Then, the only positive equilibrium point ỹ of Eq.(1.1) is given by Ei-1 di = (2.7) Li=1 provided that A < 1. Now, let us introduce a continuous function H : (0, 00)6 → (0, 0) which is defined by H(uo, ..., uz) = Auo + (2.8) Therefore, it follows that H(uo,..us) А, Ong H(u0..,u5) H(40,..,us) duz a2 (B1u1+E(Biu;)] – B2 [a141+E_3(a;u;)] H(uo,...us) duz H(uo,...us) dug as E (B;u;)+Bgus] – Ba [C1(@;u;)+a5us] H(u0,...,u5) dug as E(B:u;)] – Bs EL(@;u;)]
Then, the only positive equilibrium point ỹ of Eq.(1.1) is given by Ei-1 di = (2.7) Li=1 provided that A < 1. Now, let us introduce a continuous function H : (0, 00)6 → (0, 0) which is defined by H(uo, ..., uz) = Auo + (2.8) Therefore, it follows that H(uo,..us) А, Ong H(u0..,u5) H(40,..,us) duz a2 (B1u1+E(Biu;)] – B2 [a141+E_3(a;u;)] H(uo,...us) duz H(uo,...us) dug as E (B;u;)+Bgus] – Ba [C1(@;u;)+a5us] H(u0,...,u5) dug as E(B:u;)] – Bs EL(@;u;)]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Show me the steps of determine yellow and inf is here
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Step 1
We have a function H defined as
which implies that H is the function of 6 variables.
We have to explain the partial derivatives of the function H with respect
to these 6 variables and .
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