## 104 Reputation

14 years, 317 days

## Wow.  That did exactly what...

Wow.  That did exactly what I wanted.  Is the 'factor' just telling collect to get all the derivatives together?

Thanks a lot,

Ryan

## Wow.  That did exactly what...

Wow.  That did exactly what I wanted.  Is the 'factor' just telling collect to get all the derivatives together?

Thanks a lot,

Ryan

## I'm not sure what you mean...

I'm not sure what you mean complete.  The question I have is in regards to what I am calling ExpandedEqn5.  I am trying to figure out a way to basically collect terms.  I want to collect p and it's derivatives, and then q and its derivatives.  Could you be more specific?

It is correct.

Ryan

## I'm not sure what you mean...

I'm not sure what you mean complete.  The question I have is in regards to what I am calling ExpandedEqn5.  I am trying to figure out a way to basically collect terms.  I want to collect p and it's derivatives, and then q and its derivatives.  Could you be more specific?

It is correct.

Ryan

## Yes, but I need it for later...

Yes, but I need it for later calculations.  I have a governing equation that contains an unkown function I am approximating.  To solve for the coefficents I am using the real-valuedness of some of the other functions.

However, some of these other functions are defined as the z and zbar derivatives, so I need to be working with complex variables.  For example, I have a term S_z*u_{zbar}, but both u and S are real.  I am trying a work around, but in cartesian the equations are absurd.

I will try your suggestion and see what works.

Thanks

## Yes, but I need it for later...

Yes, but I need it for later calculations.  I have a governing equation that contains an unkown function I am approximating.  To solve for the coefficents I am using the real-valuedness of some of the other functions.

However, some of these other functions are defined as the z and zbar derivatives, so I need to be working with complex variables.  For example, I have a term S_z*u_{zbar}, but both u and S are real.  I am trying a work around, but in cartesian the equations are absurd.

I will try your suggestion and see what works.

Thanks

## Okay, almost...

Okay, almost there.

Something funny happened when I tried to simplify the notation.  Below is a list of subs that gave me a nice looking function.  But when I tried to take the derivative wrt x, the function phi lost its dependancies, that is, phi lost its dependence on fx and fx, and so the chain rule is not correct.

Any ideas?

> restart; with(PDEtools); f := 'f'; g := 'g'; u := 'u'; S := 'S'; x := 'x'; y := 'y'; z := 'z'; w := 'w'; Z := 'Z';

> u := proc (x, y, z) options operator, arrow; (diff(f(x, y, z), x))*(diff(f(x, y, z), y)) end proc; fx := diff(f(x, y, z), x); fy := diff(f(x, y, z), y); diff(u(x, y, z), z); Eqn310 := simplify(subs(diff(f(x, y, z), x, z) = diff(phi(x, y, z, fx, fy), x), diff(f(x, y, z), y, z) = diff(phi(x, y, z, fx, fy), y), %));
print(`output redirected...`); # input placeholder
>
print(`output redirected...`); # input placeholder
This is completely correct up to this point for KM's paper.  This is equation 3.10
> ee := subs((D[2](f))(x, y, z) = f[y], Eqn310); subs((D[1](f))(x, y, z) = f[x], ee); subs((D[1, 1](f))(x, y, z) = f[xx], %); subs((D[1, 2](f))(x, y, z) = f[xy], %); subs((D[2, 2](f))(x, y, z) = f[yy], %); subs((D[1](phi))(x, y, z, f[x], f[y]) = phi[x], %); subs((D[2](phi))(x, y, z, f[x], f[y]) = phi[y], %); subs((D[4](phi))(x, y, z, f[x], f[y]) = phi[f[x]], %); Eqn := subs((D[5](phi))(x, y, z, f[x], f[y]) = phi[f[y]], %);
print(`output redirected...`); # input placeholder
f[y] phi[x] + f[y] phi[f[x]] f[xx] + f[y] phi[f[y]] f[xy] + f[x] phi[y]

+ f[x] phi[f[x]] f[xy] + f[x] phi[f[y]] f[yy]
>

> diff(Eqn(x, y, z), x);
print(`output redirected...`); # input placeholder
/ d               \                                 / d                 \
|--- f[y](x, y, z)| phi[x](x, y, z) + f[y](x, y, z) |--- phi[x](x, y, z)|
\ dx              /                                 \ dx                /

/ d               \
+ |--- f[y](x, y, z)| phi[f[x]](x, y, z) f[xx](x, y, z)
\ dx              /

/ d                    \
+ f[y](x, y, z) |--- phi[f[x]](x, y, z)| f[xx](x, y, z)
\ dx                   /

/ d                \
+ f[y](x, y, z) phi[f[x]](x, y, z) |--- f[xx](x, y, z)|
\ dx               /

/ d               \
+ |--- f[y](x, y, z)| phi[f[y]](x, y, z) f[xy](x, y, z)
\ dx              /

/ d                    \
+ f[y](x, y, z) |--- phi[f[y]](x, y, z)| f[xy](x, y, z)
\ dx                   /

/ d                \
+ f[y](x, y, z) phi[f[y]](x, y, z) |--- f[xy](x, y, z)|
\ dx               /

/ d               \                                 / d                 \
+ |--- f[x](x, y, z)| phi[y](x, y, z) + f[x](x, y, z) |--- phi[y](x, y, z)|
\ dx              /                                 \ dx                /

/ d               \
+ |--- f[x](x, y, z)| phi[f[x]](x, y, z) f[xy](x, y, z)
\ dx              /

/ d                    \
+ f[x](x, y, z) |--- phi[f[x]](x, y, z)| f[xy](x, y, z)
\ dx                   /

/ d                \
+ f[x](x, y, z) phi[f[x]](x, y, z) |--- f[xy](x, y, z)|
\ dx               /

/ d               \
+ |--- f[x](x, y, z)| phi[f[y]](x, y, z) f[yy](x, y, z)
\ dx              /

/ d                    \
+ f[x](x, y, z) |--- phi[f[y]](x, y, z)| f[yy](x, y, z)
\ dx                   /

/ d                \
+ f[x](x, y, z) phi[f[y]](x, y, z) |--- f[yy](x, y, z)|
\ dx               /
>

## Okay, almost...

Okay, almost there.

