Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exact solution of a 4th order DE

  1. Oct 26, 2011 #1
    Hi guys,

    I am trying to find the exact solution of:

    u'''' + 5u =f on (0,1) f is a constant >0

    where

    u(0)=u'(0) =0
    u''(1) =k constant >0
    u'''(1)=0

    I assumed I should look at solving the homogeneous equation and got the following

    uh = A*exp(Zx)*sin(Zx) + B*exp(ZX)*cos(Zx) + C*exp(-Zx)*sin(Zx) +D*exp(-Zx)*cos(Zx)

    Where Z is (sqrt(2)/2)*sqrt(sqrt(5)) and A,B,C,D are constants.

    My questions are:

    1) Am I on the right track
    2) How would I go about solving for the particular solution
    3) Does the exact solution actually exist?

    Thanks
     
  2. jcsd
  3. Oct 27, 2011 #2
    Just by inspection, u=f/5 is a particular solution right?
     
  4. Oct 27, 2011 #3
    surely the particular solution must satisfy the boundary conditions?
     
  5. Oct 27, 2011 #4

    lurflurf

    User Avatar
    Homework Helper

    You are on the right track and the problem is well posed.
    ^depends how particular

    The solution of the equation can be written as a sum by the superposition principle. The final answer needs to satisfy the differential equation and the boundary conditions (and is a particular solution. The terms we are adding up need not satisfy all (or any) of the conditions. Just let each term satisfy certain conditions and keep track that the sum will satisfy all conditions.

    basically either take
    u = A*exp(Zx)*sin(Zx) + B*exp(ZX)*cos(Zx) + C*exp(-Zx)*sin(Zx) +D*exp(-Zx)*cos(Zx) +f/5
    and fit the boundary conditions or take
    u = A*exp(Zx)*sin(Zx) + B*exp(ZX)*cos(Zx) + C*exp(-Zx)*sin(Zx) +D*exp(-Zx)*cos(Zx) +E
    and fit the boundary conditions and u'''' + 5u =f

    It is the same in the end
     
  6. Oct 27, 2011 #5
    What did we do before Mathematica?

    Code (Text):

    z = (Sqrt[2]/2)*Sqrt[Sqrt[5]];

    u[x_] := a*Exp[z*x]*Sin[z*x] + b*Exp[z*x]*Cos[z*x] + c*Exp[(-z)*x]*Sin[z*x] + d*Exp[(-z)*x]*Cos[z*x]+f/5;

    myequations = {0 == u[0], 0 == Derivative[1][u][0], k == Derivative[2][u][1], 0 == Derivative[3][u][1]}

    Solve[myequations, {a, b, c, d}]
     
     
  7. Oct 27, 2011 #6
    Hmmm and if I don't have mathematica?

    I tried wolfram alpha but it cant solve it with all 4 boundary conditions.
    If I only use 2 I end up with a massive equation (18+ terms) that I still have to solve for 2 constants.

    I tried solving it by hand from scratch and get the wrong answer. (well wrong according to my finite element program - and I know it is correct!)
     
  8. Oct 27, 2011 #7

    lurflurf

    User Avatar
    Homework Helper

    wolfram alpha can be wonky but I entered

    u''''(x) + 5u(x) =f;u(0)=0,u'(0)=0,u''(1)=k,u'''(1)=0

    and got a reasonable answer
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Exact solution of a 4th order DE
  1. 2nd Order De Solution (Replies: 2)

  2. 4th order DE (Replies: 4)

  3. 4th order DE (Replies: 1)

Loading...