Automatically replacing derivatives in Mathematica
HEREMAN WILLY
Dear Colleagues:
If in Mathematica you have built up an equation, such as
eq=D[f[x,t],t]+6*f[x,t]*D[f[x,t],x]+D[f[x,t],{x,3}] (1)
and you THEN want to replace f[x,t] and ITS DERIVATIVES by an
explicit expression, such as,
f[x,t] = u[0][x,t]*g[x,t]^(alpha) (2)
how do you tell Mathematica to replace the function and its derivatives,
without having to write replacement rules for the derivatives of f[x,t]
that occur in the equation (1)?
Note that we do NOT want to give the form of f[x,t] first and then build up
the equation. That, of course, would be trivial and also besides the point.
Thanks in advance for your help and suggestions,
Willy
------------
Willy A. Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, CO 80401-1887
Phone: (303) 273-3881 (O) or 440-6089 (H)
Fax: (303) 273-3875 (mention for Dr. Hereman)
Email: WHER...@FLINT.MINES.COLORADO.EDU or
WHER...@LIE.MINES.COLORADO.EDU
-------------
Jorma Virtamo
wher...@slate.Mines.Colorado.EDU (HEREMAN WILLY) wrote:
>
> If in Mathematica you have built up an equation, such as
>
> eq=D[f[x,t],t]+6*f[x,t]*D[f[x,t],x]+D[f[x,t],{x,3}] (1)
>
> and you THEN want to replace f[x,t] and ITS DERIVATIVES by an
> explicit expression, such as,
>
> f[x,t] = u[0][x,t]*g[x,t]^(alpha) (2)
>
> how do you tell Mathematica to replace the function and its derivatives,
> without having to write replacement rules for the derivatives of f[x,t]
> that occur in the equation (1)?
> Note that we do NOT want to give the form of f[x,t] first and then build up
> the equation. That, of course, would be trivial and also besides the point.
>
The problem you have is revealed by evaluating
In[]:= eq//FullForm
Out[]= Plus[Derivative[0, 1][f][x, t],
Times[6, Power[g[x, t], alpha], u[0][x, t],
Derivative[1, 0][f][x, t]], Derivative[3, 0][f][x, t]]
The derivatives operate on the function f itself, not on its value f[x,t].
To get the desired behavior write
f = Function[{x,t},u[0][x,t]*g[x,t]^(alpha)]
or equivalently
f = u[0][#1,#2]*g[#1,#2]^(alpha)&
Then eq evaluates to what you probably want.
-- Jorma Virtamo
============================================================
Jorma Virtamo
VTT Information Technology / Telecommunications
P.O. Box 1202, FIN-02044 VTT, Finland
phone: +358 0 456 5612 fax: +358 0 455 0115
email: jorma....@vtt.fi web: http://www.vtt.fi/tte/
============================================================
Robert Villegas
> If in Mathematica you have built up an equation, such as
>
> eq=D[f[x,t],t]+6*f[x,t]*D[f[x,t],x]+D[f[x,t],{x,3}] (1)
>
> and you THEN want to replace f[x,t] and ITS DERIVATIVES by an
> explicit expression, such as,
>
> f[x,t] = u[0][x,t]*g[x,t]^(alpha) (2)
>
> how do you tell Mathematica to replace the function and its derivatives,
> without having to write replacement rules for the derivatives of f[x,t]
> that occur in the equation (1)?
Dave Wagner shows that if you replace 'f' with a formula wrapped in an
& pure function, then the derivatives will automatically evaluate:
> You need to define f as a pure function:
>
> In[1]:=
> eq=D[f[x,t],t]+6*f[x,t]*D[f[x,t],x]+D[f[x,t],{x,3}]
> Out[1]=
> (0,1) (1,0) (3,0)
> f [x, t] + 6 f[x, t] f [x, t] + f [x, t]
> In[2]:=
> f = u[0][#1,#2]*g[#1,#2]^(alpha)&
> Out[2]=
> alpha
> u[0][#1, #2] g[#1, #2] &
> In[3]:=
> eq
> Out[3]=
> -1 + alpha (0,1)
> alpha g[x, t] u[0][x, t] g [x, t] +
>
>
> [...much output deleted...]
>
>
> alpha (3,0)
> g[x, t] (u[0]) [x, t]
The reason this works is that all those superscripted expressions like
(0,1)
f [x, t]
are partial derivative operators acting on f, and just waiting for
f to become something they can act upon. If you look at the FullForm of
a D result, you'll see a Derivative operator wrapped around f:
In[9]:= D[f[x, t], x] //FullForm
Out[9]//FullForm= Derivative[1, 0][f][x, t]
Derivative[1, 0] is acting on a meaningless symbol f here, but if
you give it a meaningful function, say using #1, #2, & notation
as Dave suggests, it will differentiate it. Let's take a simpler example
where we want f[x_, t_] := Sin[x] Cos[t] to be the formula.
