0
$\begingroup$

(* I am trying to find the null space of a matrix I used NUllspace command but I am getting the empty matrix and I also tried gauss elimination. All I am doing is I am finding out a green function a fourth order differential equation. I Used GreenFunction command in Mathematica I have given 4 boundary condition still I could not able to get the Green function solution *)

        W1 = C2*Sin[\[Beta]*x] + C4*Sinh[\[Beta]*x]  
        W2 = B2*Sin[\[Beta]*(1 - x)] + B4*Sinh[\[Beta]*(1 - x)]  
        (*Compatability condition*)  
        ccd1 = (W1 /. {x -> z}) - (W2 /. {x -> z});  
        ccs1 = ((D[W1, {x}]) /. {x -> z}) - ((D[W2, {x}]) /. {x -> z})  
        ccm1 = ((D[W1, {x, 2}]) /. {x -> z}) - ((D[W2, {x, 2}]) /. {x -> z})  
        ccsh1 = ((D[W1, {x, 3}]) /. {x -> z}) - ((D[W2, {x, 3}]) /. {x -> z})  
        R = Normal@
          CoefficientArrays[{ccd1, ccs1, ccm1, ccsh1}, {C2, C4, B2, B4}][[2]]  
        S = R // MatrixForm  
        NN1 = [NullSpace[R]]  
        (*Greenfunction *)  
        GreenFunction[{(u''''[x] - \[Beta]^4*u[x]), u[0] == 0, u''[0] == 0, 
        u[1] == 0, u''[1] == 0}, u[x], {x, 0, 1}, y]  
$\endgroup$
  • 2
    $\begingroup$ It's not surprising that NullSpace returns an empty matrix since there's no reason that any of the eigenvalues would be identically zero for all beta and z. $\endgroup$ – aardvark2012 Sep 10 '17 at 11:13
5
$\begingroup$

As pointed out by aardvark2012, your matrix is invertible and thus its null space is empty. To see for which parameter values it is not invertible, you can do Simplify[Det[S] == 0] to get the simple answer β Sin[β] Sinh[β] == 0. So, for example:

NullSpace[S /. β -> π]

gives

{{Csc[π z] Sin[π (1 - z)], 0, 1, 0}}

and

NullSpace[S /. β -> 8 I π]

gives

{{-((Csch[8 π z] (Cos[8 π (1 - z)] +Cot[8 π z] Sin[8 π (1 - z)]) Sinh[8 π (1 - z)])/(Cosh[8 π (1 - z)] + Coth[8 π z] Sinh[8 π (1 - z)])), 
  Csc[8 π z] Sin[8 π (1 - z)], 
  -((Cos[8 π (1 - z)] Sinh[8 π z] + 
  Cot[8 π z] Sin[8 π (1 - z)] Sinh[8 π z])/(Cosh[8 π z] Sinh[8 π (1-z)] + Cosh[8 π (1 - z)] Sinh[8 π z])), 
  1}}
| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.