Re: Can Mathematica solve this differential equation ?

• To: mathgroup at smc.vnet.net
• Subject: [mg107139] Re: [mg107076] Can Mathematica solve this differential equation ?
• From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
• Date: Wed, 3 Feb 2010 06:28:30 -0500 (EST)

```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
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```

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