# Linearization of differential equations

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:

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

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


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

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_] :=
Module[{ϵ},
Normal@Series[
functions ->
Map[Function[{t, x}, #] &,
equilibrium + ϵ Through[
1}] /. ϵ -> 1
]

With[{
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]


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

• 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]? Mar 3 '15 at 10:40
• 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?
– Jens
Mar 3 '15 at 19:05
• I meant X and Y, sorry, anyway I think I fixed it. I just define X[t_,x_]:=constant; after the calculation Mar 3 '15 at 20:34
• OK - I added a version where I wrap everything in a function called op.
– Jens
Mar 3 '15 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)] /.