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I'm trying to solve a linear system of ODEs:

DSolve[{D[c1m[y], y] == 0, D[c2m[y], y] == 2 I kf c3p[y], 
  D[c2p[y], y] == 2 I kf c3m[y], -D[c2m[y], y] - A == 0,
  D[c1m[y], y] - D[c3m[y], y] == 2 I kf c2p[y],
  2 D[c0[y], y] - D[c2p[y], y] - B == 0,
  D[c1p[y], y] - D[c3p[y], y] == 2 I kf c2m[y]},
 {c1m[y], c1p[y], c2m[y], c2p[y], c3m[y], c3p[y], c0[y]}, {y}]

Note that A, B and kf have dimensions of 1/space, and that the different c's are dimensionless whereas y has a space dimension so that this set of ODEs is valid. However, the answer I get for c1p is dimensionally wrong (involves adding space dimension with space squared).

What am I missing?

Thanks!

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  • $\begingroup$ The dimensional analysis also carries through if y and the parameters are treated as dimensionless. Indeed, the system is autonomous and therefore mathematically invariant under the transformation $y\rightarrow y+3$, where $3$ could be any other real number. Try Expand[soln]. One of the terms you identified cancels out (at least what I identified from your description). However, you'll still have a constant term 6 I kf C[6] that is supposed to be dimensionless. Try Simplify[soln] and maybe you can justify 3 + y as having the same dimensions as y. Not sure it can be fixed. $\endgroup$
    – Michael E2
    Dec 13, 2020 at 19:29
  • $\begingroup$ Indeed, the problem is having the y+3 as 3 is obviously dimensionless and can't be justified from the equations (which keep the dimensional requirement). Also, it's unclear to me where this 3 stems from. $\endgroup$
    – Oren
    Dec 13, 2020 at 20:38
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    $\begingroup$ Substituting the DSolve result into the ODEs yields {True, True, True, True, True, True, True}, so the result appears to be correct. $\endgroup$
    – bbgodfrey
    Dec 13, 2020 at 21:15
  • $\begingroup$ How did you check? I used FullSimplify[eqn/.soln] and it doesn't yield as you write. Regardless, I agree that Mathematica thinks it's correct. However, I'm trying to figure out how or why because it's uncanny to give up dimensional analisys like that. $\endgroup$
    – Oren
    Dec 13, 2020 at 22:05
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    $\begingroup$ Use DSolve[{D[c1m[y], y] == 0, D[c2m[y], y] == 2 I kf c3p[y], D[c2p[y], y] == 2 I kf c3m[y], -D[c2m[y], y] - A == 0, D[c1m[y], y] - D[c3m[y], y] == 2 I kf c2p[y], 2 D[c0[y], y] - D[c2p[y], y] - B == 0, D[c1p[y], y] - D[c3p[y], y] == 2 I kf c2m[y]}, {c1m, c1p, c2m, c2p, c3m, c3p, c0}, y] // Flatten to obtain Functions as solutions. Then use Simplify[{ ... }/.{ ... }] where first ... is your list of equations and the second ... is the solution you just obtained. Hope this helps. $\endgroup$
    – bbgodfrey
    Dec 13, 2020 at 22:20

1 Answer 1

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One approach that avoids the units ambiguity is to solve the first 5 equations and then solve the other 2 equations, like this

deqs = {
   D[c1m[y], y] == 0,
   D[c2m[y], y] == 2 I kf c3p[y],
   D[c2p[y], y] == 2 I kf c3m[y],
   -D[c2m[y], y] - A == 0,
   D[c1m[y], y] - D[c3m[y], y] == 2 I kf c2p[y],
   2 D[c0[y], y] - D[c2p[y], y] - B == 0,
   D[c1p[y], y] - D[c3p[y], y] == 2 I kf c2m[y]};

soln = First@DSolve[deqs[[1 ;; 5]],
    {c1m[y], c2m[y], c2p[y], c3m[y], c3p[y]}, y];

dsoln := soln /. Rule[lhs_, rhs_] :> (D[lhs, y] -> D[rhs, y]);

soln = Join[soln, First@DSolve[deqs[[{6, 7}]] /. soln /. dsoln,
     {c1p[y], c0[y]}, y]];

soln // ColumnForm // TeXForm

$\begin{array}{c} \text{c1m}(y)\to c_1 \\ \text{c2p}(y)\to c_2 \cosh (2 \text{kf} y)+i c_3 \sinh (2 \text{kf} y) \\ \text{c3m}(y)\to c_3 \cosh (2 \text{kf} y)-i c_2 \sinh (2 \text{kf} y) \\ \text{c2m}(y)\to c_4-A y \\ \text{c3p}(y)\to \frac{i A}{2 \text{kf}} \\ \text{c0}(y)\to \frac{B y}{2}+c_5+\frac{1}{2} c_2 \cosh (2 \text{kf} y)+\frac{1}{2} i c_3 \sinh (2 \text{kf} y) \\ \text{c1p}(y)\to c_6-2 i \text{kf} \left(\frac{A y^2}{2}-y c_4\right) \\ \end{array}$

Besides having no ambiguity in the solution for c1p[y], this approach yields hyperbolic trig functions, which can be easier to interpret than several exponentials.

Checking this solution

Replace the solutions and the derivatives of the solutions and check

deqs /. soln /. dsoln // Simplify

(*  {True, True, True, True, True, True, True}  *)

Original ambiguity

Note: the constants C[5] and C[6] found below are not the same as those above.

When we solve all 7 DEs at once, the solution for c1p[y] contains an ambiguity

amb = c1p[y] /. First@DSolve[deqs,
                {c1m[y], c1p[y], c2m[y], c2p[y], c3m[y], c3p[y], c0[y]}, y];

amb // Simplify 

(*  A (I/(2 kf) - I kf y^2) + C[5] + 2 I kf (3 + y) C[6]  *)

The y+3 term looks ambiguous, since $y$ has units and 3 does not, but the 3 can be consolidated into the constant $c_5$ by introducing the initial condition c1p[0] like this

amb /. First@First@Solve[
         (Last[amb] /. y -> 0) == c1p[0], C[5]] // Simplify

(*  c1p[y] -> -I A kf y^2 + 2 I kf y C[6] + c1p[0]  *)

Or by introducing a new constant, C[7]

amb[[-1]] = amb[[-1]] + C[7] - (amb[[-1]] /. y -> 0) // Simplify;
amb

(*  c1p[y] -> -I A kf y^2 + 2 I kf y C[6] + C[7]  *)

How many ODEs are there?

We count 7 DEs in the list, but there are only 6 arbitrary constants. Taken as a separate system, the second and fourth DE's give an algebraic solution for c3p[y] without a constant of integration.

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