Derivative of a convolution with respect to another variable

Question

I have two 2-variable functions $u=u(x,t), g=g(x,t)$

I want to simplify the expression:

$ \frac{\partial}{\partial t} ( u*g)$

Where $*$ is the convolution of $u$ and $g$ with respect to $x$. My understanding of Wikipedia https://en.wikipedia.org/wiki/Convolution#Properties under the title Differentiation yields that

$ \frac{\partial}{\partial t} ( u*g)= \frac{\partial u}{\partial t}*g $

But when I try to prove that I get:

$\frac{\partial}{\partial t} ( u*g) =\frac{\partial}{\partial t} \int_{-\infty}^\infty u(x-\alpha,t)g( \alpha,t) d\alpha = \int_{-\infty}^\infty \frac{\partial}{\partial t} u(x-\alpha,t)g(\alpha,t) d\alpha =\int_{-\infty}^\infty u'_t(x-\alpha,t)g( \alpha,t)+u(x-\alpha,t)g'_t( \alpha,t) d\alpha = u'_t*g+u*g'_t \ne u'_t*g $

Where am I'm going wrong?

Answer 1

The formula you got from Wikipedia holds if are differentiating w.r.t. $x$, not w.r.t. $t$. For differentiation w.r.t $t$ what you have obtained is correct.

Answer 2

As the convolution is with respect to $x$ and the differentiation is with respect to $t$, you can't use the formulas from the Wikipedia article. Your derivation looks correct to me.