A temperature field T(x, t) is determined by the following governing equation:

$$\frac 1\alpha\frac {dT}{dt} = \frac {d^2T}{dx^2}$$ (Eq 1)

T(x,t) can be expressed as a form of expansion of T(x,t) = (Eq 2)

$$\sum_{n=0}^\infty cos(\lambda_nx)\bar T(\lambda_n,t)$$

where $\lambda_n=n*pi(n=0,1,2...)$ are known. using the orthogonality property of $cos(\lambda_nx)$ in domain $x \epsilon [a,b]$, one obtains

$$\bar T(\lambda_n,t)=\frac {\int_a^bT(x,t)cos(\lambda_nx)dx} {\int_a^bcos^2(\lambda x)dx}$$

One needs to solve for $\bar T(\lambda_n,t)$ to finally determine the temperature field T(x,t).

Write the expression for

$$\frac {d \bar T(\lambda_n,t)}{dt}$$

from Eq 2, and then replace $\frac {dT}{dt}$ inside this equation with $\alpha \frac {d^2x}{dt^2}$.

You then rearrange using integration by parts twice and you should end up with an ordinary ODE for $\bar T(\lambda_n,t)$.

I just typed the whole problem because I am really not sure where to start. It looks like I need to differentiate equation 2 but I don't know how because of the integrals in there. If anyone could even just tell me which topics I should look for in which textbooks, or a general guideline to how to go about solving, I'd really appreciate it.

  • $\begingroup$ What's the question? $\endgroup$ – Jean-Claude Arbaut Sep 1 '14 at 18:53
  • $\begingroup$ how do you rearrange the expression to get an ODE for $\bar T (\lambda_n,t)$. But if I had an idea how to differentiate the equation with the integrals that would help. So how do you find $\frac {d \bar T(\lambda_n,t)} {dt}$ $\endgroup$ – lori Sep 1 '14 at 20:48

I am not really sure what you require here, but here goes. $$ \bar{T}\left(\lambda_n,t\right) = \dfrac{\int_a^b T(x,t)\cos\left(\lambda_n x\right)dx}{\int_a^b\cos^2\left(\lambda_n x\right)dx} $$

$$ \dfrac{\partial}{\partial t}\bar{T}\left(\lambda_n,t\right) = \dfrac{\partial}{\partial t}\dfrac{\int_a^b T(x,t)\cos\left(\lambda_n x\right)dx}{\int_a^b\cos^2\left(\lambda_n x\right)dx} $$ since $x$ and $t$ are independent we can swap around the integral and differentiation to yield

$$ \dfrac{\partial}{\partial t}\dfrac{\int_a^b T(x,t)\cos\left(\lambda_n x\right)dx}{\int_a^b\cos^2\left(\lambda_n x\right)dx} = \dfrac{\int_a^b \left(\dfrac{\partial }{\partial t}T(x,t)\right)\cos\left(\lambda_n x\right)dx}{\int_a^b\cos^2\left(\lambda_n x\right)dx} $$

now you can replace $$ \dfrac{\partial T}{\partial t} = \alpha\dfrac{\partial^2 T}{\partial x^2} $$

and then integrate by parts of something..

I do not understand where $\alpha\dfrac{d^2 x}{dt^2}$ comes from? Also think about using partial derivative notation :). $$ $$

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  • $\begingroup$ OK so what I have so far after the substitution is... $$ \frac{\int_a^b \alpha \frac{\delta ^2 T}{\delta x^2} cos(\lambda _nx)dx}{\int_a^b cos^2(\lambda _n x) dx} $$ And I simplified by dividing out the cosines...but I wasn't sure if that was right. I got: $$ \frac{\int_a^b \alpha \frac{\delta ^2 T}{\delta x^2}dx}{\int_a^b cos(\lambda _n x) dx} $$ By what rule do you integrate this equation? Using integration by parts which functions do you use for u and v? $\endgroup$ – lori Sep 5 '14 at 20:03

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