Results 1 to 5 of 5

Math Help - Heat Equation without Fourier Series

  1. #1
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943

    Heat Equation without Fourier Series

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,346
    Thanks
    29
    Quote Originally Posted by redsoxfan325 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Just one quick question: Is it the fixed boundary conditions that say it's steady-state?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,346
    Thanks
    29
    Quote Originally Posted by redsoxfan325 View Post
    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!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Thanks, I was never so good with physical interpretations of problems.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Heat equation, Fourier Transform
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: August 12th 2010, 07:41 PM
  2. Solving the Heat Equation with the Fourier Integral
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 4th 2010, 10:34 AM
  3. Fourier Transform Of Heat Equation Solution?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 2nd 2010, 11:39 AM
  4. Heat problem with a heat equation
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: November 25th 2009, 09:40 AM
  5. poisson equation by fourier series
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 25th 2009, 05:24 AM

Search Tags


/mathhelpforum @mathhelpforum