Heat equation/Solution to the 2-D Heat Equation

The solution to the 2-dimensional heat equation (in rectangular coordinates) deals with two spatial and a time dimension, ${\displaystyle u(x,y,t)}$. The heat equation, the variable limits, the Robin boundary conditions, and the initial condition are defined as:

${\displaystyle {\begin{aligned}u_{t}=k\left[u_{xx}+u_{yy}\right]+h(x,y,t)\\(x,y,t)\in [0,L]\times [0,M]\times [0,\infty )\\\alpha _{1}u(0,y,t)-\beta _{1}u_{x}(0,y,t)=b_{1}(y,t)\\\alpha _{2}u(L,y,t)+\beta _{2}u_{x}(L,y,t)=b_{2}(y,t)\\\alpha _{3}u(x,0,t)-\beta _{3}u_{y}(x,0,t)=b_{3}(x,t)\\\alpha _{4}u(x,M,t)+\beta _{4}u_{y}(x,M,t)=b_{4}(x,t)\\u(x,y,0)=f(x,y)\end{aligned}}}$

The solution is just an advanced version of the solution in 1 dimension. If you have questions about the steps shown here, review the 1-D solution.

Just as in the 1-D solution, we partition the solution into a "steady-state" and a "variable" portion:

${\displaystyle u(x,y,t)=\underbrace {s(x,y,t)} _{\text{steady-state}}+\underbrace {v(x,y,t)} _{\text{variable}}}$

We substitute this equation into the initial boundary value problem (IBVP):

${\displaystyle {\begin{cases}s_{t}+v_{t}=k\left[s_{xx}+v_{xx}+s_{yy}+v_{yy}\right]+h(x,y,t)\\\alpha _{1}s(0,y,t)+\alpha _{1}v(0,y,t)-\beta _{1}s_{x}(0,y,t)-\beta _{1}v_{x}(0,y,t)=b_{1}(y,t)\\\alpha _{2}s(0,y,t)+\alpha _{2}v(L,y,t)+\beta _{2}s_{x}(L,y,t)+\beta _{2}v_{x}(L,y,t)=b_{2}(y,t)\\\alpha _{3}s(x,0,t)+\alpha _{3}v(x,0,t)-\beta _{3}s_{y}(x,0,t)-\beta _{3}v_{y}(x,0,t)=b_{3}(x,t)\\\alpha _{4}s(x,M,t)+\alpha _{4}v(x,M,t)+\beta _{4}s_{y}(x,M,t)+\beta _{4}v_{y}(x,M,t)=b_{4}(x,t)\\s(x,y,0)+v(x,y,0)=f(x,y)\end{cases}}}$

We want to set some conditions on s and v:

1. Let ${\displaystyle s}$ satisfy the Laplace equation: ${\displaystyle s_{xx}+s_{yy}=0~.}$
2. Let ${\displaystyle s}$ satisfy the non-homogeneous boundary conditions.
3. Let v satisfy the non-homogeneous equation and homogeneous boundary conditions.

We end up with 2 separate IBVPs:

${\displaystyle {\begin{cases}s_{xx}+s_{yy}=0\\\alpha _{1}s(0,y,t)-\beta _{1}s_{x}(0,y,t)=b_{1}(y,t)\\\alpha _{2}s(0,y,t)+\beta _{2}s_{x}(L,y,t)=b_{2}(y,t)\\\alpha _{3}s(x,0,t)-\beta _{3}s_{y}(x,0,t)=b_{3}(x,t)\\\alpha _{4}s(x,M,t)+\beta _{4}s_{y}(x,M,t)=b_{4}(x,t)\\\end{cases}}}$

${\displaystyle {\begin{cases}v_{t}=k\left[v_{xx}+v_{yy}\right]+h(x,y,t)-s_{t}(x,y,t)\\\alpha _{1}v(0,y,t)-\beta _{1}v_{x}(0,y,t)=0\\\alpha _{2}v(L,y,t)+\beta _{2}v_{x}(L,y,t)=0\\\alpha _{3}v(x,0,t)-\beta _{3}v_{y}(x,0,t)=0\\\alpha _{4}v(x,M,t)+\beta _{4}v_{y}(x,M,t)=0\\v(x,y,0)=f(x,y)-s(x,y,0)\end{cases}}}$

Solving for the steady-state portion is exactly like solving the Laplace equation with 4 non-homogeneous boundary conditions. Using that technique, a solution can be found for all types of boundary conditions.

