Can Mathematica solve this differential equation ?
in [Mathematica]
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From: Ashok on 2 Feb 2010 03:27 In beam bending, we have the following situation: y '' [x] = M(x), boundary conditions specified for y[0] and y[1] Simple enough, but the problem arises as M is a piecewise-defined function (linear in all pieces though) i.e., M(x) = M1(x) for 0<x<1 and M(x) = M2(x) for 1<x<2 M(x) = 0 for all other values of x Obviously, we are only interested in the interval 0<x<1 This leads to two separate equations: y1'' [x] = M1(x) in 0<x<1 y2 '' [x] = M2(x) in 1 <x <2 Now we will have 4 constants of integration. We therefore need 4 equations to solve for them. Two of them are obtained from the specified boundary conditions for y[0] and y[1]. The other two come from continuity equations: y1[1] = y2[1] and y1 ' [1] = y2 ' [1]. It is these last two that totally throw me off. I do not understand how to put them into Mathematica. Any help is appreciated. Thank you Ashok
From: Nasser M. Abbasi on 3 Feb 2010 06:08 "Ashok" <nils_von_nostrand(a)yahoo.com> wrote in message news:hk8nlh$8h4$1(a)smc.vnet.net... > In beam bending, we have the following situation: > > y '' [x] = M(x), boundary conditions specified for y[0] and y[1] > > Simple enough, but the problem arises as M is a piecewise-defined > function (linear in all pieces though) > > i.e., M(x) = M1(x) for 0<x<1 > and M(x) = M2(x) for 1<x<2 > M(x) = 0 for all other values of x > > Obviously, we are only interested in the interval 0<x<1 > > This leads to two separate equations: > > y1'' [x] = M1(x) in 0<x<1 > y2 '' [x] = M2(x) in 1 <x <2 > > Now we will have 4 constants of integration. We therefore need 4 > equations to solve for them. Two of them are obtained from the > specified boundary conditions for y[0] and y[1]. The other two come > from continuity equations: > y1[1] = y2[1] and y1 ' [1] = y2 ' [1]. It is these last two that > totally throw me off. I do not understand how to put them into > Mathematica. > > Any help is appreciated. > > Thank you > > Ashok > One way to solve the first ode and use its results as initial conditions for the second ode? Here is an example: Clear[x,M1,M2,y1,y2] M1=x; M2=2 x; First(a)DSolve[{y1''[x]==M1,y1[0]==0,y1[1]==1},y1[x],x]; y1=y1[x]/.% Out[44]= 1/6 (5 x+x^3) First(a)DSolve[{y2''[x]==M2,y2[1]==y1/.x->1,y2'[1]==D[y1,x]/.x->1},y2[x],x]; y2=y2[x]/.% Out[47]= 1/3 (1+x+x^3) --Nasser
From: JH on 3 Feb 2010 06:13 On 2 feb, 09:27, Ashok <nils_von_nostr...(a)yahoo.com> wrote: > In beam bending, we have the following situation: > > y '' [x] = M(x), boundary conditions specified for y[0] and y[1] > > Simple enough, but the problem arises as M is a piecewise-defined > function (linear in all pieces though) > > i.e., M(x) = M1(x) for 0<x<1 > and M(x) = M2(x) for 1<x<2 > M(x) = 0 for all other values of x > > Obviously, we are only interested in the interval 0<x<1 > > This leads to two separate equations: > > y1'' [x] = M1(x) in 0<x<1 > y2 '' [x] = M2(x) in 1 <x <2 > > Now we will have 4 constants of integration. We therefore need 4 > equations to solve for them. Two of them are obtained from the > specified boundary conditions for y[0] and y[1]. The other two come > from continuity equations: > y1[1] = y2[1] and y1 ' [1] = y2 ' [1]. It is these last two that > totally throw me off. I do not understand how to put them into > Mathematica. > > Any help is appreciated. > > Thank you > > Ashok Hi, Ashok, Why do you include the second domain (1<x<2) if you say 'Obviously, we are only interested in the interval 0<x<1'? I don't understand the reason for this last domain. Or else, could you give the particular forms of the functions M1(x) and M2(x), so that we can try to find a solution? Thanks. JH
