I was wondering if one could define an operator such that, when I give a certain number of (differential) equations as an output, and an "equilibrium" value for each of the variables, it returns the linear part in the perturbative expansion of the above-mentioned equations about the reference

For example, given:

$$\partial_t X + \partial_x Y+X\partial_xY+XY=0$$

and equilibrium values: $X=X_0+\delta X, Y=0+\delta Y$

Output would be, the linearized:

$$\partial_t (\delta X) + \partial_x (\delta Y)+X_0\partial_x(\delta Y)+X_0\ \delta Y=0$$


2 Answers 2


How about this:

Clear[x, t, X, Y, X0, δ, ϵ];
  {functions = {X, Y},
   equilibrium = {X0, 0}
     D[X[t, x], t] +
        D[Y[t, x], x] +
        X[t, x] D[Y[t, x], x] +
        X[t, x] Y[t, x] == 0 /. 
      Thread[functions -> 
        Map[Function[{t, x}, #] &, 
         equilibrium + ϵ Through[
            Thread[δ[functions]][t, x]]]
       ], {ϵ, 0, 1}] /. ϵ -> 1
  ] // TraditionalForm

$$\delta (X)^{(1,0)}(t,x)+\text{X0}\, \delta (Y)(t,x)+\text{X0} \,\delta (Y)^{(0,1)}(t,x)+\delta (Y)^{(0,1)}(t,x)=0$$

In the differential equations, X and Y are functions, so that a replacement must substitute a Function in their place. I do the linearization by the common trick of expanding linearly with respect to a dummy parameter $\epsilon$ and setting $\epsilon = 1$ at the end.

Here I used $\delta(X)$, $\delta(Y)$ for the linear terms. They are functions of x and t, whereas X0 is a constant. Since these two objects appear in a sum, I have to wrap that sum by Function in the replacement that is applied to the differential equation.

Here is a function that does the same as above:

op[eqn_, functions_, equilibrium_] := 
     eqn /. Thread[
       functions -> 
        Map[Function[{t, x}, #] &, 
         equilibrium + ϵ Through[
            Thread[δ[functions]][t, x]]]], {ϵ, 0, 
      1}] /. ϵ -> 1

  eqn = D[X[t, x], t] + D[Y[t, x], x] + X[t, x] D[Y[t, x], x] + 
     X[t, x] Y[t, x] == 0,
  functions = {X, Y},
  equilibrium = {X0, 0}
 op[eqn, functions, equilibrium]

$$\delta (X)^{(1,0)}(t,x)+\text{X0}\, \delta (Y)(t,x)+\text{X0} \,\delta (Y)^{(0,1)}(t,x)+\delta (Y)^{(0,1)}(t,x)=0$$

  • $\begingroup$ Great! Thanks. Two more questions: 1) Can this be generalized also to the case where X, and T are constants? Because doesn't work in this specific case as I see; 2) Is there a way to define this as an operator, i.e., such that I can assign my equation to the variable EQ1, and then call OP[EQ1,X0,Y0,etc]? $\endgroup$ Commented Mar 3, 2015 at 10:40
  • $\begingroup$ I don't understand exactly what you mean by X and T are constants. What is T? Do you mean that you add some constant to the differential equation? By "doesn't work," do you mean you want such constants to be dropped? $\endgroup$
    – Jens
    Commented Mar 3, 2015 at 19:05
  • $\begingroup$ I meant X and Y, sorry, anyway I think I fixed it. I just define X[t_,x_]:=constant; after the calculation $\endgroup$ Commented Mar 3, 2015 at 20:34
  • $\begingroup$ OK - I added a version where I wrap everything in a function called op. $\endgroup$
    – Jens
    Commented Mar 3, 2015 at 20:42

Beginning with

D[X[t, x], t] + D[Y[t, x], t] + X D[Y[t, x], x] + X Y == 0

First replace the derivative by some other function to simplify manipulations.

%[[1]] /. Derivative[z1__][z2__][z3__] -> W[{z1}, z2, {z3}]

Make the first order substitution.

%/. {X -> X0 + d X1, Y -> Y0 + d Y1}

Expand the second argument of W. (Depending on the complexity of the second argument, more transformations may be needed.)

%/. W[z__] :> Thread[W[z], Plus] /. W[z1__, d z2__, z3__] :> d W[z1, z2, z3]

Perform Series and extract the Coefficient of d.

Coefficient[Normal[Series[%, {d, 0, 1}]], d]

Finally, replace W by Derivative

(% /. W[{z1__}, z2_, {z3__}] :> Derivative[z1][z2][z3]) == 0
(* X1*Y0 + X0*Y1 + X1*Derivative[0, 1][Y0][t, x] + X0*Derivative[0, 1][Y1][t, x] 
   + Derivative[1, 0][X1][t, x] + Derivative[1, 0][Y1][t, x] == 0 *)

This process is laborious because such functions as Expand do not work with Derivative. Perhaps, in a future release of Mathematica.


Note that a much simpler approach is available, if the equation to be linearized exists in an Unevaluated form.

Unevaluated[(D[X[t, x], t] + D[Y[t, x], t] + X[t, x] D[Y[t, x], x] + X[t, x] Y[t, x] == 0)] /. 
  {X[t, x] -> X0[t, x] + d X1[t, x], Y[t, x] -> Y0[t, x] + d Y1[t, x]};
Coefficient[Normal[Series[%[[1]], {d, 0, 1}]], d] == 0

yields the desired result.


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