I need to solve this D. E. (( plz help )

**CaptainBlack** · Jun 28th 2009, 07:55 AM

Originally Posted by amro05

Are you sure? Where did this ODE come from? It clearly has no real solutions and I suspect the complex solutions are all constant derivative solutions.

CB

**shawsend** · Jun 28th 2009, 08:01 AM

I get $y(t)=c_1-t\text{W}(-1)$

where W is the Lambert W-function. May I ask what's wrong with a complex solution?

**CaptainBlack** · Jun 28th 2009, 09:31 AM

Originally Posted by shawsend

I get $y(t)=c_1-t\text{W}(-1)$

where W is the Lambert W-function. May I ask what's wrong with a complex solution?

Except the Lambert-W is mulitivalued (and $W(-1)$ is complex for all branches IIRC), you do not know which branch you are on without an initial condition, and not all initial conditions give solutions.
CB

**chisigma** · Jun 28th 2009, 09:39 AM

The equation...

$e^{z}=z$ (1)

... in the complex domain is equilvalent to a couple of equations in real domain. Setting $z= \sigma + i\cdot \omega$ the (1) became...

$e^{\sigma} \cdot \cos \omega = \sigma$

$e^{\sigma}\cdot \sin \omega = \omega$ (2)

If $\sigma^{*} + i\cdot \omega^{*}$ is a solution of (2) then the proposed D.E. becomes...

$y^{'}= \sigma^{*} + i\cdot \omega^{*}$ (3)

... and its solution is...

$y^{'}= \{\sigma^{*} + i\cdot \omega^{*}\} \cdot x + c$ (4)

The question is now: does (2) have solutions?...

Kind regards

$\chi$ $\sigma$

**shawsend** · Jun 28th 2009, 09:55 AM

Originally Posted by CaptainBlack

Except the Lambert-W is mulitivalued (and $W(-1)$ is complex for all branches IIRC), you do not know which branch you are on without an initial condition, and not all initial conditions give solutions.
CB

If you might, which initial conditions do not give solutions?

**amro05** · Jun 28th 2009, 02:26 PM

Originally Posted by CaptainBlack

Are you sure? Where did this ODE come from? It clearly has no real solutions and I suspect the complex solutions are all constant derivative solutions.

CB

let meput it this way
this ODE came from the solution of

the last one gives this ODE

the last one is to hard to solve so I put it = 0 < jest hopping to find same idees

**chisigma** · Jul 1st 2009, 05:27 AM

Originally Posted by chisigma

The equation...

$e^{z}=z$ (1)

... in the complex domain is equilvalent to a couple of equations in real domain. Setting $z= \sigma + i\cdot \omega$ the (1) became...

$e^{\sigma} \cdot \cos \omega = \sigma$

$e^{\sigma}\cdot \sin \omega = \omega$ (2)

...

... the question is now: does (1) have solutions?...

The answers is done by the follwing lemma of complex variable functions theory...

Given the complex variable function...

$w = f(z) = c_{1} z + c_{2} z^{2} + c_{3} z^{3} + ...$ , $c_{1} \ne 0$ (1)

... holomorphic in a circle $|z|< \epsilon$ , its inverse function is done by the Taylor expansion...

$z= f^{-1} (w) = d_{1} w + d_{2} w^{2} + d_{3} w^{3} + ...$ , $d_{n} = \frac{1}{n!} \lim _{z \rightarrow 0} \frac {d^{n-1}}{dz^{n-1}} \{\frac{z}{f(z)}\}^{n}$ (2)

Now we use this lemma to find the solution $z=z(w)$ of the trascendental equation...

$w= z\cdot e^{-z}$ (3)

With the formulas (2) we obtain the Taylor expansion...

$z= \sum_{n=1}^{\infty} \frac {n^{n-1}}{n!} w^{n}$ (4)

... which converges only for $w=0$ . But our equation was...

$z\cdot e^{-z}=1$ (5)

... and then it has been demonstrated it has no solutions...

Kind regards

$\chi$ $\sigma$

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