From: Alexei Boulbitch on 3 Feb 2010 06:29 Dear Ashok, are you sure that this is the beam equation? Should not it be something as y''(x)=M(x)y(x) or even y''''(x)=M(x)y''(x), rather than y''(x)=M(x) ??? OK, assuming that you know what you ask, the solution of the equation y''(x)=M(x) you are asking about is simple. So simple that you should not waste time in thinking how to put it into Mathematica, but should write down the solution right away in its explicit analytical form. Evaluate the code below in Mathematica to visualize this solution: Style["\!\(\*SubscriptBox[\"y\", \ \"1\"]\)'[x]=\!\(\*SubsuperscriptBox[\"\[Integral]\", \"0\", \"x\"]\)\ \!\(\*SubscriptBox[\"M\", \ \"1\"]\)[\[Xi]]\[DifferentialD]\[Xi]+\!\(\*SubscriptBox[\"y\", \"1\"]\ \)'[0]", Italic, 16] Style["\!\(\*SubscriptBox[\"y\", \"1\"]\)[x]=\!\(\*SubsuperscriptBox[\ \"\[Integral]\", \"0\", \ \"x\"]\)\[DifferentialD]\[Xi]\!\(\*SubsuperscriptBox[\"\[Integral]\", \ \"0\", \"\[Xi]\"]\)\!\(\*SubscriptBox[\"M\", \"1\"]\)[\[Zeta]]\ \[DifferentialD]\[Zeta]+ \!\(\*SubscriptBox[\"xy\", \"1\"]\)'[0]+\!\(\ \*SubscriptBox[\"y\", \"1\"]\)[0]", Italic, 16] Style["\!\(\*SubscriptBox[\"y\", \ \"2\"]\)'[x]=\!\(\*SubsuperscriptBox[\"\[Integral]\", \"1\", \"x\"]\)\ \!\(\*SubscriptBox[\"M\", \ \"1\"]\)[\[Xi]]\[DifferentialD]\[Xi]+\!\(\*SubscriptBox[\"y\", \"1\"]\ \)'[0]+\!\(\*SubsuperscriptBox[\"\[Integral]\", \"0\", \ \"1\"]\)\!\(\*SubscriptBox[\"M\", \ \"1\"]\)[\[Xi]]\[DifferentialD]\[Xi]", Italic, 16] Style["\!\(\*SubscriptBox[\"y\", \ \"2\"]\)[x]=-\!\(\*SubsuperscriptBox[\"\[Integral]\", \"x\", \"2\"]\)\ \[DifferentialD]\[Xi]\!\(\*SubsuperscriptBox[\"\[Integral]\", \"0\", \ \"\[Xi]\"]\)\!\(\*SubscriptBox[\"M\", \"2\"]\)[\[Zeta]]\ \[DifferentialD]\[Zeta]+ (x-2)\!\(\*SubscriptBox[\"y\", \ \"1\"]\)'[0]+(x-2)\!\(\*SubsuperscriptBox[\"\[Integral]\", \"0\", \"1\ \"]\)\!\(\*SubscriptBox[\"M\", \ \"1\"]\)[\[Xi]]\[DifferentialD]\[Xi]+\!\(\*SubscriptBox[\"y\", \"1\"]\ \)[0]+\!\(\*SubsuperscriptBox[\"\[Integral]\", \"0\", \"2\"]\)\ \[DifferentialD]\[Xi]\!\(\*SubsuperscriptBox[\"\[Integral]\", \"0\", \ \"\[Xi]\"]\)\!\(\*SubscriptBox[\"M\", \"1\"]\)[\[Zeta]]\ \[DifferentialD]\[Zeta]+\!\(\*SubscriptBox[\"y\", \"2\"]\)[2]", \ Italic, 16] Check it once more before using, I was writing fast, and may have missed some small terms. Take care: some terms in the solution vanished and others have a specially simple form just because you have chosen the boundaries 1 and 2. It may be different in a more general case. Have fun, Alexei In beam bending, we have the following situation: y '' [x] = M(x), boundary conditions specified for y[0] and y[1] Simple enough, but the problem arises as M is a piecewise-defined function (linear in all pieces though) i.e., M(x) = M1(x) for 0<x<1 and M(x) = M2(x) for 1<x<2 M(x) = 0 for all other values of x Obviously, we are only interested in the interval 0<x<1 This leads to two separate equations: y1'' [x] = M1(x) in 0<x<1 y2 '' [x] = M2(x) in 1 <x <2 Now we will have 4 constants of integration. We therefore need 4 equations to solve for them. Two of them are obtained from the specified boundary conditions for y[0] and y[1]. The other two come from continuity equations: y1[1] = y2[1] and y1 ' [1] = y2 ' [1]. It is these last two that totally throw me off. I do not understand how to put them into Mathematica. Any help is appreciated. Thank you Ashok -- Alexei Boulbitch, Dr., habil. Senior Scientist IEE S.A. ZAE Weiergewan 11, rue Edmond Reuter L-5326 Contern Luxembourg Phone: +352 2454 2566 Fax: +352 2454 3566 Website: www.iee.lu This e-mail may contain trade secrets or privileged, undisclosed or otherwise confidential information. If you are not the intended recipient and have received this e-mail in error, you are hereby notified that any review, copying or distribution of it is strictly prohibited. Please inform us immediately and destroy the original transmittal from your system. Thank you for your co-operation.
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