Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
[parent] time-dependent example of heat equation (Example)

The initial temperature (at  $t = 0$) of a thin homogeneous plate

$\displaystyle A = \{(x,\,y)\in\mathbb{R}^2\vdots\,\,\,0 < x < a,\,\,0 < y < b\}$
in the $xy$-plane is given by the function  $f = f(x,\,y)$.  The faces of the plate are supposed completely isolating.  After the moment  $t = 0$  the boundaries of $A$ are held in the temperature 0.  Determine the temperature function

$\displaystyle u = u(x,\,y,\,t)$
on $A$ (where $t$ is the time).

Since it's a question of a two-dimensional heat flow, the heat equation gets the form

$\displaystyle \nabla^2u \equiv u''_{xx}+u''_{yy} = \frac{1}{c^2}u'_t.$ (1)
One have to find for (1) a solution function $u$ which satisfies the initial condition
$\displaystyle u(x,\,y,\,0) = f(x,\,y)\,\,\,\mathrm{in}\,\,\,A$ (2)
and the boundary condition
$\displaystyle u(x,\,y,\,t) = 0\,\,\, \mathrm{on\,\,boundary\,\,of}\,\,A\,\,\,\mathrm{for}\,\,t > 0.$ (3)

For finding a simple solution of the differential equation (1) we try the form

$\displaystyle u(x,\,y,\,t) := X(x)Y(y)T(t),$ (4)
whence the boundary condition reads
$\displaystyle X(0) = Y(0) = X(a) = Y(b) = 0.$ (5)
Substituting (4) in (1) and dividing this equation by $XYT$ give the form
$\displaystyle \frac{X''}{X}+\frac{Y''}{Y} = \frac{1}{c^2}\!\cdot\!\frac{T'}{T}.$ (6)
It's easily understood that such a condition requires that the both addends of the left side and the right side ought to be constants:
$\displaystyle \frac{X''}{X} = -k_1^2,\,\, \frac{Y''}{Y} = -k_2^2,\,\, \frac{1}{c^2}\!\cdot\!\frac{T'}{T} = -k^2,$ (7)
where  $k^2 = k_1^2+k_2^2$.  We soon explain why these constants are negative. Because the equations (7) may be written

$\displaystyle X'' = -k_1^2X,\,\,\, Y'' = -k_2^2Y,\,\,\, T' = -k^2c^2T,$
the general solutions of these ordinary differential equations are
\begin{align*}\begin{cases} X = C_1\cos{k_1x}+D_1\sin{k_1x},\ Y = C_2\cos{k_2y}+D_2\sin{k_2y},\ T = Ce^{-k^2c^2t}. \end{cases}\end{align*} (8)
Now we remark that if the right side of the third equation (7) were  $+k^2$, then we had  $T = Ce^{k^2c^2t}$  which is impossible, since such a $T$ and along with this also the temperature  $u = XYT$  would ascend infinitely when  $t\to\infty$.  And since, by symmetry, the right sides the two first equations (7) must have the same sign, also they must by (6) be negative.

The two first boundary conditions (5) imply by (8) that  $C_1 = C_2 = 0$,  and then the two last conditions (5) require that

$\displaystyle D_1\sin{k_1a} = 0,\,\,\, D_2\sin{k_2b} = 0.$
If we had  $D_1 = 0$  or  $D_2 = 0$,  then $X$ or $Y$ would vanish identically, which cannot occur.  Thus we have

$\displaystyle \sin{k_1a} = 0\,\,\, \mathrm{and}\,\,\, \sin{k_2b} = 0,$
whence only the eigenvalues
\begin{align*}\begin{cases} k_1 = \frac{m\pi}{a}\,\,\,(m = 1,\,2,\,3,\,\ldots)\ k_2 = \frac{n\pi}{b}\,\,\,(n = 1,\,2,\,3,\,\ldots) \end{cases}\end{align*}    
are possible for the obtained $X$ and $Y$.  Considering the equation  $k^2 = k_1^2+k_2^2$  we may denote
$\displaystyle q_{mn} := k^2c^2 = \left[\left(\frac{m\pi}{a}\right)^2\!+\!\left(\frac{n\pi}{b}\right)^2\right]c^2$ (9)
for all   $m,\,n \in \mathbb{Z}_+.$

Altogether we have infinitely many solutions

$\displaystyle u_{mn} = XYT = CD_1D_2e^{-q_{mn}t}\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} = c_{mn}e^{-q_{mn}t}\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}$
of the equation (1), where the coefficients $c_{mn}$ are, for the present, arbitrary constants.  These solutions fulfil the boundary condition (3).  The sum of the solutions, i.e. the double series
$\displaystyle u(x,\,y,\,t) := \sum_{m=1}^\infty\sum_{n=1}^\infty c_{mn}e^{-q_{mn}t}\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b},$ (10)
provided it converges, is also a solution of the linear differential equation (1) and fulfils the boundary condition.  In order to fulfil also the initial condition (2), one must have

$\displaystyle \sum_{m=1}^\infty\sum_{n=1}^\infty c_{mn}e^{-q_{mn}t}\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} = f(x,\,y).$
But this equation presents the Fourier double sine series expansion of  $f(x,\,y)$  in the rectangle $A$, and therefore we have the expression
$\displaystyle c_{mn} := \frac{4}{ab} \int_0^a\!\int_0^b f(x,\,y)\,\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\,dx\,dy$ (11)
for the coefficients.

The result of calculating the solution of our problem is the temperature function (10) with the formulae (9) and (11).

Bibliography

1
K. Väisälä: Matematiikka IV.  Handout Nr. 141.    Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).



"time-dependent example of heat equation" is owned by bloftin. [ full author list (2) ]

View style:


This object's parent.

Cross-references: ordinary differential equations, differential equation, heat equation, heat, two-dimensional, boundaries, function, temperature

This is version 2 of time-dependent example of heat equation, born on 2007-08-10, modified 2007-08-10.
Object id is 257, canonical name is TimeDependentExampleOfHeatEquation.
Accessed 1415 times total.

Classification:
Physics Classification: 44. (Heat transfer)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "