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[MUG] dsolve does not find second solution

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[MUG] dsolve does not find second solution
Author: Helmut Kahovec    Posted: Mon, 16 Dec 2002 12:26:41 +0100
>> From: Helmut Kahovec "helmut.kahovec" Dear Maple users, One of my students found a bug in dsolve(). It behaves the same in Maple7 and Maple8. The following Maple7 session shows that bug. > restart; > infolevel[dsolve]:=10; infolevel[dsolve] := 10 > ode1:=diff(y(x),x)+(4*x+3*y(x)+1)/(3*x+2*y(x)+1); /d \ 4 x + 3 y(x) + 1 ode1 := |-- y(x)| + ---------------- \dx / 3 x + 2 y(x) + 1 > sol1:=dsolve(ode1,y(x)); Methods for first order ODEs: Trying to isolate the derivative dy/dx... Successful isolation of dy/dx --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous C trying homogeneous types: trying homogeneous D <- homogeneous successful <- homogeneous successful 2 2 3/2 (x + 1) _C1 - 1/2 sqrt((x + 1) _C1 + 4) sol1 := y(x) = 1 - --------------------------------------------- _C1 There should be a second solution, which is missing: > sol2:=y(x)=1-(3/2*(x+1)*_C1+1/2*((x+1)^2*_C1^2+4)^(1/2))/_C1; 2 2 3/2 (x + 1) _C1 + 1/2 sqrt((x + 1) _C1 + 4) sol2 := y(x) = 1 - --------------------------------------------- _C1 > odetest(sol1,ode1); 0 > odetest(sol2,ode1); 0 We can produce both solutions if we first change coordinates appropriately: > with(PDEtools): > T:={x=t-1,y(x)=z(t)+1}; T := {y(x) = z(t) + 1, x = t - 1} > ode2:=dchange(T,ode1); /d \ 4 t + 3 z(t) ode2 := |-- z(t)| + ------------ \dt / 3 t + 2 z(t) > dsolve(ode2,z(t)); Methods for first order ODEs: Trying to isolate the derivative dz/dt... Successful isolation of dz/dt --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful 2 2 - 3/2 t _C1 - 1/2 sqrt(t _C1 + 4) z(t) = -----------------------------------, _C1 2 2 - 3/2 t _C1 + 1/2 sqrt(t _C1 + 4) z(t) = ----------------------------------- _C1 The inverse transformation gives the same two solutions as before: > Tinv:={t=x+1,z(t)=y(x)-1}; Tinv := {z(t) = y(x) - 1, t = x + 1} > sol12:=op(map(isolate,subs(Tinv,[%%]),y(x))); sol12 := 2 2 - 3/2 (x + 1) _C1 - 1/2 sqrt((x + 1) _C1 + 4) y(x) = ----------------------------------------------- + 1, _C1 2 2 - 3/2 (x + 1) _C1 + 1/2 sqrt((x + 1) _C1 + 4) y(x) = ----------------------------------------------- + 1 _C1 > map( normal, zip( `-`, map(subs,[sol2,sol1],y(x)), # you may have to use [sol1,sol2], map(subs,[sol12],y(x)) # instead ) ); [0, 0] Kind regards, Helmut

[MUG] Re: dsolve does not find second solution
Author: Maple User Group    Posted: Tue, 17 Dec 2002 13:07:28 -0500
>> From: Maple User Group "maple_gr" -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Mon, 16 Dec 2002 08:23:00 -0500 (EST) From: Stephen Forrest "saforres" To: "maple-list" Subject: dsolve does not find second solution On Mon, 16 Dec 2002, Helmut Kahovec wrote: [snip] > 2 2 > 3/2 (x + 1) _C1 - 1/2 sqrt((x + 1) _C1 + 4) > sol1 := y(x) = 1 - --------------------------------------------- > _C1 > > There should be a second solution, which is missing: > > > sol2:=y(x)=1-(3/2*(x+1)*_C1+1/2*((x+1)^2*_C1^2+4)^(1/2))/_C1; > > 2 2 > 3/2 (x + 1) _C1 + 1/2 sqrt((x + 1) _C1 + 4) > sol2 := y(x) = 1 - --------------------------------------------- > _C1 Observe that sol2 is equal to sol1 after the substitution _C1 -> -_C1, thus since _C1 can be any real, sol2 is contained in dsolve's solution. Steve -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: "Edgardo S. Cheb-Terrab" "ecterrab" To: "maple-expert" "maple-list" Subject: [maple-expert] dsolve does not find second solution Date: Mon, 16 Dec 2002 16:01:17 -0800 Hi, I see no solutions missing in dsolve's output - no bug. Given two general solutions S1(y, x, _C1) and S2(y, x, _C1), we say "these solutions are the same" (so there is only one) whenever S2(y, x, _C1) = S1(y,x, F(_C1) ) for some F. In simpler words: whenever S2 can be obtained from S1 by redefining the arbitrary constant _C1~ = F(_C1), then S2 is not a solution different than S1. For example, I can always change _C1 -> F(_C1), in the solution sol1 you show, with whatever F of my choice, and obtain as many different forms of the solution as I want - they are all"the same solution. In particular, if I change _C1 = -_C1 in the sol1 you show I obtain your sol2, from where sol2 is not another solution and therefore it is not the case that dsolve is missing one solution. Edgardo

[MUG] Re: dsolve does not find second solution
Author: Zijlstra    Posted: Thu, 19 Dec 2002 09:41:15 -0600
>> From: zijlstra "zijlstra" Dear Dr. Kahovec, The solutions are indeed (explicit) branches of one (implicit) solution. MAPLE should return either both (explicit) branches or stop at the implicit solution. The reason it omits one branch is in the method of solution: rather than realizing that the problem is exact (which makes for a very straightforward dtermination of the implicit solution), MAPLE first arrives at the type homogeneousD. Solutions of this type are generated in 'RootOf' form. But MAPLE does not stop there. Rather, it returns an closed form solution using 'allvalues' - the probable source of omission of the second branch. The bigger problem here is: why does MAPLE not recognize this ODE as exact? >ode1:=diff(y(x),x)=-(4*x+3*y(x)+1)/(3*x+2*y(x)+1); >with(DEtools): >odeadvisor(ode1,y(x),[exact]); [NONE] Hope this narrows down the problem some, Jan Zijlstra, Dept. Math. Sci., MTSU

[MUG] Re: dsolve does not find second solution
Author: Edgardo S Cheb-Terrab    Posted: Thu, 02 Jan 2003 15:52:03 -0800
>> From: "Edgardo S. Cheb-Terrab" "ecterrab" > ... (regarding) > >ode1:=diff(y(x),x)=-(4*x+3*y(x)+1)/(3*x+2*y(x)+1); >.. > The solutions are indeed (explicit) branches of one (implicit) solution. MAPLE > should return either both (explicit) branches or stop at the implicit > solution. Due to the arbitrary character of the integration constants, what seems to be two solutions are actually only one and the same - Maple is not missing a solution or a branch of it (there were previous replies to this). > ... > The bigger problem here is: why does MAPLE not recognize this ODE as exact? > >ode1:=diff(y(x),x)=-(4*x+3*y(x)+1)/(3*x+2*y(x)+1); > >with(DEtools): > >odeadvisor(ode1,y(x),[exact]); > [NONE] This is a different thing. The ode1 you show, > ode1:=diff(y(x),x)=-(4*x+3*y(x)+1)/(3*x+2*y(x)+1); 4 x + 3 y + 1 ode1 := y' = - ------------- 3 x + 2 y + 1 is not an exact differential equation. When the coefficient of y' is a constant, as in the above, the ODE is exact if and only if the right hand side does not depend on y(x), which is not the case. For an arbitrary ODE, one rapid way to check exactness in Maple (regardless of the ODE differential order) is to check whether a constant is an integrating factor. Let's see: # Verify whether "the number one" (a constant) is an integrating factor > DEtools[mutest]( 1, ode1); # Note zero, hence ode1 is not exact. x 1 ---------------- + ---------------- 2 2 (3 x + 2 y + 1) (3 x + 2 y + 1) For that reason, odeadvisor is telling ode1 is not exact. What perhaps is in your mind is that ode1 "becomes exact" if you multiply it (both sides) by the denominator of its right-hand-side, which in this example happens to be an integrating factor: # two integrating factors, the first one is just the denom of the rhs > DEtools[intfactor]( ode1) ; 3 x + 2 y + 1 3 x + 2 y + 1, --------------------- (x + y) (2 x + y + 1) So, multiplying by denom@rhs > denom(rhs(ode1)) * ode1; # <- this ODE is exact (3 x + 2 y + 1) y' = -4 x - 3 y - 1 > DEtools[mutest]( 1, %); 0 > odeadvisor( %%, y(x), [exact]); [_exact] Despite the apparent 'simplicity' of "just multiply by the denominator of the rhs" to make the ODE exact, in truth the denom@rhs is not a better or worse candidate than any other possible multiplying factor candidate and odeadvisor does not make any guess when testing: will tell the ODE is exact if an only if it is "as given". Regarding tackling an ODE using methods for homogeneous ODEs or for exact ones, Maple's dsolve has its default ordering of methods (first 'this' then 'that' one etc.) but you can always make it work as you prefer: just give the methods to be tried and in the order you want as second argument - see ?dsolve. Anyway, making the ODE exact by multiplying it by an integrating factor and then solving it as an exact equation you still see there is only one solution - the one returned when you try dsolve(ode1) - regardless of how many solutions dsolve outputs when you use optional arguments. Edgardo ----- Original Message ----- From: "zijlstra" "zijlstra" To: "maple-list" "maple-list" Sent: Thursday, December 19, 2002 7:41 AM Subject: [MUG] Re: dsolve does not find second solution >> From: zijlstra "zijlstra" Dear Dr. Kahovec, The solutions are indeed (explicit) branches of one (implicit) solution. MAPLE should return either both (explicit) branches or stop at the implicit solution. The reason it omits one branch is in the method of solution: rather than realizing that the problem is exact (which makes for a very straightforward dtermination of the implicit solution), MAPLE first arrives at the type homogeneousD. Solutions of this type are generated in 'RootOf' form. But MAPLE does not stop there. Rather, it returns an closed form solution using 'allvalues' - the probable source of omission of the second branch. The bigger problem here is: why does MAPLE not recognize this ODE as exact? >ode1:=diff(y(x),x)=-(4*x+3*y(x)+1)/(3*x+2*y(x)+1); >with(DEtools): >odeadvisor(ode1,y(x),[exact]); [NONE] Hope this narrows down the problem some, Jan Zijlstra, Dept. Math. Sci., MTSU

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