Inequality satisfied by solution
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From: Blue Venom on 4 May 2010 21:00 y'=SQRT(1+x2+y2) y(0)=0 Assuming there exists a local and unique solution defined on (-d,d), prove that the solution y is defined for all x in R and that y(x) >= sinh(x) for all x >=0. Any hint?
From: Ray Vickson on 4 May 2010 23:14 On May 4, 6:00 pm, Blue Venom <mandalayray1...(a)gmail.com> wrote: > y'=SQRT(1+x2+y2) > y(0)=0 > > Assuming there exists a local and unique solution defined on (-d,d), > prove that the solution y is defined for all x in R and that y(x) >= > sinh(x) for all x >=0. > > Any hint? I think it is false. Solving the DE numerically and plotting the results for 0 <= x <= 5 shows the exact _opposite_ inequality, at least when done using the default numeric solver in Maple 9.5. Anyway: why did you re-post the question, using incorrect notation this time? x2 should be x^2 and y2 should be y^2. R.G. Vickson
From: Blue Venom on 5 May 2010 03:30 05/05/2010 5.14, Ray Vickson: > I think it is false. Solving the DE numerically and plotting the > results for 0<= x<= 5 shows the exact _opposite_ inequality, at > least when done using the default numeric solver in Maple 9.5. > > Anyway: why did you re-post the question, using incorrect notation > this time? x2 should be x^2 and y2 should be y^2. > > R.G. Vickson Yeah, my mistake. I guess I have to prove the opposite then and there's a typo in the text. But no clue as to how to do it.
From: Alois Steindl on 5 May 2010 03:59 Blue Venom <mandalayray1985(a)gmail.com> writes: > > Yeah, my mistake. I guess I have to prove the opposite then and > there's a typo in the text. But no clue as to how to do it. Hello, since that sounds much like a homework, you should definitivly try to do it yourself. If you don't have any clue, you should read, what you have learned in the recent lectures; if you don't understand that, you have to go back, until you understand it. Newsgroups are not for cheating. Alois -- Alois Steindl, Tel.: +43 (1) 58801 / 32558 Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598 Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10
From: Torsten Hennig on 4 May 2010 23:54
> y'=SQRT(1+x2+y2) > y(0)=0 > > Assuming there exists a local and unique solution > defined on (-d,d), > prove that the solution y is defined for all x in R > and that y(x) >= > sinh(x) for all x >=0. > > > Any hint? Hint: sinh(x) satisfies the differential equation y' = sqrt(1+y^2). Best wishes Torsten. |