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?
integration differential-equations fourier-analysis laplace-transform transformation
asked Nov 17 at 22:05
Pascal
15410
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up vote
0
down vote
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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?
integration differential-equations fourier-analysis laplace-transform transformation
asked Nov 17 at 22:05
Pascal
15410
add a comment |
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?
integration differential-equations fourier-analysis laplace-transform transformation
asked Nov 17 at 22:05
Pascal
15410
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
integration differential-equations fourier-analysis laplace-transform transformation
asked Nov 17 at 22:05
Pascal
15410
asked Nov 17 at 22:05
Pascal
15410
asked Nov 17 at 22:05
Pascal
15410
asked Nov 17 at 22:05
Pascal
15410
asked Nov 17 at 22:05
Pascal
15410
15410
add a comment |
add a comment |
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