Re: 2 coupled diff. eqns
- To: mathgroup at smc.vnet.net
- Subject: [mg21169] Re: 2 coupled diff. eqns
- From: Pasquale Nardone <Pasquale.Nardone at ulb.ac.be>
- Date: Fri, 17 Dec 1999 01:22:50 -0500 (EST)
- Organization: Brussels Free Universities VUB/ULB
- References: <831vap$g58@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
just define z=f+I*g then the equations reads z'=q-c*(z/|z|) where q=a+I*b is a complex constant Now you go to a polar representation z=rho*Exp[I*theta] you also define q=mu*Exp[I*phi] (mu=Sqrt[a^2+b^2], and Tan[phi]=b/a, as usual) the equations reads: rho'+I*rho*theta'=mu*Exp[I*(phi-theta)]-c which you decompose in real and imaginary part leading to rho'=mu*Cos[(phi-theta)]-c rho*theta'=mu*Sin[(phi-theta)] you go then to rho versus theta (instead of rho(t) and theta(t)->rho(theta)) and the eq is: (1/rho)*(drho/dtheta)=Cotan[phi-theta]-(c/mu)*(1/Sin[phi-theta]) which is integrable: Log[rho]=-Log[Sin[phi-theta]]+(c/mu)*Log[Tan[(phi-theta)/2]]+Constant now you put rho[theta] in rho*theta'=mu*Sin[(phi-theta)] to find the dependance of "t" versus theta to finish the problem -- -------------------------------------------- Pasquale Nardone * Universite Libre de Bruxelles * CP 231, Sciences-Physique-Agrigation * Bld du Triomphe * 1050 Bruxelles, Belgium * tel: 650,55,15 fax: 650,57,67 (+32,2) * http://homepages.ulb.ac.be/~pnardon/ * --------------------------------------------