Solving First-Order PDE: Explaining Basics

[ahmed markhoos]

Sorry to keep the title too broad and general.

I'm starting learning pde by myself , using "linear partial differential equations for scientists and engineers"

I'm having some problems with the basics "I took ODE". The following differentiation is totally new to me, can some one explain to me?

f(x,y,z,a,b)=0, a and b are parameters

Differentiating the function with respect to x we get:
f_x+ pf_z =0

And with respect to y we get:

f_y+ qf_z =0

Where q=∂z/∂y , p=∂z/∂x.

How? Why?

 

[andrewkirk]

The formulas look wrong. Where did you get them?

 

[Simon Bridge]

linear partial differential equations for scientists and engineers

This one: http://link.springer.com/book/10.1007/978-0-8176-4560-1 [Myint-U & Debnath 2007]
There is a PDF of that reference here http://sharif.ir/~moosavi/Myint-U_D...al_Equations_for_Scientists_and_Engineers.pdf
... the equations look like they are from (4th ed) p29, starting at eq 2.3.1 ... is this correct?

 

[andrewkirk]

I think the confusion arises from the text not explaining clearly enough what it is doing.
When it says the first equation in 2.3.2 is obtained by differentiating 2.3.1 with respect to $x$, it should have added 'while keeping $f$ and $y$ but not $z$ constant'.

Formally, what they are doing is, at any point $(x_0,y_0,z_0)$, identifying the line that is the intersection of the surface $y=y_0$ with the surface defined by equation 2.3.1. They then parametrise that line with parameter $t$ by setting $t=x$. Then what they describe as 'differentiating with respect to $x$' is actually differentiating with respect to $t$. Using the formula for the total differential:

$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+
\frac{\partial f}{\partial y}\frac{dy}{dt}+
\frac{\partial f}{\partial z}\frac{dz}{dt}$$

we can get their formula when we observe that $\frac{dy}{dt}=0$ and $\frac{dx}{dt}=1$.

 

[Simon Bridge]

I think the confusion arises from the text not explaining clearly enough what it is doing.

... I agree.
I was thinking more, following 2.2.2, they have implicitly chosen z to be dependant on x and y in F(x,y,z,a,b) and applied the chain rule per the core concepts in chapter 1.

I don't think this text is intended for self-study...

 

[ahmed markhoos]

... I agree.
I was thinking more, following 2.2.2, they have implicitly chosen z to be dependant on x and y in F(x,y,z,a,b) and applied the chain rule per the core concepts in chapter 1.

I don't think this text is intended for self-study...

I choosed it because it seemed to be the best pde book for me while studying physics. A lot of people gave a good feedback to it.

Do you recommend any other books?

 

[ahmed markhoos]

I think the confusion arises from the text not explaining clearly enough what it is doing.
When it says the first equation in 2.3.2 is obtained by differentiating 2.3.1 with respect to $x$, it should have added 'while keeping $f$ and $y$ but not $z$ constant'.

Formally, what they are doing is, at any point $(x_0,y_0,z_0)$, identifying the line that is the intersection of the surface $y=y_0$ with the surface defined by equation 2.3.1. They then parametrise that line with parameter $t$ by setting $t=x$. Then what they describe as 'differentiating with respect to $x$' is actually differentiating with respect to $t$. Using the formula for the total differential:

$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+
\frac{\partial f}{\partial y}\frac{dy}{dt}+
\frac{\partial f}{\partial z}\frac{dz}{dt}$$

we can get their formula when we observe that $\frac{dy}{dt}=0$ and $\frac{dx}{dt}=1$.

Exactly, thank you very much sir.