Differentiating $u(b, t) = u(b + ch, t + h)$

Question

I'm a beginner in PDE, studying the introduction part of Strauss' Partial Differential Equations book.

I'm stuck in a trivial part that says:

$u(b, t) = u(b + ch, t + h)$

Differentiating this with respect to $h$ and putting $h = 0$, we get $0 = cu_x (b, t) + u_t (b, t)$

But I don't understand I how I can differentiate the expression $u(b, t) = u(b + ch, t + h)$ with respect to h. Can someone please help me?

Answer 1

You can use the chain rule:

$\frac{du}{dh} = \frac{\partial u}{\partial x}\frac{dx}{dh} + \frac{\partial u}{\partial y}\frac{dy}{dh}$

Where $x = b + ch$ and $y = t + h$,

This will give $\frac{du}{dh} = cu_x(b + ch, t + h ) + u_y(b + ch, t + h)$