The solution to the 2-dimensional heat equation (in rectangular coordinates) deals with two spatial and a time dimension,
. The heat equation, the variable limits, the Robin boundary conditions, and the initial condition are defined as:
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19fc00e49a597f5be3745e0b5979e737a5ed6c4b)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08f85e22315bb728e6739091b3ad15a21c911088)
We want to set some conditions on s and v:
- Let
satisfy the Laplace equation: ![{\displaystyle s_{xx}+s_{yy}=0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37b7e229c16eb23cfaa770b200cf0590db6bdd44)
- Let
satisfy the non-homogeneous boundary conditions.
- 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32263affe8b45ed139af92f38cba23d4d458de59)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/830da48b7b54ba447f0e42a5fdbba2c0afac9491)
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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c270f9b64a336a850431eb8bd57a3b39c957f5e)
The boundary conditions for v are the ones in the IBVP above.
![{\displaystyle v(x,y,t)=X(x)Y(y)T(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5916d161161bd918a822b42b7b11f97545e74cfb)
![{\displaystyle \Rightarrow XYT'=k\lbrack X''YT+XY''T\rbrack }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4109212cc0b67b4973d609251f94ff8eed04489c)
![{\displaystyle \Rightarrow {\frac {T'}{kT}}={\frac {X''}{X}}+{\frac {Y''}{Y}}=\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eac1fd97abe9e0bbf41679b8f2b0d89ee56dc831)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5de110395a451716542634efdc64fce2a56b71ae)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/069725cc9f2f5ed7bc59781131af1292faf6e621)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4acd0d814c0f881995883c6294bb4d0aea379c4a)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bee0bdc6ee099bd565c3a5eeeb3cd872ebbccb2)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/676df8faa708d46d4e0ede6b2bf794d455e2836b)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2431837933432bb2bfc1ee80b44d5d25dbbb1fa)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c17fd5a9184d237d70e94568bd88af403bbfcc1f)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57234e7bc17a51eb025be75a42ee46e3f331906f)
This implies that
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faa29b4860bd141f0d76c5e4d485e4c047a16bbe)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32d44b909d6e265496e798653aec90b0e0625e8f)
Solving for our coefficient we get:
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a834faa6987476ec444b95135d73ed99cae4a87c)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e131a1f0f9844a97a12d55fa9cf63e0e004e454)
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.