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$$U_{t} - D U_{xx}= -kU$$ where

BC: $U_{x}(0,t)=0$, $U_{x}(l,t)=0$

where

$0 < x < l$, $t > 0$

IC: $U(x,0)=A + B cos \big(\frac{2πx}{l}\big)$

where

$ 0<x<l$

where $D$ and $k$ are positive constants and $A$ and $B$ are constants.

I have been unable to find an example i can understand and apply to this problem.

Matthew Cassell
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Max
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  • Is $u\equiv u\left(x,t\right)$, or is that a typo? (i.e. was it meant to be an upper case $U$?) – parsiad Dec 17 '14 at 07:08
  • Also, you mention constants $A,B$, but they never appear in your conditions. – parsiad Dec 17 '14 at 07:09
  • I'm sorry i haven;t used latex in a really long time and i tried to copy and edit a post to make it easier and help me remember. It's fixed now – Max Dec 17 '14 at 07:24
  • You should be able to do the separation of variables I.e $$U = X(x)T(t)$$

    Can you put the rest together?

     – Chinny84 Dec 17 '14 at 07:28
  • I know using that i would end up with $T'(t)/T(t) - D*X''(x)/X = -k$

    (Also does Latex work in the comments or just in asking a question? I'm new to this site) Thank you.

     – Max Dec 17 '14 at 07:35
  • Ah never mind it does work. Good i can type cleaner comments now. – Max Dec 17 '14 at 07:36
  • Are you sure about the initial condition? – Chinny84 Dec 17 '14 at 08:37
  • $U(x,0)=A + B cos (2πx/l)$. This is what i meant sorry – Max Dec 17 '14 at 12:02

1 Answers1

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This will get you started. Suppose $$ U\left(x,t\right)=X\left(x\right)T\left(t\right) $$ with $X,T$ smooth. Substituting into the PDE, $$ XT^{\prime}-DX^{\prime\prime}T=-kXT. $$ Further suppose that $U\neq0$ so that dividing through by $XT$ yields $$ \frac{T^{\prime}}{T}+k=D\frac{X^{\prime\prime}}{X}. $$ Since the left-hand side depends only on $t$ and the right-hand side depends only on $x$, this reveals that $$ \frac{T^{\prime}}{T}+k=\text{constant} $$ and $$ \frac{X^{\prime\prime}}{X}=\text{constant}. $$

parsiad
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