transform equation with Mathematica - wolfram-mathematica

Question

I'm try to use Mathematica to transform a couple of komplex equations, but I didn't get the right syntax.
Solve[{EI*w[x] == Piecewise[{{(x - a)^3, x >= a}, {0, x < a}}]*F*(1/6) +
ca*(x³)*(1/6) + cb*(x^2)*0.5 + cc*x + cd,
EI*w'[x] == Piecewise[{{(x - a)², x >= a}, {0, x < a}}]*F*0.5 + ca*(x^2)*0.5 + cb*x + cc,
EI*w''[x] == Piecewise[{{(x - a)^1, x >= a}, {0, x < a}}]*F + ca*x*0.5 + cb,
EI*w'''[x] == Piecewise[{{(x - a)^0, x >= a}, {0, x < a}}]*F + ca,
w[0] == -EIw'''[0]/ka, w[l] == EIw'''[0]/kb, w'[0] == 0, w'[l] == 0},
{ca, cb, cc, cd}]
Can someone give me a hint how to solve these equations with Mathematica.

Answer 1

Begin by cleaning up some typesetting issues.
We simplify this by looking only at the domain x>=a
Solve[{EI*w[x] == (x-a)^3*F*(1/6)+ca*(x^3)*(1/6)+cb*(x^2)*(1/2)+cc*x+cd,
EI*w'[x] == (x-a)^2*F*(1/2)+ca*(x^2)*(1/2)+cb*x+cc,
EI*w''[x] == (x-a)^1*F+ca*x*(1/2)+cb,
EI*w'''[x] == (x-a)^0*F+ca,
w[0] == -EI*w'''[0]/ka, w[1] == EI*w'''[0]/kb, w'[0] == 0, w'[1] == 0},
w[x], x]
Your second equation is just derivative of first equation and provides no additional information
In[1]:= D[(x-a)^3*F*(1/6)+ca*(x^3)*(1/6)+cb*(x^2)*(1/2)+cc*x+cd, x] ==
(x-a)^2*F*(1/2)+ca*(x^2)*(1/2)+cb*x+cc
Out[1]= True
Your third and fourth equations give information about ca
In[2]:= Simplify[D[(x-a)^1*F+ca*x*(1/2)+cb, x] == (x-a)^0*F+ca]
Out[2]= ca == 0
OR it seems more and more likely that this is the result of a typo in the equations from the original poster. Without more information from the original poster it isn't possible to decide whether this is correct or not. But the person should be able to look at the method used and get the solution needed.
Thus
w[x] == ((x-a)^3*F*(1/6)+cb*(x^2)*(1/2)+cc*x+cd)/EI
In[3]:= w'[x] == D[((x-a)^3*F*(1/6)+cb*(x^2)*(1/2)+cc*x+cd)/EI, x]
Out[3]= w'[x] == (cc+cb x+1/2 F (-a+x)^2)/EI
In[4]:= w'''[x] == D[((x-a)^3*F*(1/6)+cb*(x^2)*(1/2)+cc*x+cd)/EI, {x, 3}]
Out[4]= w'''[x] == F/EI
We already know ca, solve for the remaining three variables using the first three boundary conditions
In[5]:= Simplify[Solve[{((0-a)^3*F*(1/6)+cb*(0^2)*(1/2)+cc*0+cd)/EI == -EI*(F/EI)/ka,
((l-a)^3*F*(1/6)+cb*(l^2)*(1/2)+cc*l+cd)/EI == EI*(F/EI)/kb,
(cc+cb x+1/2 F (-a+0)^2)/EI == 0}, {cb, cc, cd}]]
Out[5]= {{cb -> (F (6 EI (ka+kb)+ka kb (3 a-l) l^2))/(3 ka kb l (l-2 x)),
cc -> -((F (3 a^2 ka kb l (l-2 x)+6 a ka kb l^2 x+
2 (6 EI (ka+kb)-ka kb l^3) x))/(6 ka kb l (l-2 x))),
cd -> 1/6 F (a^3-(6 EI)/ka)}}
Look at the final boundary condition
In[7]:= Simplify[(cc+cb x+1/2 F (-a+l)^2)/EI == 0 /. Out[5]]
Out[7]= {(F l (-2 a+l))/EI == 0}
Check every step of this carefully and verify that no errors were made.
Then go back and see if you can use these same methods for the domain x < a