@tomleslie 

I have read your attachment. Before posting here, I had posted on a math forum to ask if anyone else has seen the total differential dF(x,y) being "divided" by dy, thereby forming a "total" derivative of F w/respect to y.

Whether this is mathematically correct needs to be seen by delving further into the calculus. So far, I have received only comments about dy/du needing to be nonzero as well as about the invertibility of x=x(u) and y=y(u). Once this is resolved, it should help on how it is to be used in Maple.

Judging by the above comments, it seems it was a bit premature to ask a math question which looks like a question on Maple syntax.

Thanks for your replies.

@tomleslie 

F(x,y) is considered a function of x and y only, doesn't depend explicitly on u. Why the question? I thought I didn't enter the expression properly, or there is syntax I am missing.

I am enetering dF/dy, Maple interprets it as \partial F / \partial y

because F depends on 2 arguments. I stumbled on this, so decided to post about it. Please go to Maple, 2D Math and type

(1):

where we silently assume "a" is the ratio

Then Maple will reply:

since it assumes that a derivative of F(x,y) w/respect to y s/be be expressed this way. This is an interesting situation when x and y depend on parameter.

@Z1493 

Edit: the "dy" in the second line, end cancels out.

@rlopez 

We have the function F(x,y) and the curve x=x(u), y=y(u). The total differential is

"divide" by dy:

where

and dy/du can be assumed nonzero.

@Preben Alsholm 

The above PDE system was put together without reference to physics... I thought the x=x(t) could be interpreted as constraint of some sort. Else, I tweaked the PDE system until it was simple and its solution would fit in a small image file :) 

@Preben Alsholm 

Thank you for pointing the x=e^t. I replaced x with e^t, but the solutions are still different. 

You said:>>>>

Question: Are you asking if given one system of (linear) pdes with coefficients depending on t and another where t is replaced by ln(x) in the cofficients only then the general solution for the two different systems will agree if we replace x by exp(t) in both?

Yes. But it is becoming clear that this would depend on information coming from some physics model and has little to do with Maple, right? I only wanted to know if I have a PDE system with composite functions (as in the above case, f=f(x(t),t) ) how would I proceed to integrate it.

Thank you for your input!

@Preben Alsholm 

Thank you for you input.

Finally, let's compare Solution 2 with Solution 1. 

In Solution 2, substitute ln(x)=t;

Then take Solution 2 and test if it satisfies PDE system #1 :

It doesn't. 

THe question then is - we have the same PDE system, but depending on how we introduce x=x(t), we get 2 different solutions. Which ne is correct?

@Preben Alsholm 

Here is a worked example. The picture below shows Solution 2:

In this example, since x=x(t)=e^t --> t=ln(x)

a=a(t)=a(t(x))=1/t(x)=1/ln(x)

b=b(t(x))=2

c=c(t(x))=t(x)=ln(t)

d=d(t(x))=2

@Preben Alsholm 

Here is a worked example. The picture below shows Solution 1:

In this example,

a=a(t)=1/t

b=b(t)=1

c=c(t)=t

d=d(t)=2

@Carl Love 

It can certainly be done; I am confused because if the coefficients before the partial derivatives are functions of t (as written now, all of them are just 1 or -1) , then t=ln(x). Plugging this in the coefficients and then solving with Maple gives a different solution. Compared to solving as-is and then substituting x=e^t; 

Which is "correct" then:

(1) We have the PDE system

Ai(t) * f,i (x(t),t)  = 0 ,

Bi(t) * f,i (x(t),t)  = 0 , i = 1..2 where x(t)=e^t  (f,i is the partia derivative of w/respect to x or t)

Solve with Maple, plug in x=e^t --> Yields Solution 1.

(2) t=ln(x), then Ai(t) = Ai(ln(x))

The PDE system becomes

Ai(ln(x)) * f,i (x(t),t)  = 0

Bi(ln(x)) * f,i (x,t)  = 0 , i = 1..2

Solve with Maple, convert back to x=e^t --> Yields Solution 2.

@Carl Love 

Thank you for your input.

@Preben Alsholm 

THanks, I will consider your suggestion.

@Rouben Rostamian  

Thank you for the helpful answer!

@Rouben Rostamian  

The ics give a well defined solution with C1*C2=1 and c1=1. THis is the reason I asked the question (unless I am missing saomething). Maple should find this or something in the syntax is wrong

@Rouben Rostamian  

Thank you, your solution works. I tried using it for a PDE system, but cannot make it to work. Maple doesn't give an error message; it does not give an answer either: