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We have two variables $u_{1}$ and $u_{2}$ which are functions of $x,t$. That is, $u_{1}=u_{1}(x,t)$ and $u_{2}=u_{2}(x,t)$. Consider the the system of partial differential equations $$\frac{\partial u_{1}}{\partial x}=u_{2},\\\frac{\partial u_{2}}{\partial x}=\frac{\partial u_{1}}{\partial t}.$$

Next, consider the transformation $$w_{11}=u_{1},\quad w_{12}=\frac{u_{1}}{u_{2}}.$$ We need to compute $\frac{\partial w_{1k}}{\partial x}$ for $k=1,2$.

In Mathematica my process is very manual. I am basically using Mathematica to simplify the partial derivatives that I have found using pen and paper. The code is as follows.

(*f[k] denotes the partial derivative of u[k] with respect to x*)
(*g[1] denotes the partial derivative of u[1] with respect to t*)
f[1] = u[2];
f[2] = g[1];

trans = {u[1] -> w[1, 1], u[2] -> w[1, 1] w[1, 2]};

(*Denote y[1,1] as the partial derivative of w[1,1] with respect to x*)
z[1, 1] = f[1] /. trans

(*Denote y[1,2] as the partial derivative of w[1,2] with respect to x*)
(*Denote z[1,1] as the partial derivative of w[1,1] with respect to t*)
z[1, 2] = Factor[(u[2]^2 - u[1] g[1])/u[2]^2 /. trans] /. g[1]->g[1,1]

I would like to compute the partial derivatives of $w_{1k}$ using a single code as opposed to doing it separately. I tried using the built-in function $D[\;]$, but run into problems when defining, for example, $\frac{\partial u_{1}}{\partial x}=u_{2}$ using $D[\;]$. In Mathematica:

D[u[1][x, t], x] = u[2][x, t];

Set::write: Tag D in \!\(\*SubscriptBox[\(\[PartialD]\), \(x\)]\(\(u[1]\)[x, t]\)\) is Protected. >>

I guess my main problem is how to define the system $\frac{\partial u_{1}}{\partial x}=u_{2},\;\;\frac{\partial u_{2}}{\partial x}=\frac{\partial u_{1}}{\partial t}$ in Mathematica that will allow me to use $D[\;]$. Any help/suggestions will be appreciated.

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First we define your system of equations:

eqns = {D[u1[x,t],x] == u2[x,t], D[u2[x,t],x] == D[u1[x,t],t]};

Then we solve your substitution for u1 and u2:

sub = Solve[{w11[x, t] == u1[x, t], w12[x, t] == u1[x, t]/u2[x, t]},
            {u1[x, t], u2[x, t]}]

(* {{u1[x, t] -> w11[x, t], u2[x, t] -> w11[x, t]/w12[x, t]}} *)

Then we transform the variables into the correct functional form:

sub = sub /. {(f_[var__] -> e_) :> (f -> Function[{var}, e])}

(* {{u1 -> Function[{x, t}, w11[x, t]],
     u2 -> Function[{x, t}, w11[x, t]/w12[x, t]]}} *)

Finally we can make the substitution:

eqns /. sub

( *{{Derivative[1, 0][w11][x, t] == w11[x, t]/w12[x, t], 
     Derivative[1, 0][w11][x, t]/w12[x, t] -
      (w11[x, t]*Derivative[1, 0][w12][x, t])/w12[x, t]^2 == 
      Derivative[0, 1][w11][x, t]}} *)

In order to interpret this result, you'll need to know a little about the Derivative operator. In a form like this:

Derivative[3][f]

It represents the third derivative of a function f with one argument. It does partial derivatives like this:

Derivative[3,2][g]

This represents g differentiated three times with respect to its first argument, and differentiated two times with respect to its second argument.

So, to take an example from your transformed system:

Derivative[1, 0][w11][x, t]

Is equivalent to:

$$ \frac{\partial w_{11}}{\partial x} $$

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  • $\begingroup$ Thanks. For the UpSet how can I write the output produced by D[w11[x, t], x] (* u2[x, t] *) purely in terms of w11 and w12? $\endgroup$
    – Jack
    Apr 28 '15 at 2:35
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    $\begingroup$ @Jack It wasn't clear to me that you wanted the whole system in terms of w11 and w12... I've changed my answer to hopefully do the transformation you want. $\endgroup$ Apr 28 '15 at 3:14

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