Something funny happened when I tried to simplify the notation.  Below is a list of subs that gave me a nice looking function.  But when I tried to take the derivative wrt x, the function phi lost its dependancies, that is, phi lost its dependence on fx and fx, and so the chain rule is not correct.

Any ideas?

> restart; with(PDEtools); f := 'f'; g := 'g'; u := 'u'; S := 'S'; x := 'x'; y := 'y'; z := 'z'; w := 'w'; Z := 'Z';

> u := proc (x, y, z) options operator, arrow; (diff(f(x, y, z), x))*(diff(f(x, y, z), y)) end proc; fx := diff(f(x, y, z), x); fy := diff(f(x, y, z), y); diff(u(x, y, z), z); Eqn310 := simplify(subs(diff(f(x, y, z), x, z) = diff(phi(x, y, z, fx, fy), x), diff(f(x, y, z), y, z) = diff(phi(x, y, z, fx, fy), y), %));
print(`output redirected...`); # input placeholder
>
print(`output redirected...`); # input placeholder
This is completely correct up to this point for KM's paper.  This is equation 3.10
> ee := subs((D[2](f))(x, y, z) = f[y], Eqn310); subs((D[1](f))(x, y, z) = f[x], ee); subs((D[1, 1](f))(x, y, z) = f[xx], %); subs((D[1, 2](f))(x, y, z) = f[xy], %); subs((D[2, 2](f))(x, y, z) = f[yy], %); subs((D[1](phi))(x, y, z, f[x], f[y]) = phi[x], %); subs((D[2](phi))(x, y, z, f[x], f[y]) = phi[y], %); subs((D[4](phi))(x, y, z, f[x], f[y]) = phi[f[x]], %); Eqn := subs((D[5](phi))(x, y, z, f[x], f[y]) = phi[f[y]], %);
print(`output redirected...`); # input placeholder
f[y] phi[x] + f[y] phi[f[x]] f[xx] + f[y] phi[f[y]] f[xy] + f[x] phi[y]

+ f[x] phi[f[x]] f[xy] + f[x] phi[f[y]] f[yy]
>

> diff(Eqn(x, y, z), x);
print(`output redirected...`); # input placeholder
/ d               \                                 / d                 \
|--- f[y](x, y, z)| phi[x](x, y, z) + f[y](x, y, z) |--- phi[x](x, y, z)|
\ dx              /                                 \ dx                /

/ d               \
+ |--- f[y](x, y, z)| phi[f[x]](x, y, z) f[xx](x, y, z)
\ dx              /

/ d                    \
+ f[y](x, y, z) |--- phi[f[x]](x, y, z)| f[xx](x, y, z)
\ dx                   /

/ d                \
+ f[y](x, y, z) phi[f[x]](x, y, z) |--- f[xx](x, y, z)|
\ dx               /

/ d               \
+ |--- f[y](x, y, z)| phi[f[y]](x, y, z) f[xy](x, y, z)
\ dx              /

/ d                    \
+ f[y](x, y, z) |--- phi[f[y]](x, y, z)| f[xy](x, y, z)
\ dx                   /

/ d                \
+ f[y](x, y, z) phi[f[y]](x, y, z) |--- f[xy](x, y, z)|
\ dx               /

/ d               \                                 / d                 \
+ |--- f[x](x, y, z)| phi[y](x, y, z) + f[x](x, y, z) |--- phi[y](x, y, z)|
\ dx              /                                 \ dx                /

/ d               \
+ |--- f[x](x, y, z)| phi[f[x]](x, y, z) f[xy](x, y, z)
\ dx              /

/ d                    \
+ f[x](x, y, z) |--- phi[f[x]](x, y, z)| f[xy](x, y, z)
\ dx                   /

/ d                \
+ f[x](x, y, z) phi[f[x]](x, y, z) |--- f[xy](x, y, z)|
\ dx               /

/ d               \
+ |--- f[x](x, y, z)| phi[f[y]](x, y, z) f[yy](x, y, z)
\ dx              /

/ d                    \
+ f[x](x, y, z) |--- phi[f[y]](x, y, z)| f[yy](x, y, z)
\ dx                   /

/ d                \
+ f[x](x, y, z) phi[f[y]](x, y, z) |--- f[yy](x, y, z)|
\ dx               /
>

## Right, so now what if f_z is...

Right, so now what if f_z is some function phi?  How do I get Maple to recognize it?

I tried using subs with no luck.  In your example, what if phi = diff(f(x,y,z),z)?

Thanks,

Ryan

## Right, so now what if f_z is...

Right, so now what if f_z is some function phi?  How do I get Maple to recognize it?

I tried using subs with no luck.  In your example, what if phi = diff(f(x,y,z),z)?

Thanks,

Ryan

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