# First order partial diff eqn

1. Apr 17, 2010

### jam12

Hi there, how can i solve this first order partial differential equation:

where p=p(x,y,z,t) is pressure
where ρ=ρ(x,y,z,t) is density
where grad p is in three dimensional Cartesian coordinates

can i just separately solve the three differential equations?
ie dpx/dx = 0
dpy/dy = 0
dpz/dz=-ρg

But im confused as p is not a vector function itself. so the final answer cannot be in the form p=(px,py,pz)

2. Apr 17, 2010

### Redbelly98

Staff Emeritus
Since p is a scalar, instead you would have
∂p/∂x = 0
∂p/∂y = 0
∂p/∂z=-ρg​

3. Apr 17, 2010

### jam12

I see, so how do i combine these solutions for p?
i get three solutions p=A (constant), p=B (constant) and p=-ρgz + C.
so what does my solution look like for p?

4. Apr 17, 2010

### Redbelly98

Staff Emeritus
5. Apr 17, 2010

### jam12

6. Apr 18, 2010

### HallsofIvy

$$\frac{\partial p}{\partial x}= 0$$
and
$$\frac{\partial p}{\partial y}= 0$$
say that p is not a function of x or y but a function of z only.

To solve
$$\frac{dp}{dz}= -\rho g$$
just intgrate to get $p(x, y, z)= -\rho gz+ C$.

Partial derivatives do NOT give three different solutions. If you had, for example,
$$\frac{\partial f}{\partial x}= x+ y$$
and
$$\frac{\partial f}{\partial y}= e^y+ x$$
From the first equation, you would get $f(x,y)= (1/2)x^2+ xy+ C(y)[/tex]. Because, in taking the partial derivative with respect to x, we treat y as a constant, the "constant of integration" may be a function of y. Now differentiate that f with respect to y: $$\frac{\partial f}{\partial y}= x+ C'(y)= e^y+ x$$ The "x" terms cancel (that had to happen for this system to have a solution) giving just [itex]C'(y)= e^y$ so $C(y)= e^y+ D$ where D now really is a constant.

That gives $f(x,y)= (1/2)x^2+ xy+ e^y+ D$.

$$\frac{\partial p}{\partial x}= 0$$
$$\frac{\partial p}{\partial y}= 0$$
do NOT give "p= A" or "p= B". Since differentiation with respect to x treats both y and z as constant, the "constant of integration" may be a function of y and z: p= A(y,z). Then
$$\frac{\partial p}{\partial y}= \frac{0\partial A}{\partial y}= 0$$
which says that A is a "constant" with respect to y- but may be a function of z.

Last edited by a moderator: Apr 18, 2010
7. Apr 20, 2010

### jam12

Thanks this is very well explained, i understand it now