# Math Help - Heat Equation without Fourier Series

1. ## Heat Equation without Fourier Series

The problem:

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

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

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

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

I started setting up the data:

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

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

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

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

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

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

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!

2. Originally Posted by redsoxfan325
The problem:

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

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

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

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

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

I started setting up the data:

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

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

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

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

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

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

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 $x = L_1$ and the total rod is $L_1+L_2$.

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

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

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

3) the temp is continuous at the joint

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

4) So are the fluxes so

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

You now have 4 equations for the unknowns $a_1, b_1, a_2 \; \text{and}\; b_2$.

3. Just one quick question: Is it the fixed boundary conditions that say it's steady-state?

4. 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!

5. Thanks, I was never so good with physical interpretations of problems.