I am currently struggling with the chain rule for PDE. Let's suppose that we have a function $u(t,x_1,x_2)$ invariant by translation, i.e. :
$$\forall s, u(t,x_1,x_2) = u(t,x_1+s,x_2+s).$$ I define a new variable $y=x_1-x_2$ and I am interested in the study of the "difference function" defined as :
$$d(t,y) := u(t,0,x1-x2).$$
What would be the derivative $\partial_y d$ of $d$ w.r.t. $y$ ? (A combination of $\partial_{x_1}u$ and $\partial_{x_2}u$ I guess...)
Thank you very much !
It's a bit difficult for me to write this in the comments, so I'm writing here: either you have $y$ and have $$\frac{\partial d(t,y)}{\partial y} = \frac{\partial u(t,0,y)}{\partial y} = u'_3(t,0,y)$$
or you have a complex function $u(t,x_1,x_1-x_2)$ and then $$\frac{\partial u(t,x_1,x_1-x_2)}{\partial x_1} = u'_2(t,x_1,x_1-x_2)+u'_3(t,x_1,x_1-x_2)$$ $$\frac{\partial u(t,x_1,x_1-x_2)}{\partial x_2}=-u'_3(t,x_1,x_1-x_2)$$