Inequality satisfied by solution
in [Math]
|
Prev: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Next: Chapt 3, tried the experiment with blue tinted fiberglass sheet; Quasars support a Refraction redshift, not a Doppler redshift #59; ATOM TOTALITY
From: Ray Vickson on 5 May 2010 17:55 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? Here, finally, is an approach to the proof. It is based on the Picard Method; see, eg., http://math.fullerton.edu/mathews/n2003/picarditeration/PicardIterationProof/Links/PicardIterationProof_lnk_2.html or http://www-math.mit.edu/~dav/picard.pdf . Let f(x,y) = sqrt(1+x^2+y^2) and h(x,y) = sqrt(1+y^2). Let y(x) solve dy/dx = f(x,y), y(0)=0 and z(x) solve dz/dx = h(x,z), z(0)=0. Of course, z(x) = sinh(x). In Picard's method we would convert the DEs to integral equations and then solve them iteratively. We have y(x) = int{f(t,y(t)) dt, t=0..x}, while z(x) = int{ h(t,z(t)) dt, t=0..x} for x > 0. Start with y_0(x) = z_0(x) = 0 for all x >= 0, and consider the iterations y_{n+1}(x) = int{f(t,y_n(t)) dt, t=0..x}, z_{n+1}(x) = int{h(t,z_n(t)) dt, t=0..x}. In the references for the Picard method there are theorems about the convergence of y_n(x) to y(x) and of z_n(x) to z(x) as n --> infinity. Assume we have such convergence (which WILL apply at least in a finite interval 0 <= t <= T0; at t = T0 you can re-start the method and extend the result to T0 + T1, etc.) Anyway, notice that for t > 0 we have h(t,w) < f(t,w). We have z_1(x) = int{h(t,0) dt, t=0..x}= integral(1 dt, t=0..x) = x and y_1(x) = int{sqrt(1+t^2) dt, t=0..x) > z_1(x) for x > 0. Thus, z_2(x) = int{g(t,z_1(t)) dt, t=0..x} < int{f(t,z_1(t)) dt, t=0..x} < int{f(t,y_1(t)) dt, t=0..x) = y_2(x), since f(t,a) < f(t,b) if a < b. Thus, z_2(x) < y_2(x) for x > 0. Continuing gives z_n(x) < y_n(x) for x > 0, so in the limit (assuming it is OK) we get y(x) < z(x). You can clean up the argument a bit, worrying about convergence, etc., but that is the basic idea. R.G. Vickson
From: Torsten Hennig on 5 May 2010 22:41 > 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? > > Here, finally, is an approach to the proof. It is > based on the Picard > Method; see, eg., > http://math.fullerton.edu/mathews/n2003/picarditeratio > n/PicardIterationProof/Links/PicardIterationProof_lnk_ > 2.html > or http://www-math.mit.edu/~dav/picard.pdf . > > Let f(x,y) = sqrt(1+x^2+y^2) and h(x,y) = > sqrt(1+y^2). Let y(x) solve > dy/dx = f(x,y), y(0)=0 and z(x) solve dz/dx = h(x,z), > z(0)=0. Of > course, z(x) = sinh(x). In Picard's method we would > convert the DEs to > integral equations and then solve them iteratively. > We have y(x) = > int{f(t,y(t)) dt, t=0..x}, while z(x) = int{ > h(t,z(t)) dt, t=0..x} for > x > 0. Start with y_0(x) = z_0(x) = 0 for all x >= 0, > and consider the > iterations y_{n+1}(x) = int{f(t,y_n(t)) dt, t=0..x}, > z_{n+1}(x) = > int{h(t,z_n(t)) dt, t=0..x}. In the references for > the Picard method > there are theorems about the convergence of y_n(x) to > y(x) and of > z_n(x) to z(x) as n --> infinity. Assume we have such > convergence > (which WILL apply at least in a finite interval 0 <= > t <= T0; at t = > T0 you can re-start the method and extend the result > to T0 + T1, etc.) > Anyway, notice that for t > 0 we have h(t,w) < > f(t,w). We have z_1(x) > = int{h(t,0) dt, t=0..x}= integral(1 dt, t=0..x) = x > and y_1(x) = > int{sqrt(1+t^2) dt, t=0..x) > z_1(x) for x > 0. Thus, > z_2(x) = > int{g(t,z_1(t)) dt, t=0..x} < int{f(t,z_1(t)) dt, > t=0..x} < > int{f(t,y_1(t)) dt, t=0..x) = y_2(x), since f(t,a) < > f(t,b) if a < b. > Thus, z_2(x) < y_2(x) for x > 0. Continuing gives > z_n(x) < y_n(x) for > x > 0, so in the limit (assuming it is OK) we get > y(x) < z(x). > > You can clean up the argument a bit, worrying about > convergence, etc., > but that is the basic idea. > > R.G. Vickson > > > Why so difficult ? Since the initial conditions for both ODEs y' = sqrt(1+y^2) and y' = sqrt(1+x^2+y^2) are identical, sqrt(1+y^2) <= sqrt(1+x^2+y^2) directy implies sinh(x) <= (solution of the ODE y'=sqrt(1+x^2+y^2)). Best wishes Torsten.
