This is my Mathematica code:

X[z_] := {U[z], V[z], Subscript[T, zz][z], Subscript[T, rz][z]};
A[z_] := {{0, -λ*ξ*σ^-1, σ^-1, 0}, {ξ, 0, 0, μ^-1}, {0, 0, 0, -ξ}, {0, 4 μ*η*ξ^2*σ^-1, λ*ξ*σ^-1, 0}};
system = X'[z] == A[z].X[z];
DSolve[system /. {σ -> λ + 2 μ, η -> λ + μ}, {U, V, Subscript[T, zz], Subscript[T, rz]}, z] // FullSimplify

But it's return is the general solutions.

I just want to get the four independent solutions (see my picture)

The four indepenten solutions is here.

Can one use Mathematica to get the four linearly independent solutions as I given in the below picture?

And if given boundary = {U[-Infinity] == 0, V[-Infinity] == 0, Subscript[T, zz][0] == -1/(2 Pi), Subscript[T, rz][0] == 0}, how to use the conditions at -infinity?


2 Answers 2


I guess what you're looking for is a set of any four independent solutions. As it's a linear system, any linearly independent set of vectors for {C[1], C[2], C[3], C[4]} will give independent solutions.

sol = First@ DSolve[system /. {σ -> λ +  2 μ, η -> λ + μ},
    {U, V, Subscript[T, zz], Subscript[T, rz]}, z];

basis = sol /. (Thread[Array[C, 4] -> #] & /@ IdentityMatrix[4]);
Through[{U, V, Subscript[T, zz], Subscript[T, rz]}[z]] /. basis // Simplify
(* long output omitted *)
  • $\begingroup$ Thanks very much! Follow you method, I can get 4 independent solutions. $\endgroup$
    – tanghe2014
    Commented Nov 27, 2016 at 1:52
  • $\begingroup$ If I want to get the specific solution, some boundary conditions are needed. They are ` boundary = {U[-Infinity] == 0, V[-Infinity] == 0, Subscript[T, zz][0] == -1/(2 Pi), Subscript[T, rz][0] == 0};; but Mathematica's return is ****contains an infinite object infinity`; how to mark it works? $\endgroup$
    – tanghe2014
    Commented Nov 27, 2016 at 1:53
  • $\begingroup$ @tanghe2014 Use Limit for BCs at infinity. Your solution seems to have terms of the form Exp[-ξ z] and Exp[ξ z], I don't think you can find a solution that vanishes at -Infinity. Here's the basic idea: boundary = {Limit[U[z], z -> -Infinity] == 0, Limit[V[z], z -> -Infinity] == 0, Subscript[T, zz][0] == -1/(2 Pi), Subscript[T, rz][0] == 0}; c34sol = Solve[boundary[[{3, 4}]] /. sol, {C[3], C[4]}]; Assuming[Variables[{U[z], V[z]} /. sol] \[Element] Reals && \[Xi] > 0, boundary[[{1, 2}]] /. sol /. c34sol] but you get infinity in the equations. As I said, I think you can't get rid of it. $\endgroup$
    – Michael E2
    Commented Nov 27, 2016 at 13:44
  • $\begingroup$ Because $ξ$ is positive, some terms involved $Exp[-ξ z]$ can vanish at -Infinity. I cannot get the 4 independent solutions same as given in the picture even use eigen values and vectors. $\endgroup$
    – tanghe2014
    Commented Nov 28, 2016 at 10:47

Actually, the problem is find the 4 linearly independent solutions for system: $X'(z) = AX(z)$. Note that $A$ is not dependent on $z$. So we can use Eigensystem[A] to get it's eigenvalues and eigenvector:

Eigensystem[A[z]/.{\[Sigma] -> \[Lambda] + 
   2 \[Mu], \[Eta] -> \[Lambda] + \[Mu]}]//FullSimplify//MatrixForm

then eigenvalues and eigenvectors are: $a1=\xi$, $a2=\xi$, $a3=-\xi$, $a4=-\xi$;$u1=\left\{-\frac{1}{2 \mu \xi },\frac{1}{2 \mu \xi },-1,1\right\}$, $u2=\{0,0,0,0\}$, $u3=\left\{-\frac{1}{2 \mu \xi },-\frac{1}{2 \mu \xi },1,1\right\}$, $u4=\{0,0,0,0\}$. So the solution for this system is: $$c1*u1*\exp(a1*z)+c2*u2*\exp(a2*z)+c3*u3*\exp(a3*z)+c4*u4*\exp{(a4*z)}$$ With the boundary condition at infinity, I get $\left\{-\frac{c1 e^{\xi z}}{2 \mu \xi },\frac{c1 e^{\xi z}}{2 \mu \xi },-c1 e^{\xi z},c1 e^{\xi z}\right\}$, but it is unable to satisfy another boundary condition.

Maybe the eigenvectors cannot be zeros?


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