time-dependent example of heat equation

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

A={(x,y)2   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   t=0  the boundaries of A are held in the temperature 0.  Determine the temperature function


on A (where t is the time).

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

2uuxx′′+uyy′′=1c2ut. (1)

One have to find for (1) a solution function u which satisfies the initial conditionMathworldPlanetmath

u(x,y, 0)=f(x,y)inA (2)

and the boundary condition

u(x,y,t)=0onboundaryofAfort>0. (3)

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

u(x,y,t):=X(x)Y(y)T(t), (4)

whence the boundary condition reads

X(0)=Y(0)=X(a)=Y(b)=0. (5)

Substituting (4) in (1) and dividing this equation by XYT give the form

X′′X+Y′′Y=1c2TT. (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:

X′′X=-k12,Y′′Y=-k22,1c2TT=-k2, (7)

where  k2=k12+k22.  We soon explain why these constants are negative. Because the equations (7) may be written


the general solutions of these ordinary differential equations are

{X=C1cosk1x+D1sink1x,Y=C2cosk2y+D2sink2y,T=Ce-k2c2t. (8)

Now we remark that if the right side of the third equation (7) were  +k2, then we had  T=Cek2c2t  which is impossible, since such a T and along with this also the temperature  u=XYT  would ascend infinitely when  t.  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  C1=C2=0,  and then the two last conditions (5) require that


If we had  D1=0  or  D2=0,  then X or Y would vanish identically, which cannot occur.  Thus we have


whence only the eigenvalues

{k1=mπa(m=1, 2, 3,)k2=nπb(n=1, 2, 3,)

are possible for the obtained X and Y.  Considering the equation  k2=k12+k22  we may denote

qmn:=k2c2=[(mπa)2+(nπb)2]c2 (9)

for all   m,n+.

Altogether we have infinitely many solutions


of the equation (1), where the coefficients cmn are, for the present, arbitrary constants.  These solutions fulfil the boundary condition (3).  The sum of the solutions, i.e. the double series

u(x,y,t):=m=1n=1cmne-qmntsinmπxasinnπyb, (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


But this equation presents the Fourier double sine series of  f(x,y)  in the rectangle A, and therefore we have the expression

cmn:=4ab0a0bf(x,y)sinmπxasinnπybdxdy (11)

for the coefficients.

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


  • 1 K. Väisälä: Matematiikka IV.  Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).
Title time-dependent example of heat equation
Canonical name TimedependentExampleOfHeatEquation
Date of creation 2013-03-22 16:32:47
Last modified on 2013-03-22 16:32:47
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 10
Author pahio (2872)
Entry type Example
Classification msc 35Q99
Synonym solving a time-dependent heat equation
Related topic ExampleOfSolvingTheHeatEquation
Related topic EigenvalueProblem