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Heat Conduction Equation


A partial differential diffusion equation of the form

 (partialU)/(partialt)=kappadel ^2U.
(1)

Physically, the equation commonly arises in situations where kappa is the thermal diffusivity and U the temperature.

The one-dimensional heat conduction equation is

 (partialU)/(partialt)=kappa(partial^2U)/(partialx^2).
(2)

This can be solved by separation of variables using

 U(x,t)=X(x)T(t).
(3)

Then

 X(dT)/(dt)=kappaT(d^2X)/(dx^2).
(4)

Dividing both sides by kappaXT gives

 1/(kappaT)(dT)/(dt)=1/X(d^2X)/(dx^2)=-1/(lambda^2),
(5)

where each side must be equal to a constant. Anticipating the exponential solution in T, we have picked a negative separation constant so that the solution remains finite at all times and lambda has units of length. The T solution is

 T(t)=Ae^(-kappat/lambda^2),
(6)

and the X solution is

 X(x)=Bcos(x/lambda)+Csin(x/lambda).
(7)

The general solution is then

U(x,t)=T(t)X(x)
(8)
=Ae^(-kappat/lambda^2)[Bcos(x/lambda)+Csin(x/lambda)]
(9)
=e^(-kappat/lambda^2)[Dcos(x/lambda)+Esin(x/lambda)].
(10)

If we are given the boundary conditions

 U(0,t)=0
(11)

and

 U(L,t)=0,
(12)

then applying (11) to (10) gives

 Dcos(x/lambda)=0=>D=0,
(13)

and applying (12) to (10) gives

 Esin(L/lambda)=0=>L/lambda=npi=>lambda=L/(npi),
(14)

so (10) becomes

 U_n(x,t)=E_ne^(-kappa(npi/L)^2t)sin((npix)/L).
(15)

Since the general solution can have any n,

 U(x,t)=sum_(n=1)^inftyc_nsin((npix)/L)e^(-kappa(npi/L)^2t).
(16)

Now, if we are given an initial condition U(x,0), we have

 U(x,0)=sum_(n=1)^inftyc_nsin((npix)/L).
(17)

Multiplying both sides by sin(mpix/L) and integrating from 0 to L gives

 int_0^Lsin((mpix)/L)U(x,0)dx=int_0^Lsum_(n=1)^inftyc_nsin((mpix)/L)sin((npix)/L)dx.
(18)

Using the orthogonality of sin(nx) and sin(mx),

sum_(n=1)^(infty)c_nint_0^Lsin((npix)/L)sin((mpix)/L)dx=sum_(n=1)^(infty)1/2Ldelta_(mn)c_n
(19)
=L/2c_m
(20)
=int_0^Lsin((mpix)/L)U(x,0)dx,
(21)

so

 c_m=2/Lint_0^Lsin((mpix)/L)U(x,0)dx.
(22)

If the boundary conditions are replaced by the requirement that the derivative of the temperature be zero at the edges, then (◇) and (◇) are replaced by

 (partialU)/(partialx)|_((0,t))=0
(23)
 (partialU)/(partialx)|_((L,t))=0.
(24)

Following the same procedure as before, a similar answer is found, but with sine replaced by cosine:

 U(x,t)=1/2c_0+sum_(n=1)^inftyc_ncos((npix)/L)e^(-kappa(npi/L)^2t),
(25)

where

 c_n=2/Lint_0^Lcos((mpix)/L)U(x,0)dx.
(26)

See also

Heat Conduction Equation--Disk

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Cite this as:

Weisstein, Eric W. "Heat Conduction Equation." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/HeatConductionEquation.html

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