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

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

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

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

Originally Posted by shawsend

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

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

Except the Lambert-W is mulitivalued (and $\displaystyle W(-1)$ 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

The equation...

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

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

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

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

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

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

... and its solution is...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Originally Posted by CaptainBlack

Except the Lambert-W is mulitivalued (and $\displaystyle W(-1)$ 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?

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

Originally Posted by chisigma

The equation...

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

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

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

$\displaystyle e^{\sigma}\cdot \sin \omega = \omega$ (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...

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

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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$