Integral Transforms and Partial DEs: Heat Equation of a rod











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Consider a rod of length $l$ whose lateral surface is insulated. Let its initial temperature be zero, then maintain one of its end faces as at zero temperature and the other at $u_0$.

A. Determine the rod as a function of time.

B. Complete (A) by mode of separation of variables and compare answers.



--

First, I will define some of the equations and the given information.
$frac{d^2u}{dt^2}=frac{1}{alpha}frac{du}{dt}$ where $0 leq x$, $0 <t$, $0leq x leq l$.
$mathscr{L}{frac{d^2u}{dx^2}}=int_{0}^{infty}e^{-st}frac{d^2u}{dx^2}dx$ $=frac{d^2}{du^2}int_{0}^{-infty}e^{-st}u(x,t)dt=frac{d^2 hat{U}(x,s)}{dx^2}$

I believe that this solves A. But how would I approach the solving via seperation of variables?










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    Consider a rod of length $l$ whose lateral surface is insulated. Let its initial temperature be zero, then maintain one of its end faces as at zero temperature and the other at $u_0$.

    A. Determine the rod as a function of time.

    B. Complete (A) by mode of separation of variables and compare answers.



    --

    First, I will define some of the equations and the given information.
    $frac{d^2u}{dt^2}=frac{1}{alpha}frac{du}{dt}$ where $0 leq x$, $0 <t$, $0leq x leq l$.
    $mathscr{L}{frac{d^2u}{dx^2}}=int_{0}^{infty}e^{-st}frac{d^2u}{dx^2}dx$ $=frac{d^2}{du^2}int_{0}^{-infty}e^{-st}u(x,t)dt=frac{d^2 hat{U}(x,s)}{dx^2}$

    I believe that this solves A. But how would I approach the solving via seperation of variables?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Consider a rod of length $l$ whose lateral surface is insulated. Let its initial temperature be zero, then maintain one of its end faces as at zero temperature and the other at $u_0$.

      A. Determine the rod as a function of time.

      B. Complete (A) by mode of separation of variables and compare answers.



      --

      First, I will define some of the equations and the given information.
      $frac{d^2u}{dt^2}=frac{1}{alpha}frac{du}{dt}$ where $0 leq x$, $0 <t$, $0leq x leq l$.
      $mathscr{L}{frac{d^2u}{dx^2}}=int_{0}^{infty}e^{-st}frac{d^2u}{dx^2}dx$ $=frac{d^2}{du^2}int_{0}^{-infty}e^{-st}u(x,t)dt=frac{d^2 hat{U}(x,s)}{dx^2}$

      I believe that this solves A. But how would I approach the solving via seperation of variables?










      share|cite|improve this question













      Consider a rod of length $l$ whose lateral surface is insulated. Let its initial temperature be zero, then maintain one of its end faces as at zero temperature and the other at $u_0$.

      A. Determine the rod as a function of time.

      B. Complete (A) by mode of separation of variables and compare answers.



      --

      First, I will define some of the equations and the given information.
      $frac{d^2u}{dt^2}=frac{1}{alpha}frac{du}{dt}$ where $0 leq x$, $0 <t$, $0leq x leq l$.
      $mathscr{L}{frac{d^2u}{dx^2}}=int_{0}^{infty}e^{-st}frac{d^2u}{dx^2}dx$ $=frac{d^2}{du^2}int_{0}^{-infty}e^{-st}u(x,t)dt=frac{d^2 hat{U}(x,s)}{dx^2}$

      I believe that this solves A. But how would I approach the solving via seperation of variables?







      integration differential-equations fourier-analysis laplace-transform transformation






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      asked Nov 17 at 22:05









      Pascal

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