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I have been trying to solve a physical problem during which I reach the following third-order, Linear O.D.E.

The solution I get using this expression is really messy. Is there any way to simplify it in the form of trig functions or some alternative ?


$$f'''_n(x) + \alpha f''_n(x)-\Bigg(\bigg(\frac{n\pi}{d}\bigg)^2 + \beta\Bigg) f_n'(x)-\alpha \bigg(\frac{n\pi}{d}\bigg)^2 f_n(x)=-\frac{2 \alpha \beta \gamma d}{(md)^2 + (n\pi)^2} \tag 2$$

with the following boundary conditions $$f'_n(0)=f'_n(L)=0\\ f''_n(0)- \Bigg(\bigg(\frac{n\pi}{d}\bigg)^2 + \beta\Bigg)f_n(0)=0$$

ALSO $$\beta = m^2$$ Any help on solving $(2)$ is really appreciated


$(2)$

DSolve[{y'''[x] == -\[Alpha]*y''[x] + ((n*\[Pi]/d)^2 + \[Beta])*
     y'[x] + \[Alpha]*((n*\[Pi]/d)^2)*
     y[x] - ((2*\[Alpha]*\[Beta]*\[Gamma]*d)/((m*d)^2 + (n*\[Pi])^2)),
   y'[0] == 0, y'[L] == 0, 
  y''[0] - ((n*\[Pi]/d)^2 + \[Beta])*y[0] == 0}, y[x], x]
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    $\begingroup$ The solution I get using this expression is really messy solution is messy because the ODE is messy. You could try sol=ToRadicals[[sol] to remove roots. But simplifying the few pages of such solution is not going to be easy. You can try FullSimplify[sol] and see what it does. What are you expecting the solution to look like? If you have values for beta and alpha, then it is different matter. $\endgroup$ – Nasser Nov 22 '19 at 7:41
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    $\begingroup$ Third order DE with constant coefficients and a constant non-homogenous term should not be too difficult. If you can put your DE into the form deq = y'''[x] - (a + b + c) y''[x] + (a b + a c + b c) y'[x] - a b c y[x] == d, then DSolve[{deq}, y[x], x] will give you a nice answer. It seems feasible to solve for $a, b, c$ in terms of $\alpha, \beta$, but it will be messy. Consider the case of repeated roots of the characteristic polynomial by setting c=b before using DSolve. Solve for $d$ to get your non-homogeneous term after putting in the BCs. $\endgroup$ – LouisB Nov 23 '19 at 6:52
  • $\begingroup$ @LouisB Thanks for the suggestion. But I am afraid, I made some mistakes while deriving these equations and posted them in a wrong form earlier. I have now corrected them and the code. Can you have a look again ? Meanwhile, I will try to incorporate the suggestions from your last comment. $\endgroup$ – Indrasis Mitra Nov 23 '19 at 7:39
  • $\begingroup$ If you call the roots a,b,c or something then it's not that bad. FullSimplify[% /. Root[-β + α β + α #1^2 + #1^3 &, n_] :> {a, b, c}[[n]]] $\endgroup$ – Simon Woods Nov 23 '19 at 13:47
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    $\begingroup$ In the ODE $$y''' + \alpha y''+\beta y'+\gamma y + \delta = 0$$ changing variable to $Y = y -\frac{\delta}{\gamma}$ we get the new ODE $$Y'''+\alpha Y''+\beta Y'+\gamma Y = 0$$ and also $Y'(0) = 0$,$ Y'(L) = 0$, $Y''(0)+\zeta\left(Y(0)+\frac{\delta}{\gamma}\right) = 0$. The solution is not so lengthy. $\endgroup$ – Cesareo Nov 24 '19 at 20:54
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Echoing Nasser, the solution is messy because the ODE is messy. There is not much you can do. You can massage the solution into a slightly more compact form as follows: let $q_i$ be the solutions to the algebraic equation $$ q^3+ \alpha q^2- \left(\beta +\frac{\pi ^2 n^2}{d^2}\right)q-\frac{\pi ^2 \alpha n^2}{d^2}=0 $$ i.e.,

q[i_] :> Root[(n^2 π^2 α)/d^2 + ((n^2 π^2)/d^2 + β) #1 - α #1^2 - #1^3 &, i]

Let also $q_{i,j}$ denote the "other root", that is, $q_{1,2}=q_3$ together with cyclic permutations, $q_{2,3}=q_1$ and $q_{1,3}=q_2$. Finally, let $s_{ij}$ denote the sign of the permutation, that is, $s_{ij}=\operatorname{sign}(q_i,q_j,q_{ij})$:

q[1, 2] -> q[3]
q[1, 3] -> q[2]
q[2, 3] -> q[1]
s[i_, j_] := Signature[{i, j, q[i, j][[1]]}]

With this, the solution can be expressed as follows: $$ y(x)\propto \sum_{i,j} q_i q_j s_{ij} \left(e^{L q_j} \left(\beta d^2-\pi ^2 n^2 e^{x q_{i,j}}\right)+e^{L q_i} \left(\beta d^2 \left(2 e^{x q_{i,j}}-1\right)+2 d^2 q_{i,j}^2+\pi ^2 n^2 \left(e^{x q_{i,j}}-2\right)\right)\right) $$ as given by

Sum[s[i, j] q[i] q[j] (E^(L q[j]) (-E^(x q[i, j]) n^2 π^2 + d^2 β) + E^(L q[i]) ((-2 + E^(x q[i, j])) n^2 π^2 + d^2 (-1 + 2 E^(x q[i, j])) β + 2 d^2 q[i, j]^2)), {i, 1, 3}, {j, 1, 3}]

The constant of proportionality can easily be fixed by demanding that the ODE is satisfied, say, at $x=0$. I don't think one can find an expression for $y(x)$ much more compact than this. The problem is messy, its solution is messy. The presence of $s_{ij}$ suggests to me that perhaps it can be expressed as a determinant, but in practice I believe the expression above is your best bet. This, or do numerics only.

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  • $\begingroup$ Thanks for this insight. I could follow your answer till you defined the rules and the sign of permutation. Could you please elaborate a little more on how you reached the $y(x)$. I am afraid this might be trivial, but I have been unable to follow it through. $\endgroup$ – Indrasis Mitra Nov 25 '19 at 4:02
  • $\begingroup$ Awarded, the bounty because your representation is surely the most concise form there is But Still unable to reproduce it. $\endgroup$ – Indrasis Mitra Nov 27 '19 at 20:51

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