From: Ray Vickson on 6 May 2010 09:06
On May 5, 11:41 pm, Torsten Hennig <Torsten.Hen...(a)umsicht.fhg.de> wrote: > > 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? > > > Here, finally, is an approach to the proof. It is > > based on the Picard > > Method; see, eg., > >http://math.fullerton.edu/mathews/n2003/picarditeratio > > n/PicardIterationProof/Links/PicardIterationProof_lnk_ > > 2.html > > orhttp://www-math.mit.edu/~dav/picard.pdf. > > > Let f(x,y) = sqrt(1+x^2+y^2) and h(x,y) = > > sqrt(1+y^2). Let y(x) solve > > dy/dx = f(x,y), y(0)=0 and z(x) solve dz/dx = h(x,z), > > z(0)=0. Of > > course, z(x) = sinh(x). In Picard's method we would > > convert the DEs to > > integral equations and then solve them iteratively. > > We have y(x) = > > int{f(t,y(t)) dt, t=0..x}, while z(x) = int{ > > h(t,z(t)) dt, t=0..x} for > > x > 0. Start with y_0(x) = z_0(x) = 0 for all x >= 0, > > and consider the > > iterations y_{n+1}(x) = int{f(t,y_n(t)) dt, t=0..x}, > > z_{n+1}(x) = > > int{h(t,z_n(t)) dt, t=0..x}. In the references for > > the Picard method > > there are theorems about the convergence of y_n(x) to > > y(x) and of > > z_n(x) to z(x) as n --> infinity. Assume we have such > > convergence > > (which WILL apply at least in a finite interval 0 <= > > t <= T0; at t = > > T0 you can re-start the method and extend the result > > to T0 + T1, etc.) > > Anyway, notice that for t > 0 we have h(t,w) < > > f(t,w). We have z_1(x) > > = int{h(t,0) dt, t=0..x}= integral(1 dt, t=0..x) = x > > and y_1(x) = > > int{sqrt(1+t^2) dt, t=0..x) > z_1(x) for x > 0. Thus, > > z_2(x) = > > int{g(t,z_1(t)) dt, t=0..x} < int{f(t,z_1(t)) dt, > > t=0..x} < > > int{f(t,y_1(t)) dt, t=0..x) = y_2(x), since f(t,a) < > > f(t,b) if a < b. > > Thus, z_2(x) < y_2(x) for x > 0. Continuing gives > > z_n(x) < y_n(x) for > > x > 0, so in the limit (assuming it is OK) we get > > y(x) < z(x). > > > You can clean up the argument a bit, worrying about > > convergence, etc., > > but that is the basic idea. > > > R.G. Vickson > > Why so difficult ? > Since the initial conditions for both ODEs > y' = sqrt(1+y^2) and y' = sqrt(1+x^2+y^2) are > identical, > sqrt(1+y^2) <= sqrt(1+x^2+y^2) > directy implies > sinh(x) <= (solution of the ODE y'=sqrt(1+x^2+y^2)). Yes, but how do we KNOW this? Of course, this is quite intuitive. However, this still needs a _proof_. One can either quote a theorem from some source (which I could not find easily) or _prove_ the result. The proof I gave is actually quite general and does not depend on the actual forms of f(x,y) and g(x,y), only on their ordering and monotonicity. R.G. Vickson > > Best wishes > Torsten. |