Assuming steady state, no heat generation, constant heat conduction, the following partial differential equation is obtained governing two dimensional conduction.

$$ \frac{\partial^2T}{\partial x^2}+\frac{\partial^2T}{\partial y^2}=0 $$

A fully general solution is difficult to obtain, but if we have three homogenous boundary conditions on a rectangular domain then we can solve by separation of variables. These boundary conditions may specify the temperature as zero (Dirichlet Boundary Condition) or prescribe insulation, specifying the gradient of temperature as zero (Neumann Boundary Condition). It may be helpful to apply a coordinate transformation in order to establish these boundary conditions, as below.

$$ \theta=T-T_b~~\Rarr~~\frac{\partial^2\theta}{\partial x^2}+\frac{\partial^2\theta}{\partial y^2}=0 $$

Solving by separation of variables involves assuming the solution to be a product of functions of $x,y$. It is critical that the negative sign be placed in the equation for the direction which has two homogenous boundary conditions, which from here on out is assumed to be the $x$ direction.

$$ \theta=X(x)~Y(y)\Rarr\theta_{xx}=YX_{xx},~\theta_{yy}=XY_{yy}\RarrYX_{xx}+XY_{yy}=0~~\Rarr $$

$$ X_{xx}-\lambda^2X=0 $$

$$ X=C_1\cosh(\lambda x)+C_2\sinh(\lambda x) $$

$$ Y_{yy}+\lambda^2Y=0 $$

$$ Y=C_3\cos(\lambda y)+C_4\sin(\lambda y) $$

At this point, application of the boundary conditions will vary the solution, depending on if these are zero-value or zero-gradient boundary conditions. In the duel homogenous-boundary condition direction, one constants will go to zero, and eigenvalues, $\lambda$, may be obtained. In the other direction, if the spatial coordinates are properly transformed, one of the constants will go to zero.

The final boundary condition may then be applied, by equating our solution evaluated are the non-homogenous boundary, $w_y$, to the prescribed boundary condition, $f_b(x)$, as indicated by the below, where $C_5=C_xC_y$, the two remaining constants, and $X_s,Y_s$ equal $C_xX,C_yY$ respectively.

$$ f_b(x)=(C_5X_sY_s)|_{y=w_y} $$

Values for $C_5$ may commonly be obtained by using the below integration technique. By multiplying by a particularly crafted factor and integrating over the $x$ domain, an expression is rendered which may be rearranged to obtain a value for $C_5$. A common example of this factor is $\sin(\frac{m\pi}{w_y}x)$, which results in an integral that simply goes to $\frac{w_x}2$ when $n=m=\N$, and zero otherwise.

$$ \int_{0}^{w_x}f_b(x)\sin(\frac{m\pi}{w_x}x)dx=(C_5Y_s)|{y=w_y}\int{0}^{w_x}X_s\sin(\frac{m\pi}{w_x}x)dx $$

$$ \int_{0}^{w_x}\sin(\frac{n\pi}{w_x}x)\sin(\frac{m\pi}{w_x}x)dx=\frac{w_x}2 $$

$$ C_5(n)=\frac{\int_{0}^{w_x}f_b(x)\sin(\frac{m\pi}{w_x}x)dx}{Y_s|{y=w_y}\int{0}^{w_x}X_s\sin(\frac{m\pi}{w_x}x)dx} $$

Since $n$ may vary, many solutions of $C_5$ are obtained by the above process. Our final analytically obtained temperature field thus takes the form of a summation over possible values of $n$.

$$ T(x,y)=\Sigma_{n=1}^\infin C_5(n)X_sY_s $$