Transformation/change of variables in this differential equation!? Maybe my background is just weak...I was thinking about this for almost 1.5 hours already, but I still end up totally confused. Perhaps this is because I was never able to understand the ideas of a function and change of variables completely...perhaps I have a serious conceptual flaw. 1. The problem statement, all variables and given/known data Consider the following partial differential equation with initial values(IV) and boundary conditions(BC): ut - k uxx = x + 2t, 1<x<7, t>0 BC: u(1,t) = u(7,t) = 0 IV: u(x,0) = x+5 Our goal is to transform the above to the interval [0,6]. Let w = x-1. Transform the whole problem to the interval w E [0,6]. (write in terms of w) 2. Relevant equations/concepts No knowledge of partial differential equation is needed. Anyone with a solid background in mutlivariable calculus and change of variables should be able to answer this question. (I think) 3. The attempt at a solution ux=uw dw/dx = (uw) (1) = uw uxx = ...(apply chain rule again) = uww [On the left side, think of u as u(x,t). On the right side, think of u as u(w,t)] BC: x=1 <=> w=0 x=7 <=> w=6 So the boundary conditions get transformed to u(0,t)=u(6,t)=0 [here think of u as u(w,t)] IV: We know u(x,0) = x+5 => u(w,0) = w+5 [I believe the logic in this step cannot be wrong, consider e.g. f(4z)=cos(4z), now how do we find f(4z-y)? Of course, f(4z-y)=cos(4z-y). How do we find f(z)? Surely, f(z)=cos(z). Right??] So my final answer is: [here think of u as u(w,t)] ut - k uww = w+1+2t, 0<w<6, t>0 BC: u(0,t) = u(6,t) = 0 IV: u(w,0) = w+5 However, I really have some bad feeling that the result u(w,0) = w+5 is wrong, but I don't know where the mistake is. I tried to calculate it in a different way and the answer is the same. u(w,0) = u(x-1,0) = (x-1)+5 => u(w,0) = w+5 Can someone please kindly explain why and where my mistake is? What is the correct answer? Any help is greatly appreciated!