Thread: Method of Transformations for Partial Diff. Eqns

1. Method of Transformations for Partial Diff. Eqns

Hey guys,

I have a problem I am trying to solve by method of transformation. This is the problem:

PDE --- U(t) = Uxx

B.C's --- U(0,t) = 0 0<t<infinity
U(1,t) = 1

I.C. --- U(x,0) = x^2

Ok, so I know that I can't use method of tranformation yet, b/c in order to use this method, the B.C's have to be = 0. I have to first do the transformation to make the B.C's = 0. All goes well with the transformation until I get to the I.C. transformation. This is what I get:

I.C. U(x,0) = x^2 - x b/c U(x,t) = f(x) - k1 (1 - x/L) - k2 (x/L)

Thing is I don't know if this is right or not. I don't know if U(x,t) will always equal the above equation or will it always change. If so, how do I figure out the new I.C. in my example above. I am really confused on this part.

Thanks for the help!

2. Originally Posted by spearfish
Hey guys,

I have a problem I am trying to solve by method of transformation. This is the problem:

PDE --- U(t) = Uxx

B.C's --- U(0,t) = 0 0<t<infinity
U(1,t) = 1

I.C. --- U(x,0) = x^2

Ok, so I know that I can't use method of tranformation yet, b/c in order to use this method, the B.C's have to be = 0. I have to first do the transformation to make the B.C's = 0. All goes well with the transformation until I get to the I.C. transformation. This is what I get:

I.C. U(x,0) = x^2 - x b/c U(x,t) = f(x) - k1 (1 - x/L) - k2 (x/L)

Thing is I don't know if this is right or not. I don't know if U(x,t) will always equal the above equation or will it always change. If so, how do I figure out the new I.C. in my example above. I am really confused on this part.

Thanks for the help!
If $\displaystyle u = v + x$, then the PDE becomes

$\displaystyle v_t = v_{xx}$

The BC's are $\displaystyle v(0,t) = 0, v(x,t) = 0$ and the IC $\displaystyle v(x,0) = x^2 - x$. Once you solve this PDE then use your tranformation to find $\displaystyle u$.

3. got it! Thanks Danny!