What is the Fourier series for a constant temperature in a box?

[josftx]

Hello i have a problem with a laplace equation in a box, because the problem says that in the six face the tempeture is constant and given by the function f(x,y). and the other faces the temperature its zero.

My teacher says that i must express the f(x,y) as a Fourier series, but i can't understand why i must do that.

Homework Statement

$\dfrac{\partial^{2}T}{\partial x^{2}}+\dfrac{\partial^{2}T}{\partial y^{2}}+\dfrac{\partial^{2}T}{\partial z^{2}}$

$T T(0,x,z)=T(L,x,z)=T(x,0,z)=T(x,L,z)=T(x,y,0)=0$
$T T(x,y,L)=F(x,y)

Homework Equations

The given after this

The Attempt at a Solution

The solution , that i can fint its

$T T(x,y,z)={\displaystyle \sum_{m,n=1}^{\infty}A_{mn}sin(\frac{n\pi x}{L})sin\left(\frac{m\pi y}{L}\right)sinh\left(\pi\sqrt{\frac{n^{2}+m^{2}}{L^{2}}}z\right)}$

and with the last boundary conditions

$F F(x,y)={\displaystyle \sum_{m,n=1}^{\infty}A_{mn}sin(\frac{n\pi x}{L})sin\left(\frac{m\pi y}{L}\right)sinh\left(\pi\sqrt{\frac{n^{2}+m^{2}}{L^{2}}}L\right)}$

 

[fzero]

Hello i have a problem with a laplace equation in a box, because the problem says that in the six face the tempeture is constant and given by the function f(x,y). and the other faces the temperature its zero.

My teacher says that i must express the f(x,y) as a Fourier series, but i can't understand why i must do that.

Homework Statement

$\dfrac{\partial^{2}T}{\partial x^{2}}+\dfrac{\partial^{2}T}{\partial y^{2}}+\dfrac{\partial^{2}T}{\partial z^{2}}$

$T T(0,x,z)=T(L,x,z)=T(x,0,z)=T(x,L,z)=T(x,y,0)=0$
$T T(x,y,L)=F(x,y)

Homework Equations

The given after this

The Attempt at a Solution

The solution , that i can fint its

$T T(x,y,z)={\displaystyle \sum_{m,n=1}^{\infty}A_{mn}sin(\frac{n\pi x}{L})sin\left(\frac{m\pi y}{L}\right)sinh\left(\pi\sqrt{\frac{n^{2}+m^{2}}{L^{2}}}z\right)}$

and with the last boundary conditions

$F F(x,y)={\displaystyle \sum_{m,n=1}^{\infty}A_{mn}sin(\frac{n\pi x}{L})sin\left(\frac{m\pi y}{L}\right)sinh\left(\pi\sqrt{\frac{n^{2}+m^{2}}{L^{2}}}z\right)}$

I think you mean (evaluate at z=L)

$F F(x,y)={\displaystyle \sum_{m,n=1}^{\infty}A_{mn}sin(\frac{n\pi x}{L})sin\left(\frac{m\pi y}{L}\right)sinh\left(\pi\sqrt{\frac{n^{2}+m^{2}}{L^{2}}}L\right)}$

In order to complete the solution, you will need to determine the A_{mn} in terms of F(x,y). If you think about it, you can accomplish this by taking the Fourier transform of both sides of this equation. If F(x,y) is already expressed as a Fourier series, most of the work is already done for you.

 

[josftx]

Yes, sorry i evaluate z in L in my last step. But i need determinate $A A_{mn}$ in terms of $F F(x,y)$. But i didn't finish the problem because i don't understand the reason of my teacher for that solution. and i don't know how complete the solution with that thougth.

can u help me?

 

[fzero]

Well verify that

\int_0^{2\pi} \sin(m u) \sin(n u) du = \delta_{mn}.

Then compute

\int_0^{ L} dx \int_0^{L} dy~ F(x,y) \sin\left(\frac{n'\pi x}{L}\right) \sin\left(\frac{m'\pi y}{L}\right)

to derive an expression for the A_{mn} in terms of the above integrals of F(x,y) (these integrals are the coefficients of the Fourier series for F).

 

[josftx]

and how i can derive an expresión for find the function, using some equality?
i'm really confused-

 

[fzero]

and how i can derive an expresión for find the function, using some equality?
i'm really confused-

Is F(x,y) specified in the problem? If not, the most you can do is derive the expression for the A_{mn} in terms of those integrals of F.

 

[josftx]

It's not specified in the problem, only the problem says that the temperature is maintained constant in the face six and this given by $T T(x,y,L)=f(x,y)$ , and my teacher says that i must find that function =S

 

[fzero]

It's not specified in the problem, only the problem says that the temperature is maintained constant in the face six and this given by $T T(x,y,L)=f(x,y)$ , and my teacher says that i must find that function =S

If the temperature is constant, then f(x,y) = f_0 and you can compute the A_{mn} by solving <br /> \int_0^{ L} dx \int_0^{L} dy~f_0 \sin\left(\frac{n&#039;\pi x}{L}\right) \sin\left(\frac{m&#039;\pi y}{L}\right) = \int_0^{ L} dx \int_0^{L} dy~T(x,y,L) \sin\left(\frac{n&#039;\pi x}{L}\right) \sin\left(\frac{m&#039;\pi y}{L}\right).<br />