Help please ! Its a heat conduction problem(the heat equation problem) use separation of variables to solve please provide a detailed explanation for how to solve this problem pls :)

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ijlal · Answer

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ijlal · Answer

Please do a step by step guide on how to solve this problem will be really helpful (:
@mathmale  @ganeshie8  @Preetha  @Mehek14

ijlal · Answer

can some 1 help??

irishboy123 · Answer

@ijlal hi Chris Tisdell can help!! this vid is so on-point, plus it's got some little adjustments to keep it interesting. https://www.youtube.com/watch?v=Jfn6A1jZF5g Watch it all but it takes off at 10:50 and the bit in 11:30 - 14:30 is the guts of the separation of variables issue. of course, it's a big hand-wavey fudge.... the thing he doesn't say enough for me is that $u(x,t)$ in the heat equation is temperature. so $X(x)$ is distance along the 1-D rod and $T(t)$ is the time function. i have watched a lot of his work and he is really good. [ he is a maths guy teaching scientists.] if you want someone to do this is parallel with you, i am quite happy to tag along, and post my own workings, working through the IV's. but i am not sufficiently fluent anymore to just go online and start working the problem. my sense is that you won't need the Fourier series, just that one sine function you have been given. plus, i think once you break through the justification for arguing that $u(x,t) = X(x) T(t)$ and that $\dfrac{X''}{X} = \dfrac{T'}{\alpha^2 T} = const$, it's really just process. i hope this helps.

ijlal · Answer

thank you @IrishBoy123  watching his video right now (: and ya you are right i wont need the fourier series just the sine function thanks alot (:

irishboy123 · Answer

cool!!!!


if you wanna compare results, do ask.

ijlal · Answer

$$u(x,t)=3\sin(5x/2)\exp(-(5\alpha/2)^2t)$$

ijlal · Answer

i got this as my answer @IrishBoy123  well i got alot help from my teacher too whats your result  :D?

irishboy123 · Answer

Yes, think i agree.   i've written it out as i don't do many of these.

if we write the pde, once separated, as 

$\dfrac{T'}{\alpha^2 T} = \dfrac{X'}{X} =  \gamma$

then we have 

$X = A \sin \sqrt{- \gamma}  x + B \cos \sqrt{- \gamma} x$ as they seem to be steering us down the complex solution to the DE

we also have $T = C e^{\alpha^2 \gamma t} $

and when combined, $u (x,t) =  e^{\alpha^2 \gamma t} (A \sin \sqrt{- \gamma}  x + B \cos \sqrt{- \gamma}  x)$


first BC

$u (o,t) =  e^{\alpha^2 \gamma t} ( B \cos \sqrt{- \gamma}  x) = 0 \implies B = 0$

leaving 

$u (x,t) =  A e^{\alpha^2 \gamma t}  \sin \sqrt{- \gamma}  x $

next BC:

$\dfrac{\partial u}{\partial x} = \sqrt{- \gamma} A e^{\alpha^2 \gamma t}  \cos \sqrt{- \gamma}  x = 0$ for $x = \pi$

so that works for $\sqrt{- \gamma} x = {\pi \over 2}, { 3\pi \over 2} \dots$ so $\sqrt{- \gamma}  = {1 \over 2}, {3 \over 2} \dots$


next BC

$u (x,0) =  A e^{0}  \sin \sqrt{- \gamma}  x  = 3 \sin \dfrac{5x}{2} \implies A = 3, \sqrt{- \gamma}  = \dfrac{5}{2} \implies \gamma = - (\frac{5}{2})^2$

so we do seem to have $u (x,t) = 3 e^{- (\frac{5 \alpha}{2})^2  t}  \sin \dfrac{5}{2}x $

strictly speaking, i think you would need to look at $\gamma = 0, \gamma > 0$ also to make a complete answer.  in fact, i think that's the real purpose of the 2nd BC.  

  PS Tisdell's vids on pde's are all extremely good!