first order PDE , verification of one solution

Hello

the solution of the PDE:

df(x,y)/dx + df(x,y)/dy =0

should be f= f(x-y) , with f an arbitrary function of the argument which is the compound variable (x-y).

Wanting to verify this symbolic solution I would like to execute these commands

1) syms x y real

2) syms f(x-y)

and then do the calculation

3) ver=diff(f(x-y),x)+diff(f(x-y),y)

and simplifying it should give

ver=zero

but unfortunately on the second line it

gives me an error :

---------------------------------------------------------------------

Error using symfun.parseString (line 101)

Invalid variable name.

Error in syms (line 276)

[name, vars] = symfun.parseString(x,'allowPercent');

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

-----------------------------------------------------------------------

I was kindly asking

how to proceed and be able to do this

symbolic verification

thankyou very much !

Valerio

Answer 1

[Walter Roberson]

syms x y real
syms f(x_minus_y)
v1 = diff(f(x-y),x)

v1 = 

v2 = diff(f(x-y),y)

v2 = 

ver = v1 + v2

ver = 

●

simplify(ver)

ans = 

Which is not zero.

I do not understand why you think it should be 0. Surely df(x,y)/dx + df(x,y)/dy = 0 is a condition, rather than a truism.

[Val]

Thankyou

I corrected the line 2) , and now i get

>> syms x y real

>> syms f(x_minus_y)

>> v1 = diff(f(x-y),x)

v1 =

diff(f(x - y), x)

>> v2 = diff(f(x-y),y)

v2 =

-D(f)(x - y)

ok I'm on the right track, but now I don't understand why deriving with respect to y i.e.

v2 = diff(f(x-y),y)

give

v2 =

-D(f)(x - y)

instead deriving with respect to x i.e.

v1 = diff(f(x-y),x)

i did not obtained

v1 =

+D(f)(x - y)

[Val]

"while deriving with respect to x " .. sorry

[Walter Roberson]

It looks like diff() is happening to create two different notations that are not immediately compatible. But substitute in any actual function and you can see that the two parts are negatives of each other.

syms x y real
syms f(x_minus_y)
F(x_minus_y) = tan(x_minus_y)^3 + exp(-x_minus_y)

F(x_minus_y) = 

v1 = diff(f(x-y),x); char(v1)

ans = 'diff(f(x - y), x)'

v2 = diff(f(x-y),y); char(v2)

ans = '-D(f)(x - y)'

simplify(v1 == -v2)

ans = 

●

V1 = subs(v1, f, F)

V1 = 

V2 = subs(v2, f, F)

V2 = 

simplify(V1 == -V2)

ans = 

symtrue

[Val]

"It looks like diff() is happening to create two different notations that are not immediately compatible"

That is the point

I think that the right answer to the command:

v1 = diff(f(x-y),x)

should be

D(f)(x - y)

not "diff(f(x - y), x)" , which is a repetition of the command line.

Consider this workaround:

----------------------------------------

>> syms a positive

>> v1 = diff(f(a*x-y),x)

v1 =

a*D(f)(a*x - y)

>> v1=subs(v1,a,1)

v1 =

D(f)(x - y)

-----------------------------------------

that get the right form of the answer, so may immediately verify the solution f(x-y)...

...may be that is a bug to be fixed ?

Thankyou very much Mr Roberson for interest in this matter

Valerio

[Val]

Dear Mr Roberson , let me have Your opinion

is a bug to be fixed ?

thankyou sincerely valerio

Answer 2

[Torsten]

You shouldn't waste your time here with symbolic manipulations.

What is important is that solutions f(x,y) of the PDE

df/dx + df/dy = 0

are constant on lines in the x-y-plane with slope equal to 1.

That means that f is not really a function of two variables, but it is already given by its values on e.g. the x- or the y-axis:

f(x,y) = f(x-y,0) = f(0,y-x)

More generally, f(x,y) as a function on IR^2 is fixed by its values on any line in the x-y-plane that has not a slope equal to 1.

The lines with slope equal to 1 are called the characteristics of the partial differential equation.

Answer 3

[Val]

many thanks to all !!

anyway i think that the right answer to the command:

v1 = diff(f(x-y),x)

should be

D(f)(x - y)

not "diff(f(x - y), x)" , which is a repetition of the command line.

consider this workaround:

----------------------------------------

>> syms a positive

>> v1 = diff(f(a*x-y),x)

v1 =

a*D(f)(a*x - y)

>> v1=subs(v1,a,1)

v1 =

D(f)(x - y)

-----------------------------------------

that get the right form of the answer, so may verify the solution f(x-y)...

...may be that is a bug to be fixed ?

Thankyou very much for interest in this matter

kind regards Valerio (i hope i comment in the correct box)

[Torsten]

I don't understand why you work with f(x-y). The solution of the differential equation

df/dx + df/dy = 0

is a function of two variables: f(x,y).

As I wrote in my answer, the thing you have to show is that f(x,y) = f(x-y,0) holds if f satisfies the partial differential equation.