How do you actually calculate this....?

[SansaStark]

Hello!
In my course on physical transport processes time and again these type of equations, containing partial derivatives, appear:

But how do you actually calculate such an equation? I know how to take the partial derivative from a function like x²tz³.
Would it be that I am given a function like x²yz³ and then I just do it like so:
x²z³+ u*2xtz³ = D*1

Or how woould calculating with this equation look like?

Thank you for an answer! Regards, Vera

 

[Mark44]

Hello!
In my course on physical transport processes time and again these type of equations, containing partial derivatives, appear:

View attachment 94908

But how do you actually calculate such an equation? I know how to take the partial derivative from a function like x²tz³.
Would it be that I am given a function like x²yz³ and then I just do it like so:
x²z³+ u*2xtz³ = D*1

Or how woould calculating with this equation look like?

Thank you for an answer! Regards, Vera

Solving (rather than calculating) a PDE is a nontrivial process. I took a 400-level math class in which we spent the entire quarter solving equations like the above. The process is too involved for me to attempt to explain it in a paragraph or three.

 

[HallsofIvy]

What, exactly, do you mean by "calculating" it? Solving such an equation, finding a function c that will satisfy that equation (for given u and D) is, as Mark44 says, a rather difficult thing. But the example you give is just "checking" whether or not a given function satisfies the equation. That is much easier. If, as you give, $c= x^2yz^3$ then $\frac{\partial c}{\partial t}= 0$ because there is no "t" in the formula (unless you are thinking of x, y, and z as functions of t which you did not say), since the derivative of $x^2$ is 2x and its second derivative is 2, $\frac{\partial c}{\partial x}= 2xyz^3$ and $\frac{\partial^2 c}{\partial x^2}= 2yz^3$ so that the equation becomes $0+ 2xyz^2u= 2yz^3D$. Whether that is true or not, whether $c= x^2yz^3$ satisfies the equation depends upon what u ad D are. I have no idea how you got what you give.

 

[SansaStark]

I'm rather clueless on the whole thing so no meaning behind the formula (just invented something ... :/).
But, Hallsoflvy, that's exactly what I wanted to know. How the term below the fraction line is related to the term above. So whatever there is below the fraction line, this determines what is to be derived in a given equation/function and the expression above the fraction is the thing that normally is the f(y or so) ( I know, not mathematical at all..).

 

[S.G. Janssens]

As was already stipulated, solving this type of equation (known as a partial differential equation or PDE) amounts to finding a function $c$, depending on $t$ and $x$, such that if you substitute $c(t,x)$ in the equation, the equality holds.

For example, if we assume that $u$ and $D$ are constants, then you can check for yourself that any function
$$
c(t,x) = A + B e^{\frac{u x}{D}}
$$
satisfies the equation. Here $A$ and $B$ are constants that you are free to choose. Note that the function does not depend on time. It is therefore called a stationary solution.

Usually, just writing down the equation is not enough to completely specify the problem. You need to also impose conditions on the boundary of the $x$-domain (which we left unspecified at the moment), as well as initial conditions.

The topic is very beautiful but, as Mark44 said, requires preparation. At the minimum, you need a good understanding of multivariable calculus. After that, I would opt for one of the better introductory books on PDE. Surely people here can advise you about books that are appropriate to learn about linear advection-diffusion equations from an applied viewpoint.

 

[BvU]

I start to suspect that Sansa has difficulty interpreting the partial derivative as such ? (especially after seeing post #4)

 

[SansaStark]

Uh ok. Then I'll probably head over to an introduction to this multivariable calculus ;) (Thought it would be easier ;)

 

[Chestermiller]

Uh ok. Then I'll probably head over to an introduction to this multivariable calculus ;) (Thought it would be easier ;)

If u is a constant, this equation is much easier to solve. By a change of variables, you can eliminate the advection term in the equation. It is equivalent to converting to a frame of reference that moves at a constant velocity u in the x direction.

 

[HallsofIvy]

I'm rather clueless on the whole thing so no meaning behind the formula (just invented something ... :/).
But, Hallsoflvy, that's exactly what I wanted to know. How the term below the fraction line is related to the term above. So whatever there is below the fraction line, this determines what is to be derived in a given equation/function and the expression above the fraction is the thing that normally is the f(y or so) ( I know, not mathematical at all..).

Have you never take Calculus? The derivative $\frac{dy}{dx}$, or partial derivative, $\frac{\partial y}{\partial x}$ is NOT a fraction and there is no "term above the fraction line" or below it. This is a notation for a single operation on a function. Other notations for the same operation, on f, are f' and Df.