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| Science Forum Index » Physics Forum » Explain heat equation/diffusion to an electrical engineer: G |
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| Author |
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| Thomas Richter |
 Posted: Mon May 09, 2005 6:12 am |
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Guest
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Hi,
[quote:5ff7df4743]Assume that I understand the numerical solution of Poisson equation.
L u = rhs
where rows of L contain a discretization of the Laplacian,
and "known" points get subtracted from both sides and appear
on the right hand side (rhs).
[/quote:5ff7df4743]
What do no mean by "known points"? Do you mean that you want to
solve the Poisson equation
\Laplace u = \rho
with boundary conditions, i.e. you prescribe the values of u
on some initial set, where u and \rho are maps from R^n to R?
[quote:5ff7df4743]Suppose a 1D problem, with a "point heat source" that holds the
temperature
at u[0] to a value "c", and the temperature is held to some other
value at the edges of the 1D domain u, e.g. u[100] = u[-100] = 1.
[/quote:5ff7df4743]
But this is then a different equation, namely the head
equation
u : R^n x R -> R, (x,t) |--> u(x,t)
\Laplace u = u'
with initial conditions. Or do you look to a "steady state"
equation of the above, i.e. a solution u(x,t) such that u
does not depend on t?
[quote:5ff7df4743]Please tell me what is wrong with the following statements
(something must be, because they are contradictory):
1) The solution u can be obtained by solving the Poisson equation.
[/quote:5ff7df4743]
Depends on your equation. The heat equation is by definition
time dependent, and so you need to solve the time dependent
equation. The Poisson equation is the steady-state equation
looking for time-independent solutions.
[quote:5ff7df4743]If the Laplacian is discretized as the kernel 1,-2,1, the point
source will appear in three equations, like
u[-2] -2u[1] + c = 0
^^^^^[/quote:5ff7df4743]
Shouldn't this be -2u[-1]?
[quote:5ff7df4743]u[-1] -2c + u[1] = 0
c -2u[1] + u[2] = 0
then subtract the known "c" terms so they appear on the rhs:
u[-2] -2u[1] = -c
u[-1] + u[1] = 2c
-2u[1] + u[2] = -c
the rest of the L matrix (except for the first and last rows
corresponding to the boudnary conditions) will have
shifted versions of the 1,-2,1 kernel.
[/quote:5ff7df4743]
[quote:5ff7df4743]2) with a point source of heat, the solution over time is a
Gaussian whose variance increases over time.
[/quote:5ff7df4743]
This is the solution of the heat equation with an initial
condition given as a "delta peak" at the origin. Different
equation, different solution.
[quote:5ff7df4743]3) As a linear differential equation,
[/quote:5ff7df4743]
Both, heat equation and poisson equation are linear differential
equations.
[quote:5ff7df4743]the Lu=rhs equation
can be solved with the Greens function approach,
[/quote:5ff7df4743]
Yes, correct.
[quote:5ff7df4743]in which the solution is obtained by convolving thh Green's function
with the rhs:
u = G * rhs
[/quote:5ff7df4743]
Right, though the Green's function depends on the boundary
conditions, and on the equation.
[quote:5ff7df4743]4) The Greens function for diffusion (in an infinite volume) is
proportional to 1/radius.
[/quote:5ff7df4743]
This is the Green's function for the three-dimensional
Poisson equation and not for the one-dimensional heat-equation.
[quote:5ff7df4743]5) (minor) Tthe resulting u in this case is the temperature
distribution. (Or is it "heat", or something else?)
[/quote:5ff7df4743]
Heat is a form of energy. It is related to temperature by
the heat capacity of the matter that is "heated". If the
matter has constant heat capacity (i.e. a solid body made
of a single material), then the two will be proportial to
each other. Example: You need more energy to heat up one
1kg of aluminum by ten degrees than to heat up 1kg of
lead by the same amount.
[quote:5ff7df4743]To me points #2 and #3,#4 seem contradictory, in that convolving
a 1/r function with the result of a point source will _not_ generate
a Gaussian.
[/quote:5ff7df4743]
Truely not, but we're talking about different equations here, see
above.
[quote:5ff7df4743]As mentioned above, #2 is a dynamic solution and 3,4
reflect steady state in this case (I believe) - is that the difference?
[/quote:5ff7df4743]
Yes, that, and the dimension of base space (R^3 vs. R) and the
difference of the initial resp. boundary conditions.
So long,
Thomas |
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