# Heat Equation without Fourier Series

• Feb 22nd 2010, 12:11 PM
redsoxfan325
Heat Equation without Fourier Series
The problem:

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Two rods of length $\displaystyle L_1$ and $\displaystyle L_2$, which have heat diffusion constants $\displaystyle k_1$ and $\displaystyle k_2$, respectively, are welded together.

The left end of the left rod (length $\displaystyle L_1$) is maintained at temperature $\displaystyle 0$, the right end of the right rod is kept at temperature $\displaystyle T$. The temperature $\displaystyle u(x,t)$ and heat flux $\displaystyle ku_x$ are continuous across the weld.

I need to find the equilibrium temperature distribution across the welded rod of length $\displaystyle L_1+L_2$.

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I started setting up the data:

$\displaystyle u_t = k_1u_{xx}$
$\displaystyle v_t = k_2v_{xx}$
$\displaystyle u(0,t)=0$
$\displaystyle v(L_2,t)=T$
$\displaystyle u(L_1,t)=v(0,t)$
$\displaystyle k_1u_x(L_1,t)=k_2v_x(0,t)$

Since nothing was given for $\displaystyle u(x,0)$ and $\displaystyle v(x,0)$, I assume they're

$\displaystyle u(x,0)=f(x)$
$\displaystyle v(x,0)=g(x)$

If we call the solution $\displaystyle w(x,t)$, it seems that
$\displaystyle w(x,t)=\bigg\{\begin{array}{lr}u(x,t)&:x\leq L_1\\v(x-L_1,t)&:x>L_1\end{array}$
$\displaystyle w(0,t)=0$
$\displaystyle w(L_1+L_2,t)=T$
$\displaystyle w(x,0)=\bigg\{\begin{array}{lr}f(x)&:x\leq L_1\\g(x-L_1)&:x>L_1\end{array}$
with $\displaystyle f(L_1)=g(0)$ and $\displaystyle k_1f'(L_1)=k_2g'(0)$

We found in class that the general solution for the heat equation on a rod of infinite length with $\displaystyle u(x,0)=f(x)$ is

$\displaystyle \frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^{\infty}\exp\left(-\frac{(x-y)^2}{4kt}\right)f(y)\,dy$

However, this doesn't seem to help because we have finite length and boundary conditions. And here I get lost (before I really even begin!) because there is just a ridiculous amount of information and I'm not really sure what to do with it.

Help is appreciated!
• Feb 22nd 2010, 01:20 PM
Jester
Quote:

Originally Posted by redsoxfan325
The problem:

------------

Two rods of length $\displaystyle L_1$ and $\displaystyle L_2$, which have heat diffusion constants $\displaystyle k_1$ and $\displaystyle k_2$, respectively, are welded together.

The left end of the left rod (length $\displaystyle L_1$) is maintained at temperature $\displaystyle 0$, the right end of the right rod is kept at temperature $\displaystyle T$. The temperature $\displaystyle u(x,t)$ and heat flux $\displaystyle ku_x$ are continuous across the weld.

I need to find the equilibrium temperature distribution across the welded rod of length $\displaystyle L_1+L_2$.

------------

I started setting up the data:

$\displaystyle u_t = k_1u_{xx}$
$\displaystyle v_t = k_2v_{xx}$
$\displaystyle u(0,t)=0$
$\displaystyle v(L_2,t)=T$
$\displaystyle u(L_1,t)=v(0,t)$
$\displaystyle k_1u_x(L_1,t)=k_2v_x(0,t)$

Since nothing was given for $\displaystyle u(x,0)$ and $\displaystyle v(x,0)$, I assume they're

$\displaystyle u(x,0)=f(x)$
$\displaystyle v(x,0)=g(x)$

If we call the solution $\displaystyle w(x,t)$, it seems that
$\displaystyle w(x,t)=\bigg\{\begin{array}{lr}u(x,t)&:x\leq L_1\\v(x-L_1,t)&:x>L_1\end{array}$
$\displaystyle w(0,t)=0$
$\displaystyle w(L_1+L_2,t)=T$
$\displaystyle w(x,0)=\bigg\{\begin{array}{lr}f(x)&:x\leq L_1\\g(x-L_1)&:x>L_1\end{array}$
with $\displaystyle f(L_1)=g(0)$ and $\displaystyle k_1f'(L_1)=k_2g'(0)$

We found in class that the general solution for the heat equation on a rod of infinite length with $\displaystyle u(x,0)=f(x)$ is

$\displaystyle \frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^{\infty}\exp\left(-\frac{(x-y)^2}{4kt}\right)f(y)\,dy$

However, this doesn't seem to help because we have finite length and boundary conditions. And here I get lost (before I really even begin!) because there is just a ridiculous amount of information and I'm not really sure what to do with it.

Help is appreciated!

First let me zero the rod at the left end so the two pieces meet at $\displaystyle x = L_1$ and the total rod is $\displaystyle L_1+L_2$.

Your dealing with steady state so $\displaystyle u_{xx} = 0$. Thus, $\displaystyle u = c_1 x + c_2$ where $\displaystyle c_1$ and $\displaystyle c_2$ are constants. Let the solution be $\displaystyle u_1 = a_1x+a_2$ in the first rod and $\displaystyle u_2 = b_1x + b_2$ in the second. Thus, we have the four conditions:

1) $\displaystyle u_1(0,t) = 0$ so $\displaystyle a_2 = 0$

2) $\displaystyle u_2(L_1+L_2,t) = T$ so $\displaystyle b_1(L_1+L_2)+b_2 = T$

3) the temp is continuous at the joint

$\displaystyle u_1(L_1,t) = u_2(L_1,t)$ so $\displaystyle a_1L_1 + a_2 = b_1 L_1 + b_2$

4) So are the fluxes so

$\displaystyle ku_x(L_1,t) = k_2 u_x(L_1,t)$ so $\displaystyle k_1 a_1 = k_2 b_1$

You now have 4 equations for the unknowns $\displaystyle a_1, b_1, a_2 \; \text{and}\; b_2$.
• Feb 22nd 2010, 01:25 PM
redsoxfan325
Just one quick question: Is it the fixed boundary conditions that say it's steady-state?
• Feb 22nd 2010, 01:28 PM
Jester
Quote:

Originally Posted by redsoxfan325
Just one quick question: Is it the fixed boundary conditions that say it's steady-state?

No, you said you wanted the equilibrium temperature distribution. When in equilibrium, the temperature isn't changing - that's steady state!
• Feb 22nd 2010, 01:41 PM
redsoxfan325
Thanks, I was never so good with physical interpretations of problems. (Doh)