The associated homogeneous BVP equation is:

${\displaystyle v_{t}=k\left[v_{xx}+v_{yy}\right]}$

The boundary conditions for v are the ones in the IBVP above.

${\displaystyle v(x,y,t)=X(x)Y(y)T(t)}$

${\displaystyle \Rightarrow XYT'=k\lbrack X''YT+XY''T\rbrack }$

${\displaystyle \Rightarrow {\frac {T'}{kT}}={\frac {X''}{X}}+{\frac {Y''}{Y}}=\mu }$

By similar methods, you obtain the following ODEs:

${\displaystyle {\begin{cases}T'-\mu kT=0\\X''-\rho X=0\\Y''-\delta Y=0\\\mu =\rho +\delta \quad {\text{ (coupling equation) }}\end{cases}}}$

${\displaystyle \left.{\begin{aligned}\left[\alpha _{1}X(0)-\beta _{1}X'(0)\right]Y(y)T(t)&=0\\\left[\alpha _{2}X(L)+\beta _{2}X'(L)\right]Y(y)T(t)&=0\\\left[\alpha _{3}Y(0)-\beta _{3}Y'(0)\right]X(x)T(t)&=0\\\left[\alpha _{4}Y(M)+\beta _{4}Y'(M)\right]X(x)T(t)&=0\end{aligned}}\right\}\Rightarrow {\begin{aligned}\alpha _{1}X(0)-\beta _{1}X'(0)=0\\\alpha _{2}X(L)+\beta _{2}X'(L)=0\\\alpha _{3}Y(0)-\beta _{3}Y'(0)=0\\\alpha _{4}Y(M)+\beta _{4}Y'(M)=0\end{aligned}}}$

${\displaystyle \left.{\begin{aligned}X''-\rho X=0\\\alpha _{1}X(0)-\beta _{1}X'(0)=0\\\alpha _{2}X(L)+\beta _{2}X'(L)=0\end{aligned}}\right\}{\begin{aligned}&-\rho =\lambda ^{2}\\&{\text{Eigenvalues }}\lambda _{n}{\text{: solutions to equation }}(\alpha _{1}\alpha _{2}-\beta _{1}\beta _{2}\lambda ^{2})\sin(\lambda L)+(\alpha _{1}\beta _{2}+\alpha _{2}\beta _{1})\lambda \cos(\lambda L)=0\\&X_{n}(x)=\beta _{1}\lambda _{n}\cos(\lambda _{n}x)+\alpha _{1}\sin(\lambda _{n}x),n=0,1,2,\cdots \end{aligned}}}$

${\displaystyle \left.{\begin{aligned}Y''-\delta Y=0\\\alpha _{3}Y(0)-\beta _{3}Y'(0)=0\\\alpha _{4}Y(M)+\beta _{4}Y'(M)=0\end{aligned}}\right\}{\begin{aligned}&-\delta ={\hat {\lambda }}^{2}\ \\&{\text{Eigenvalues }}{\hat {\lambda }}_{m}{\text{: solutions to equation }}(\alpha _{3}\alpha _{4}-\beta _{3}\beta _{4}{\hat {\lambda }}^{2})\sin({\hat {\lambda }}M)+(\alpha _{3}\beta _{4}+\alpha _{4}\beta _{3}){\hat {\lambda }}\cos({\hat {\lambda }}M)=0\\&Y_{m}(x)=\beta _{3}({\hat {\lambda }}_{m})\cos({\hat {\lambda }}_{m}y)+\alpha _{3}\sin({\hat {\lambda }}_{m}y),m=0,1,2,\cdots \end{aligned}}}$

We have obtained eigenfunctions that we can use to solve the nonhomogeneous IBVP.

