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I'm finding this particular step for solving the Transport Equation confusing:

'We exploit this insight by fixing any point $(x,t) \in \mathbb{R}^n \times (0,\infty)$ and defining $$ z(s):=u(x+sb,t+s). $$ We then calculate $$ dz/ds=\nabla u(x+sb,t+s).b+\frac{\partial u(x+sb,t+s)}{\partial t}=0. $$ I don't understand why this is the derivative of $z$ with respect to $s$; is it an application of the chain rule using partial derivatives? If so could anyone explain what form of the chain rule it is and why? Throughout all of the above $x \in \mathbb{R}^n,~t,s \in \mathbb{R}$, and $u: \mathbb{R}^n \times (0,\infty) \to \mathbb{R}$.

Thanks! Dan

Dan Burrows
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1 Answers1

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Yes, it's the chain rule.

The trick is knowing what "$\nabla$" and "$\partial u / \partial t$" mean; the first is the gradient of $u$, i.e., a vector of partial derivatives, but only with respect to the spatial variables, i.e., the entries of $x$. The second is the partial with respect the the second argument.

Letting

$$ g(s) = (x+sb, t +s) = (g_1(s), g_2(s)) $$ the chain rule basically says: $$ \frac{d (u \circ g)}{ds}(s)= \frac{du}{dx}(g(s)) \cdot \frac{dg_1}{ds} (s) + \frac{du}{dt}(g(s)) \frac{dg_2(s)}{ds} $$ Filling in the details, the "$\frac{du}{dx}(g(s))$" is your gradient. $\frac{dg_1}{ds} (s)$ is just $b$ (the tangent vector to the space-curve defined by $g_1$). and $\frac{dg_2}{ds} (s)$ is the derivative of $t+s$ wrt $s$, which is 1.

John Hughes
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