# Differential manipulation

Let $v \equiv F(y, z)$. The partial derivatives of $F$ are $$F_1 \equiv \frac{\partial F(y,z)}{\partial y} = \frac{H(v)}{H(y)},$$ $$F_2 \equiv \frac{\partial F(y,z)}{\partial z} = r\frac{H(v)}{H(z)},$$ where $H$ is an arbitrary function. This is taken from the book Probability Theory by E.T. Jaynes.

The author goes on saying that the relation $\mathrm{d}v = \mathrm{d}F(y, z) = F_1\mathrm{d}y + F_2\mathrm{d}z$ takes the form of $$\frac{\mathrm{d}v}{H(v)} = \frac{\mathrm{d}y}{H(y)} + r\frac{\mathrm{d}z}{H(z)}$$

or, on integration $$w[F(y, z)] = w(v) = w(y)w^r(z),$$

where $$w(x) \equiv exp\left\{\int^x \frac{\mathrm{d}x}{H(x)}\right\}.$$

I did undergraduate calculus; however, we didn't do this kind of differential manipulation (expansion of $\mathrm{d}v$) and I find it quite confusing. Is there a way to arrive at the same result using "standard" steps like integration by substitution etc.?

• Does it make it less confusing if you think of $w(x)$ as the logarithm? – Patrick Da Silva Jun 10 '13 at 7:06
• The problem is not so much the $w$ function as the expansion of $\mathrm{d}v$. It passes from being a differential (the principal part of the change in a function) to a variable of integration. – John Manak Jun 10 '13 at 7:15
• I hope I made everything clear in my answer! – Patrick Da Silva Jun 10 '13 at 7:20

You can try thinking of the equation $$dv = F_1 dy + F_2 dz$$ in this way : if $y$ and $z$ were to depend on some variable $t$, then $v$ would depend on this variable $t$ as well because $v = F(y,z)$. Therefore, by the chain rule, $$\frac{dv}{dt} = F_1 \frac{dy}{dt} + r F_2 \frac{dz}{dt} = \frac{H(v)}{H(y)} \frac{dy}{dt} + r \frac{H(v)}{H(z)} \frac{dz}{dt}.$$ By putting $H(v)$ on the left side, you get $$\frac 1{H(v)} \frac {dv}{dt} = \frac 1{H(y)} \frac{dy}{dt} + r \frac 1{H(z)} \frac{dz}{dt}.$$ Integrate this equation with respect to $t$ : $$\int \frac 1{H(v)} \frac {dv}{dt} \, dt = \int \frac 1{H(y)} \frac{dy}{dt} \, dt + r \int \frac 1{H(z)} \frac{dz}{dt} \, dt.$$ These three integrals are all 'the same' (besides the fact that the name of the variable changes), and the change of variables is written down for you! Take $v(t)$ as the new variable : $$\int \frac 1{H(v)} \frac{dv}{dt} \, dt = \int \frac{dv}{H(v)} \overset{def}{=} w(v(t)).$$ In other words, the function $w(v)$ is defined as this integral because we have no general means of computing it. Since we now have $$w(v(t)) = w(y(t)) + r w(z(t)),$$ writing $v(t) = F(y(t), z(t))$, we are basically done. (Note that we assumed that $H$ was not arbitrary, but rather a non-zero function for which $1/H$ has primitive, and this is a very strong assumption compared to 'arbitrary'!)

Hope that helps,

• Thanks, it helps a lot. Just one more thing - you've introduced the new variable $t$ on which both $y$ and $z$ depend. However, initially, the variables $y$ and $z$ were taken to be independent. This is OK? – John Manak Jun 10 '13 at 7:32
• @John Nanak - It's okay because the final result depends on $t$ only through $y(t),z(t)$. You should think of this as a path though $y,z$ space parametrized by $t$. The above shows that along any path from one point to another, you can work out how the function changes, and it's independent of the path you choose – Sharkos Jun 10 '13 at 9:16
• @John Manak : It's exactly as Sharkos describes. The idea is to understand what happens with $v$ by taking an arbitrary 'smooth path' (i.e. a differentiable parametrization $(y(t), z(t))$) so that you can see what happens with $\frac{dv}{dt}$ to better understand $dv$. That's essentially the point. If everything works out for an arbitrary smooth path, then you're good to go. – Patrick Da Silva Jun 10 '13 at 20:42

This is purely a statement of the chain rule in differential form. See my answer on Understanding the differential $dx$ when doing $u$-substitution for a quick attempt to make people feel a bit better about differential forms.

Mathematically, the right way to think of differential forms is as a map from first-order derivatives to functions - for example, $$\mathrm d v \left(\frac\partial{\partial x}\right) \equiv \frac{\partial v}{\partial x}$$

With this interpretation, it's clear that was is meant by this is that $$v'=v_y y'+v_z z'$$ where subscripts denote partial derivatives, and prime denotes differentiation with respect to any coordinate on a line in the space. Writing $t$ for that coordinate, all that is meant is that $v(y(t),z(t))$ has the above $t$ derivative by the chain rule. I'll leave you to Patrick Da Silva's excellent answer to see how to use this to actually derive the result.

What I wanted to convey is just a sense that the notation does actually mean something rigorous, it's just that it's kept secret until differential geometry!