Just like in the 1-D case, we define v(x,y,t) and q(x,y,t) as infinite sums:

${\displaystyle v(x,y,t):=\sum _{m,n=0}^{\infty }T_{mn}(t)X_{n}(x)Y_{m}(y)}$

${\displaystyle q(x,y,t):=\sum _{m,n=0}^{\infty }Q_{mn}(t)X_{n}(x)Y_{m}(y)~,~Q_{mn}(t)={\frac {\int \limits _{0}^{L}\int \limits _{0}^{M}q(x,y,t)X_{n}(x)Y_{m}(y)dydx}{\int \limits _{0}^{L}X_{n}^{2}(x)dx\int \limits _{0}^{M}Y_{m}^{2}(y)dy}}}$

We then substitute expansion into the PDE:

${\displaystyle {\frac {\partial }{\partial t}}\left[\sum T_{mn}(t)X_{n}(x)Y_{m}(y)\right]=k\left\{{\frac {\partial }{\partial x^{2}}}\left[\sum T_{mn}(t)X_{n}(x)Y_{m}(y)\right]+{\frac {\partial }{\partial y^{2}}}\left[\sum T_{mn}(t)X_{n}(x)Y_{m}(y)\right]\right\}+\sum Q_{mn}(t)X_{n}(x)Y_{m}(y)}$

${\displaystyle \Rightarrow \sum T_{mn}'(t)X_{n}(x)Y_{m}(y)=k\left\{\sum T_{mn}(t)X_{n}''(x)Y_{m}(y)+\sum T_{mn}(t)X_{n}(x)Y_{m}''(y)\right\}+\sum Q_{mn}(t)X_{n}(x)Y_{m}(y)}$

${\displaystyle \Rightarrow \sum T_{mn}'(t)X_{n}(x)Y_{m}(y)=\sum k\left\{T_{mn}(t)[-\lambda _{n}^{2}X_{n}(x)]Y_{m}(y)+\sum T_{mn}(t)X_{n}(x)[-{\hat {\lambda }}_{m}^{2}Y_{m}(y)]\right\}+\sum Q_{mn}(t)X_{n}(x)Y_{m}(y)}$

${\displaystyle \Rightarrow \sum \left[T_{mn}'(t)+k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})\right]X_{n}(x)Y_{m}(y)=\sum Q_{mn}(t)X_{n}(x)Y_{m}(y)}$

This implies that ${\displaystyle X_{n}(x)\otimes Y_{m}(y)~}$ forms an orthogonal basis. This means that we can write the following:

${\displaystyle \Rightarrow T_{mn}'(t)+k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})=Q_{mn}(t)}$

This is a first-order ODE which can be solved using the integration factor:

${\displaystyle \mu (t)=e^{\int k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})dt}=e^{k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})t}}$

Solving for our coefficient we get:

${\displaystyle T_{mn}(t)=e^{-k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})t}\int \limits _{0}^{t}e^{k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})s}Q_{mn}(s)ds+C_{mn}e^{-k(\lambda _{n}^{2}+{\hat {\lambda }}_{m}^{2})t}}$

We apply the initial condition to our equation above:

${\displaystyle {\begin{aligned}v(x,y,0)&=f(x,y)-s(x,y,0)\\&=\sum T_{mn}(0)X_{n}(x)Y_{m}(y)\\&=\sum C_{mn}X_{n}(x)Y_{m}(y)\end{aligned}}}$

The Fourier coefficients can be solved using the inner product definition:

${\displaystyle C_{mn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{M}\left[f(x,y)-s(x,y,0)\right]X_{n}(x)Y_{m}(y)dydx}{\int \limits _{0}^{L}X_{n}^{2}(x)dx\int \limits _{0}^{M}Y_{m}^{2}(y)dy}}}$

We have all the necessary information about the variable portion of the function.

We now have solved for the "steady-state" and "variable" portions, so we just add them together to get the complete solution to the 2-D heat equation.