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I was reading

Elements of partial differential equations by IAN N. SNEDDON

and found out the following:

It is trying to generalise the function given which leads to a pde of first order. It says to consider the function

$$F(x,y,z,a,b) = 0$$

where $a$ and $b$ are constants, $x$ and $y$ are the independent variables and $z$ is the dependent variable. Then it differentiated this equation w.r.t. $x$ and $y$ to obtain two more equations:

$$ \frac{\partial F}{\partial x} + p \frac{\partial F}{\partial z} = 0 $$

$$ \frac{\partial F}{\partial y} + q\frac{\partial F}{\partial z} = 0 $$

How can this happen? Can someone please explain? How can you differentiate an unknown function as $F$.

Here $p$ and $q$ are the partial derivatives of $z$ w.r.t $x$ and $y$ respectively.

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  • $\begingroup$ Look up the multivariable chain rule $\endgroup$
    – Dylan
    Commented Mar 11, 2018 at 16:27
  • $\begingroup$ @Dylan I already know the chain rule but how can we apply it to F and how does a plus sign come here ie why do we have partial derivative of F wrt z aren't we differentiating wrt x? $\endgroup$
    – Matt
    Commented Mar 11, 2018 at 16:39
  • $\begingroup$ The multi-variable chain rule is different from the regular chain rule. If you are asking this question, I don't think you really know it. $\endgroup$
    – Dylan
    Commented Mar 11, 2018 at 16:47
  • $\begingroup$ @Dylan If f(x,y) is a function the df = (df/dx).dx + (df/dy).dy right? $\endgroup$
    – Matt
    Commented Mar 11, 2018 at 16:49

1 Answer 1

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Suppose we have the function $$ F(u,v,z) = 0 $$

where $u = u(x,y)$, $v = v(x,y)$, $z = z(x,y)$ are all functions of $x$ and $y$. The multi-variable chain rule states

$$ \frac{\partial F}{\partial x} = \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0 $$

$$ \frac{\partial F}{\partial y} = \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial y} = 0 $$

Now let $u(x,y) = x$, and $v(x,y) = y$, then

$$ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 0 $$ $$ \frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = 1 $$

Now the equations reduce down to

$$ \frac{\partial F}{\partial x} = \frac{\partial F}{\partial u} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0 $$

$$ \frac{\partial F}{\partial y} = \frac{\partial F}{\partial v} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial y} = 0 $$

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