Transformation/change of variables in differential equation

kingwinner · New Member Offline Joined: 27 Oct 2009 Posts: 3

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.
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Consider the following partial differential equation with initial values(IV) and boundary conditions(BC):
$u_t - k u_{xx} = x + 2t, 1<x<7, t>0$
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)
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$u_x = u_w dw/dx = (u_w) (1) = u_w$
apply chain rule again => $u_{xx} = u_{ww}$
[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)]
$u_t - k u_{ww} = w+1+2t, 0<w<6, t>0$
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!

JRav

You should put your math in $\LaTeX$ . Most of what you wrote is already in proper syntax-just put dollar signs around it.

$u_t - ku_xx = x + 2t$ , $1<x<7$ , $t>0$
BC: $u(1,t) = u(7,t) = 0$
IV: $u(x,0) = x+5$

(I didn't change any of your code to get that).

**Posted:** Sun Nov 01, 2009 4:56 am

kingwinner · New Member Offline Joined: 27 Oct 2009 Posts: 3

OK, I've done that! Now can somebody help me, please? I know it's supposed to be simple (just change of variables right?), but I can't figure it out... Sad

**Posted:** Sun Nov 01, 2009 1:40 pm