Solving First Order Partial Differential Equations

[jam12]

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

grad p= (0,0,-ρg)

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 dp_x/dx = 0
dp_y/dy = 0
dp_z/dz=-ρg

But I am confused as p is not a vector function itself. so the final answer cannot be in the form p=(p_x,p_y,p_z)

 

[Redbelly98]

can i just separately solve the three differential equations?
ie dp_x/dx = 0
dp_y/dy = 0
dp_z/dz=-ρg

But I am confused as p is not a vector function itself. so the final answer cannot be in the form p=(p_x,p_y,p_z)

Since p is a scalar, instead you would have

∂p/∂x = 0
∂p/∂y = 0
∂p/∂z=-ρg​

 

[jam12]

Since p is a scalar, instead you would have

∂p/∂x = 0
∂p/∂y = 0
∂p/∂z=-ρg​

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?

 

[Redbelly98]

Is this for homework? Separation of variables would work here.

http://en.wikipedia.org/wiki/Separation_of_variables#Partial_differential_equations

 

[jam12]

Is this for homework? Separation of variables would work here.

http://en.wikipedia.org/wiki/Separation_of_variables#Partial_differential_equations

no, its in my lecture notes and I am not sure how the result was obtained

 

[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 [itex]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. <br /> <br /> Now differentiate that f with respect to y:<br /> $$\frac{\partial f}{\partial y}= x+ C'(y)= e^y+ x$$<br /> The "x" terms cancel (that <b>had</b> to happen for this system to have a solution) giving just $C'(y)= e^y$ so $C(y)= e^y+ D$ where D now really is a constant.<br /> <br /> That gives $f(x,y)= (1/2)x^2+ xy+ e^y+ D$.<br /> <br /> Your first two equations, <br /> $$\frac{\partial p}{\partial x}= 0$$<br /> $$\frac{\partial p}{\partial y}= 0$$<br /> 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 <br /> $$\frac{\partial p}{\partial y}= \frac{0\partial A}{\partial y}= 0$$<br /> which says that A is a "constant" with respect to y- but may be a function of z.[/itex]

 

[jam12]

$$\frac{\partial p}{\partial x}= 0$$
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 [itex]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. <br /> <br /> Now differentiate that f with respect to y:<br /> $$\frac{\partial f}{\partial y}= x+ C'(y)= e^y+ x$$<br /> The "x" terms cancel (that <b>had</b> to happen for this system to have a solution) giving just $C'(y)= e^y$ so $C(y)= e^y+ D$ where D now really is a constant.<br /> <br /> That gives $f(x,y)= (1/2)x^2+ xy+ e^y+ D$.<br /> <br /> Your first two equations, <br /> $$\frac{\partial p}{\partial x}= 0$$<br /> $$\frac{\partial p}{\partial y}= 0$$<br /> 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 <br /> $$\frac{\partial p}{\partial y}= \frac{0\partial A}{\partial y}= 0$$<br /> which says that A is a "constant" with respect to y- but may be a function of z.[/itex]

$<br /> Thanks this is very well explained, i understand it now<img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":cool:" title="Cool :cool:" data-smilie="6"data-shortname=":cool:" />$