2014-11-28 11:28:44 +0100 received badge ● Scholar (source) 2014-11-28 08:34:39 +0100 commented answer How to define more complicated differential expressions in Sage? Thanks for your efforts. I took this as an inspiration to convert expressions using strings and subs(), but it is quite cumbersome. The real problem is that SAGE gives D0(y) as output, but does not accept it as input! Less an expression which can be solved or derived for. The problem that I asked a solution for can be solved in Sage, by constantly replacing functions by variables back and forth, depending on whether one wants either to solve/derive for or SAGE to apply rules of differentiation correctly. 2014-11-24 21:26:20 +0100 received badge ● Editor (source) 2014-11-24 19:58:57 +0100 asked a question How to define more complicated differential expressions in Sage? Dear Community, i have a very hard time doing something pretty straight foreward in Sage. In mechanics, there is a Lagrange-formalism that yields the equation of motion of a system of $n$ degrees of freedom in the following way: $$0=\frac{d}{dt}\left(\frac{\partial L}{\partial x'_i(t)}\right) - \frac{\partial L}{\partial x_i(t)}$$ where $L$ is the Lagrange function $$L=E_{Kin} (x'_{1}(t),x'_2(t)...x'_n(t); x_1(t),x_2(t)...x_n(t) ) - E _{Pot}(x_1(t),x_2(t),...,x_n(t))$$ is the kinetic energy minus the potential energy. The kinetic energy depends on the speeds of the coordinates $x'_i(t)$, while the potential energy depends on the coordinates $x_i(t)$. In Mathematica, the hack is L = Ekin[x'[t]] - Epot[x[t]]; eq = 0 == D[D[L, x'[t]], t] - D[L, x[t]];  which gives 0 == Epot'[x[t]] + Ekin''[x'[t]] x''[t]  I can easily specify something for $E_{Kin}$ and $E_{Pot}$, solve the latter equation for $x''$, and give it as a right handside to some time integration scheme. How can this be done in Sage? Assume you have complicated expressions for $E_{Pot}$ and $E_{Kin}$ which you do not want to differentiate by hand. var('L,x,t') x = function('x',t) L = function('L',x,t) allows for diff(L,t) but not diff(L,x) Thanks already!