# Solving a differential equation using DSolve: no output

Posted 4 months ago
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 Hi everyone.I try to solve the coupled differential equations that you can see in the notebook. But Mathematica can't find a solution and just returns the 2 differential equations, even though there exists a non-trivial solution. If W1[x] and W2[x] are linear and W1[x] = (2/R0)*W2[x] the equations are fulfilled. So, this is not the solution that I am interested in, but it shows that something is going wrong if Mathematica can't find it.Does anyone know what it is?Best. Luca
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Posted 4 months ago
 DSolve[]'s support for solving coupled differential equations is still somewhat limited,so don't be surprised if some things don't work yet.A workaround: ClearAll["*"]; Remove["*"]; eq1 =k Sech[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 W1[x] + 1/2 k Sech[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 Sinh[(Sqrt[k] x)/Sqrt[ Dm]] W1[x] - (2 k Sech[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 W2[x])/R0 - ( k Sech[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 Sinh[(Sqrt[k] x)/Sqrt[Dm]] W2[ x])/R0 + Dm W1''[x] == 0; eq2 =(Dc k R0 W1[x])/(4 Dm) + ( Dc k R0 Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])] W1[x])/(2 Dm) + ( Dc k R0 Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 W1[x])/(4 Dm) - ( Dc k W2[x])/(2 Dm) - (Dc k Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])] W2[x])/ Dm - (Dc k Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 W2[x])/(2 Dm) + Dc W2''[x] == 0; EQ = eq1 /. Solve[eq2, W1[x]] /. D[Solve[eq2, W1[x]], {x, 2}] // FullSimplify Solution for W2(x) :  SOL = DSolve[EQ[[1]], W2[x], x] // Simplify (*{{W2[x] -> C[3] + x C[4] + 1/6 ((3 Dm E^((Sqrt[k] x)/Sqrt[Dm]) C[1])/k - (6 Dm C[1])/( k + E^((Sqrt[k] x)/Sqrt[Dm]) k) + 6 x^2 C[1] - ( Dm C[2])/((1 + E^((Sqrt[k] x)/Sqrt[Dm]))^2 k) + (4 Dm C[2])/( k + E^((Sqrt[k] x)/Sqrt[Dm]) k) - (2 Sqrt[Dm] x C[2])/Sqrt[ k] - (6 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[Dm])]^2)/k + ( 2 Dm C[2] Log[1 + E^((Sqrt[k] x)/Sqrt[Dm])])/k + ( 6 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[ Dm])] ((3 + 4 E^((Sqrt[k] x)/Sqrt[Dm]))/(1 + E^((Sqrt[k] x)/Sqrt[ Dm]))^2 + 2 Log[1 + E^((Sqrt[k] x)/Sqrt[Dm])]))/k + ( 12 Dm C[1] PolyLog[2, -E^(((Sqrt[k] x)/Sqrt[Dm]))])/k)}}*) Solution for W1(x) :  (Solve[eq2, W1[x]] /. SOL /. D[SOL, {x, 2}])[[1, 1, 1]] // Simplify (*{W1[x] -> 1/(12 k R0) Sech[(Sqrt[k] x)/( 2 Sqrt[Dm])]^2 (-3 Dm C[1] + 12 k x^2 C[1] + Dm C[2] - 4 Sqrt[Dm] Sqrt[k] x C[2] + 12 k C[3] + 12 k x C[4] + 3 Dm C[1] Cosh[(2 Sqrt[k] x)/Sqrt[Dm]] + 6 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[Dm])] - 12 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[Dm])]^2 + 4 Dm C[2] Log[1 + E^((Sqrt[k] x)/Sqrt[Dm])] + 24 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[Dm])] Log[ 1 + E^((Sqrt[k] x)/Sqrt[Dm])] + Cosh[(Sqrt[k] x)/Sqrt[ Dm]] (12 k x^2 C[1] + 3 Dm C[2] - 4 Sqrt[Dm] Sqrt[k] x C[2] + 12 k C[3] + 12 k x C[4] - 12 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[Dm])]^2 + 4 Dm C[2] Log[1 + E^((Sqrt[k] x)/Sqrt[Dm])] + 6 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[ Dm])] (3 + 4 Log[1 + E^((Sqrt[k] x)/Sqrt[Dm])])) + 48 Dm C[1] Cosh[(Sqrt[k] x)/(2 Sqrt[Dm])]^2 PolyLog[ 2, -E^(((Sqrt[k] x)/Sqrt[Dm]))] - 12 Dm C[1] Sinh[(Sqrt[k] x)/Sqrt[Dm]] - 3 Dm C[2] Sinh[(Sqrt[k] x)/Sqrt[Dm]] - 18 Dm C[1] Log[E^((Sqrt[k] x)/Sqrt[Dm])] Sinh[(Sqrt[k] x)/Sqrt[ Dm]])}*) 
 An approach using a change of dependent variable: origEqs = {(k + k Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])] - k Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])]^2) (W1[x] - (2 W2[x])/R0) + Dm (W1^\[Prime]\[Prime])[x] == 0, (Dc k R0 (1 + Tanh[(Sqrt[k] x)/(2 Sqrt[Dm])])^2 (W1[x] - ( 2 W2[x])/R0))/(4 Dm) + Dc (W2^\[Prime]\[Prime])[x] == 0}; substitution = W1 -> Function[x, (2 W2[x])/R0 + f[x]]; newEqs = Solve[origEqs /. substitution, {f''[x], W2''[x]}][[1]] /. Rule -> Equal solF = DSolve[newEqs[[1]], f, x][[1]] solW2 = DSolve[newEqs[[2]] /. solF, W2[x], x][[1]] // PowerExpand // Simplify W1[x] -> (W1[x] /. substitution /. solF) It is disappointing that Mathematica cannot solve the two new equation together DSolve[newEqs, {f, W2}, x] so that I am forced to solve separately the first equation and then replace into the second.