# Can a differential equatio have NO solution ?

Can a differential equatio have NO solution ??

I mean in many cases as mathematician you must recall to an existence theorem for PDE or ODE but my question is can be a differential equation so complicated that has NO solution at all ? , i mean that for example there is no function f(x,y) or h(x) that satisfy a certain linear or nonlinear differential equation.

HallsofIvy
Homework Helper
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($\pi$)= 1.

lurflurf
Homework Helper
it also depends on domain
(y')^2+1=0
has no real solutions
y'=y'+1
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($\pi$)= 1.
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).

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

Even though turbulence is an everyday experience, it is extremely difficult to find solutions, quantify, or in general characterize. A US\$1,000,000 prize was offered in May 2000 by the Clay Mathematics Institute to whoever makes preliminary progress toward a mathematical theory which will help in the understanding of this phenomenon.

The numerical solution of the Navier–Stokes equations for turbulent flow is extremely difficult, and due to the significantly different mixing-length scales that are involved in turbulent flow, the stable solution of this requires such a fine mesh resolution that the computational time becomes significantly infeasible for calculation (see Direct numerical simulation). Attempts to solve turbulent flow using a laminar solver typically result in a time-unsteady solution, which fails to converge appropriately. To counter this, time-averaged equations such as the Reynolds-averaged Navier-Stokes equations (RANS), supplemented with turbulence models (such as the k-ε model), are used in practical computational fluid dynamics (CFD) applications when modeling turbulent flows. Another technique for solving numerically the Navier–Stokes equation is the Large-eddy simulation (LES). This approach is computationally more expensive than the RANS method (in time and computer memory), but produces better results since the larger turbulent scales are explicitly resolved.
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.

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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 ??

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.

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

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

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

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

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

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

Taylor expansion for ex.

In sum form:

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

Sin function:

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

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

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You forgot to add $$\infty$$ on top and do you mean that as you add more terms, the approximation for $$e^x$$ 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.

You forgot to add $$\infty$$ on top and do you mean that as you add more terms, the approximation for $$e^x$$ 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.
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.

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