In an intermediary step in the proof of the condition that $u(x,0) = \phi(x)$, Strauss Page 49, writes:
$u(x,t) = \int_{-\infty}^{\infty} \frac{\partial Q}{\partial x}(x-y,t)\phi(y) dy$.
On the next line they have somehow swapped the derivative to be on $y$ as follows:
$u(x,t) = -\int_{-\infty}^{\infty}\frac{\partial}{\partial y}[Q(x-y,t)]\phi(y)dy$.
Would someone know how this is done?
Thank you.
This seems to be just a fancy-pants application of the chain rule. It might be easier to see if we define: $a = x - y$.
Then $\frac{\partial}{\partial x}[Q(a,t)] = (\frac{\partial}{\partial a}[Q(a,t)])(\frac{d a}{dx}) = \frac{\partial}{\partial a}[Q(a,t)]$
and
$\frac{\partial}{\partial y}[Q(a,t)] = (\frac{\partial}{\partial a}[Q(a,t)])(\frac{d a}{dy}) = -\frac{\partial}{\partial a}[Q(a,t)]$.
Therefore $\frac{\partial}{\partial x}[Q(a,t)] = -\frac{\partial}{\partial y}[Q(a,t)]$.