# Homework Help: Transformation/change of variables in this differential equation?

1. Nov 1, 2009

### kingwinner

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!

2. Nov 1, 2009

### kingwinner

Re: Transformation/change of variables in this differential equation!?

Can somebody please help?? I know it's suuposed to be simple, but I can't figure it out...

In short, my problem is the following:
Suppose u(x,0) = x+5.
Let w = x-1.
Does it imply that u(w,0) = w+5?

3. Nov 2, 2009

### lanedance

Re: Transformation/change of variables in this differential equation!?

I haven't had a chance to read the top bit but say you have u(x,t) such that
u(x,0) = x+5

and you want to subsitute
w = x-1
then
x = w+1

so I would substitute in
u(w=x+1,0) = (w+1)+5 = w+6

4. Nov 2, 2009

### lanedance

Re: Transformation/change of variables in this differential equation!?

will try & have a better look tonight

5. Nov 2, 2009

### kingwinner

Re: Transformation/change of variables in this differential equation!?

Maybe it's u(w,0) = x+6. But can you please tell me where my mistake is? Also, HOW did you get u(w,0) = x+6? You said w=x+1, but I think we're supposed to have w=x-1?

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??

u(w,0)
= u(x-1,0) = (x-1)+5
=> u(w,0) = w+5 ?

6. Nov 3, 2009

### lanedance

Re: Transformation/change of variables in this differential equation!?

ok so how about we try and recude the variable confusion, by introducing another function f(w,t) defined by
$$f(w,t) = u(x(w),t)$$

so the variable change is
$$w(x) = x-1$$
and re-arranging in case we need it later
$$x(w) = 1+w$$
also differentiating gives
$$\frac{dx}{dw} = 1$$

then, chain rule differentiation gives
$$f_{ww} = u_{xx}$$
$$f_{t} = u_{t}$$

so the differential equation becomes
$$f_t- kf_{ww} = x(w) -2t = 1+w-2t$$

the boundary & initial conditions becomes
$$f(0,t) = f(6,t) = 0$$
and
$$f(w,0) = u(x(w),0) = x(w) + 5 = w+6$$

Last edited: Nov 3, 2009