boundary conditions may not be well-defined, making it impossible to find a unique solution.

Can all unsolvable differential equations be solved numerically?

Not necessarily. While numerical methods can often provide approximations to the solution of an unsolvable differential equation, there are cases where even these methods fail. This can happen if the equation is "stiff", meaning that the solution changes very rapidly over a small range of values, making it difficult to accurately approximate.

Are there any techniques for solving unsolvable differential equations?

Yes, there are some techniques that can be used to find solutions to certain types of unsolvable differential equations. For example, some equations can be transformed into a solvable form by using a change of variables or by applying a specific method, such as separation of variables or the method of integrating factors. However, these techniques may not work for all unsolvable equations and often require a certain level of mathematical expertise.

Can unsolvable differential equations have real-world applications?

Yes, many real-world problems can be described by unsolvable differential equations. For example, in physics, the motion of a pendulum can be modeled by an unsolvable differential equation. Similarly, in biology, population growth and the spread of diseases can be described by unsolvable equations. While these equations may not have a closed-form solution, numerical methods can still be used to study and understand the behavior of these systems.

Are there ongoing research efforts to solve unsolvable differential equations?

Yes, there is active research in the field of differential equations to find new techniques and methods for solving unsolvable equations. This includes developing new numerical methods, exploring new approaches to transform equations into solvable forms, and studying the behavior of unsolvable equations to gain new insights. However, it is still an area of ongoing study and there are many unsolvable equations that remain unsolved.

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