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My problem is given as

Arbitrary temperatures at ends . If the ends $x=0$ and $x=L$ of the bar in the text are kept at constant temperatures $U_1$ and $U_2$ respectively, what is the temperature $u_1(x)$ in the bar after a long time (theoretically, as $t \to \infty$)? First guess, then calculate.

My guess is that the temperature after a very long time is given as the meadian temperature. Etc

$$u(x,t) \approx (U_2-U_1)/L, \quad \text{as} \quad t\to \infty$$

Now one does assume that the temperature reaches a limit, which is not unlikely, then the solution will satisfy the laplacian $\nabla^2u=0$.

Which leads to the heat equation in one variable

$$ \frac{\mathrm{d}u}{\mathrm{d}t} = c^2 \frac{\mathrm{d}^2u}{\mathrm{d}x^2} $$

The standard way of assuming the solution is on the form $u(x,t)=X(x)T(t)$ fails for me.Begin with assuimg that the differential equation is equal to some arbitary constant $\lambda$ that is not dependant on $x$ nor $t$. Then I end up with the set of equations

$$\begin{array}{lcr} T' & = & \lambda c^2 T\\ \ddot{X} & = & \lambda X \end{array}$$ If we assume for a minute that $\lambda=0$, we end up with $$X(x) = Ax + B, \qquad T(t)=C$$ Which does not satifsy the initial values. So, what do I do to solve this bugger?

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Please clarify the conditions. The conditions described between in the title and in the question are contradictive. –  doraemonpaul Oct 1 '12 at 23:01
    
How are they wrong in the title? My english is not that good, they are correct in the cursive. Please help me clearify the title =) –  N3buchadnezzar Oct 1 '12 at 23:36

2 Answers 2

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=c^2X''(x)T(t)$

$\dfrac{T'(t)}{c^2T(t)}=\dfrac{X''(x)}{X(x)}=-\dfrac{\pi^2s^2}{L^2}$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{\pi^2c^2s^2}{L^2}\\X''(x)+\dfrac{\pi^2s^2}{L^2}X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}\\X(x)=\begin{cases}c_1(s)\sin\dfrac{\pi xs}{L}+c_2(s)\cos\dfrac{\pi xs}{L}&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\sum\limits_{s=0}^\infty C_3(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}\sin\dfrac{\pi xs}{L}+\sum\limits_{s=0}^\infty C_4(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}\cos\dfrac{\pi xs}{L}$

$u(0,t)=U_1$ :

$C_2+\sum\limits_{s=0}^\infty C_4(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}=U_1$

$\sum\limits_{s=0}^\infty C_4(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}=U_1-C_2$

$C_4(s)=\begin{cases}U_1-C_2&\text{when}~s=0\\0&\text{when}~s\neq0\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\sum\limits_{s=0}^\infty C_3(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}\sin\dfrac{\pi xs}{L}+U_1-C_2=C_1x+U_1+\sum\limits_{s=1}^\infty C_3(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}\sin\dfrac{\pi xs}{L}$

$u(L,t)=U_2$ :

$C_1L+U_1=U_2$

$C_1=\dfrac{U_2-U_1}{L}$

$\therefore u(x,t)=\dfrac{(U_2-U_1)x}{L}+U_1+\sum\limits_{s=1}^\infty C_3(s)e^{-\frac{\pi^2c^2ts^2}{L^2}}\sin\dfrac{\pi xs}{L}$

Hence $u(x,\infty)=\dfrac{(U_2-U_1)x}{L}+U_1$

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Your guess is not exactly correct. In the limit $t\to\infty$ assume that the temperature profile does not depend on $t$, which means that $$ u_t=0, $$ which means that your equation is $$ c^2 u_{xx}=0, $$ which has the solution $$ u(x,\infty)=Ax+B, $$ or, after using the boundary conditions $$ u(x,\infty)=U_1+\frac{(U_2-U_1)}{L}x. $$

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