Can a differential equatio have NO solution ?

I'm not sure why you say "so complicated". It is (fairly) easy to prove that if f is differentiable, then its derivative satisfies the "intermediate value theorem" (If f(a)= A, f(b)= B and C is some number between A and B then there exist c between a and b such that f(c)= C).
If f(x) is a function that does not have that property (f(x)= 0 if x is irrational, 1 if x is rational is an example), then there cannot be a function y that satisfies dy/dx= f(x).

More simply there exist many "boundary value problems" that have no solution. There is no function satisfying d²y/dx²+ y= 0, and y(0)= 0, y([itex]\pi[/itex])= 1.

 

it also depends on domain
(y')^2+1=0
has no real solutions
contradictory equations like
y'=y'+1
have no solution

 

Re: Can a differential equatio have NO solution ??

yea but that's given an initial value. and even if the function does not have a solution on a given interval, it CAN have solutions on other intervals and those are still considered solutions. Lurflurf has the more correct explanation although (y')^2+1=0 may have a solution in the complex plane? (i dunno i haven't taken complex variables).

 

Re: Can a differential equatio have NO solution ??

Try the Navier Stokes equation, I believe if you can prove that in three dimensions solutions always exist (existence), or that if they do exist they do not contain any infinities, singularities or discontinuities (smoothness) you get a nice amount of cash. It's a complicated partial differential.

And no I wouldn't dream of trying it myself.

http://en.wikipedia.org/wiki/Navier-Stokes_equations

Many differentials have no exact solution and other methods have to be used to approximate them, a good example is the non-linear equations that regulate the position systems on GPS Satellites, taking account of signal time at c.

 

Re: Can a differential equatio have NO solution ??

peace upon u all

i was wondering about this point exactly .. why shall differential equation have NO solution .. and supos a physical case described with a differential equation that have no solution .. what does that mean ???
does it have relation to chaotic analysis for some cases ??

 

Re: Can a differential equatio have NO solution ??

It just means that there is no precise solution. You could still find a solution that was close to it so as to make no difference by other means. In the same way some integrals have no solution but can be approximated with infinite slices of the graph. This is where the idea of the Taylor expansion comes in.

 

Re: Can a differential equatio have NO solution ??

erm ... do u mean that we can use a solution of a similar differential equation ??
or u mean trying to solve it numerically ??

 

Re: Can a differential equatio have NO solution ??

No you couldn't solve it numerically but you could approximate it as x approaches infinity say. Like the satellite example they use a more advanced form of triangulation to map the positions, that is accurate enough so as to not make much difference. Essentially that is a method that doesn't even require differentiation as such. Even the relativistic concerns are handled separately.

 

Re: Can a differential equatio have NO solution ??

peace upon u dear

mmmm

i'm thankful a lot for this hint nut do u have more illustration for this method u metioned ??

or what is it mathematically based on ??

my best wishes

 

Re: Can a differential equatio have NO solution ??

peace upon u dear

mmmm

i'm thankful a lot for this hint nut do u have more illustration for this method u metioned ??

or what is it mathematically based on ??

my best wishes

 

Re: Can a differential equatio have NO solution ??

http://en.wikipedia.org/wiki/Taylor_series

[tex]1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}+ \cdots \qquad = \qquad 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots.\![/tex]

Taylor expansion for e^x.

In sum form:

[tex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\text{ for all } x\![/tex]

Sin function:

[tex]\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}\quad = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\text{ for all } x\![/tex]

Note as you add more terms e^x and sin(x) the approximation becomes more accurate.

 

Re: Can a differential equatio have NO solution ??

You forgot to add [tex] \infty [/tex] on top and do you mean that as you add more terms, the approximation for [tex] e^x [/tex] and sin(x) become more accurate maybe? This is done a lot when you solve the DE using power series / method of Frobenius. Sometimes you are lucky to recognize a pattern and the appropriate function and sometimes you are not and just have to leave the answer in terms of an infinite sum.

 

Re: Can a differential equatio have NO solution ??

Considering I lazily cut and pasted that out of wiki, huh? Thanks.

I did indeed mean that too.

Always check your work even if you cut and paste.