In[10]:= Derivative[1, 0][ Sin[#1] Cos[#2] & ]
Out[10]= 1 Cos[#1] Cos[#2] + Sin[#1] -(0 Sin[#2]) &
(the 1 and 0 multipliers are there only because the structural
transformations that Derivative did are left unevaluated until the
function is used)
Substituting a #,& pure function for f isn't the only way to
get the derivative expression to evaluate, though. 'Derivative'
will take effect on a symbol that is defined as a function in the
usual "f[x_, t_] := formula" manner, too, so if you were going to do
that anyway, you can make use of it. All you have to do is re-evaluate
expr after defining f:
In[1]:= expr = D[f[x, t], x] + D[f[x, t], t]
(0,1) (1,0)
Out[1]= f [x, t] + f [x, t]
In[4]:= f[x_, t_] := Sin[x] Cos[t]
(* Derivative looks up f's definition and differentiates the formula: *)
In[5]:= expr
Out[5]= Cos[t] Cos[x] - Sin[t] Sin[x]
Even if you don't want to have an assignment to f in effect, you can take
advantage of Derivative's lookup of function assignments. Just make
an assignment to f within a Block and evaluate 'expr' while this is in
effect, and it will all work:
In[6]:= Clear[f]
In[7]:= formula = Block[{f}, f[x_, t_] := Sin[x] Cos[t] ; expr ]
Out[7]= Cos[t] Cos[x] - Sin[t] Sin[x]
(* f was Block'd, so it hasn't acquired the assignment permanently: *)
In[8]:= ? f
Global`f
Still another method, really the same as Dave's, but exhibiting the
ability of nameless, or pure, functions to use named variables:
In[15]:= expr /. f -> Function[{x, t}, Sin[x] Cos[t]]
Out[15]= Cos[t] Cos[x] - Sin[t] Sin[x]
If you had a formula in terms of x and y lying about already, this might
be easier to use; you don't have to substitute #1 for x and #2 for t
manually or programmatically.
Robby Villegas
Daniel Lichtblau
wher...@slate.Mines.Colorado.EDU (HEREMAN WILLY) writes:
>
>
>
> Dear Colleagues:
>
> If in Mathematica you have built up an equation, such as
>
> eq=D[f[x,t],t]+6*f[x,t]*D[f[x,t],x]+D[f[x,t],{x,3}] (1)
>
> and you THEN want to replace f[x,t] and ITS DERIVATIVES by an
> explicit expression, such as,
>
> f[x,t] = u[0][x,t]*g[x,t]^(alpha) (2)
>
> how do you tell Mathematica to replace the function and its derivatives,
> without having to write replacement rules for the derivatives of f[x,t]
> that occur in the equation (1)?
up
> the equation. That, of course, would be trivial and also besides the
point.
>
>
> Willy
>
> ------------
>
> Willy A. Hereman
> Department of Mathematical and Computer Sciences
> Colorado School of Mines
> Golden, CO 80401-1887
>
> Phone: (303) 273-3881 (O) or 440-6089 (H)
> Fax: (303) 273-3875 (mention for Dr. Hereman)
> Email: WHER...@FLINT.MINES.COLORADO.EDU or
> WHER...@LIE.MINES.COLORADO.EDU
>
> -------------
>
One method is as below.
In[2]:= eq = Hold[D[f[x,t],t]+6*f[x,t]*D[f[x,t],x]+D[f[x,t],{x,3}]]
Out[2]= Hold[D[f[x, t], t] + 6 f[x, t] D[f[x, t], x] + D[f[x, t], {x, 3}]]
In[3]:= neweq = Release[eq /. f[x,t] -> u[0][x,t]*g[x,t]^(alpha)] //
InputForm
Out[3]//InputForm=
alpha*g[x, t]^(-1 + alpha)*u[0][x, t]*Derivative[0, 1][g][x, t] +
g[x, t]^alpha*Derivative[0, 1][u[0]][x, t] +
(-2 + alpha)*(-1 + alpha)*alpha*g[x, t]^(-3 + alpha)*u[0][x, t]*
Derivative[1, 0][g][x, t]^3 +
3*(-1 + alpha)*alpha*g[x, t]^(-2 + alpha)*Derivative[1, 0][g][x, t]^2*
Derivative[1, 0][u[0]][x, t] +
6*g[x, t]^alpha*u[0][x, t]*(alpha*g[x, t]^(-1 + alpha)*u[0][x, t]*
Derivative[1, 0][g][x, t] + g[x, t]^alpha*Derivative[1, 0][u[0]][x,
t]
) + 3*(-1 + alpha)*alpha*g[x, t]^(-2 + alpha)*u[0][x, t]*
Derivative[1, 0][g][x, t]*Derivative[2, 0][g][x, t] +
3*alpha*g[x, t]^(-1 + alpha)*Derivative[1, 0][u[0]][x, t]*
Derivative[2, 0][g][x, t] +
3*alpha*g[x, t]^(-1 + alpha)*Derivative[1, 0][g][x, t]*
Derivative[2, 0][u[0]][x, t] +
alpha*g[x, t]^(-1 + alpha)*u[0][x, t]*Derivative[3, 0][g][x, t] +
g[x, t]^alpha*Derivative[3, 0][u[0]][x, t]
Daniel Lichtblau, WRI