Heat equation/Solution to the 2-D Heat Equation

Definition

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

$\begin{array}{c}{u}_t=k[u_{xx}+u_{yy}]+h(x,y,t)\\ (x,y,t)\in [0,L]\times [0,M]\times [0,\mathrm{\infty })\\ {\alpha }_{1}u(0,y,t)-{\beta }_1{u}_x(0,y,t)=b_1(y,t)\\ {\alpha }_{2}u(L,y,t)+{\beta }_2{u}_x(L,y,t)=b_2(y,t)\\ {\alpha }_{3}u(x,0,t)-{\beta }_3{u}_y(x,0,t)=b_3(x,t)\\ {\alpha }_{4}u(x,M,t)+{\beta }_4{u}_y(x,M,t)=b_4(x,t)\\ u(x,y,0)=f(x,y)\end{array}$

Solution

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.

Step 1: Partition Solution

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

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

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

$\{\begin{array}{c}{s}_t+v_t=k[s_{xx}+v_{xx}+s_{yy}+v_{yy}]+h(x,y,t)\\ {\alpha }_{1}s(0,y,t)+{\alpha }_{1}v(0,y,t)-{\beta }_1{s}_x(0,y,t)-{\beta }_1{v}_x(0,y,t)=b_1(y,t)\\ {\alpha }_{2}s(0,y,t)+{\alpha }_{2}v(L,y,t)+{\beta }_2{s}_x(L,y,t)+{\beta }_2{v}_x(L,y,t)=b_2(y,t)\\ {\alpha }_{3}s(x,0,t)+{\alpha }_{3}v(x,0,t)-{\beta }_3{s}_y(x,0,t)-{\beta }_3{v}_y(x,0,t)=b_3(x,t)\\ {\alpha }_{4}s(x,M,t)+{\alpha }_{4}v(x,M,t)+{\beta }_4{s}_y(x,M,t)+{\beta }_4{v}_y(x,M,t)=b_4(x,t)\\ s(x,y,0)+v(x,y,0)=f(x,y)\end{array}$

We want to set some conditions on s and v:

1. Let $s$ satisfy the Laplace equation: $s_{xx}+s_{yy}=0.$
2. Let $s$ 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:

$\{\begin{array}{c}{s}_{xx}+s_{yy}=0\\ {\alpha }_{1}s(0,y,t)-{\beta }_1{s}_x(0,y,t)=b_1(y,t)\\ {\alpha }_{2}s(0,y,t)+{\beta }_2{s}_x(L,y,t)=b_2(y,t)\\ {\alpha }_{3}s(x,0,t)-{\beta }_3{s}_y(x,0,t)=b_3(x,t)\\ {\alpha }_{4}s(x,M,t)+{\beta }_4{s}_y(x,M,t)=b_4(x,t)\end{array}$

$\{\begin{array}{c}{v}_t=k[v_{xx}+v_{yy}]+h(x,y,t)-s_t(x,y,t)\\ {\alpha }_{1}v(0,y,t)-{\beta }_1{v}_x(0,y,t)=0\\ {\alpha }_{2}v(L,y,t)+{\beta }_2{v}_x(L,y,t)=0\\ {\alpha }_{3}v(x,0,t)-{\beta }_3{v}_y(x,0,t)=0\\ {\alpha }_{4}v(x,M,t)+{\beta }_4{v}_y(x,M,t)=0\\ v(x,y,0)=f(x,y)-s(x,y,0)\end{array}$

Step 2: Solve Steady-State Portion

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.

Step 3: Solve Variable Portion

Step 3.1: Solve Associated Homogeneous BVP

The associated homogeneous BVP equation is:

$v_t=k[v_{xx}+v_{yy}]$

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

Separate Variables

$v(x,y,t)=X(x)Y(y)T(t)$

$\Rightarrow XY{T}^{\prime }=k[X^{″}YT+X{Y}^{″}T]$

$\Rightarrow \frac{T^{\prime }}{kT}=\frac{X^{″}}{X}+\frac{Y^{″}}{Y}=\mu$

By similar methods, you obtain the following ODEs:

$\{\begin{array}{c}{T}^{\prime }-\mu kT=0\\ X^{″}-\rho X=0\\ Y^{″}-\delta Y=0\\ \mu =\rho +\delta \text{ (coupling equation) }\end{array}$

Translate Boundary Conditions

$\begin{array}{cc}[{\alpha }_{1}X(0)-{\beta }_1{X}^{\prime }(0)]Y(y)T(t)& =0\\ [{\alpha }_{2}X(L)+{\beta }_2{X}^{\prime }(L)]Y(y)T(t)& =0\\ [{\alpha }_{3}Y(0)-{\beta }_3{Y}^{\prime }(0)]X(x)T(t)& =0\\ [{\alpha }_{4}Y(M)+{\beta }_4{Y}^{\prime }(M)]X(x)T(t)& =0\end{array}\}\Rightarrow \begin{array}{c}{\alpha }_{1}X(0)-{\beta }_1{X}^{\prime }(0)=0\\ {\alpha }_{2}X(L)+{\beta }_2{X}^{\prime }(L)=0\\ {\alpha }_{3}Y(0)-{\beta }_3{Y}^{\prime }(0)=0\\ {\alpha }_{4}Y(M)+{\beta }_4{Y}^{\prime }(M)=0\end{array}$

Solve SLPs

$\begin{array}{c}{X}^{″}-\rho X=0\\ {\alpha }_{1}X(0)-{\beta }_1{X}^{\prime }(0)=0\\ {\alpha }_{2}X(L)+{\beta }_2{X}^{\prime }(L)=0\end{array}\}\begin{array}{cc}& -\rho ={\lambda }^2\\ & \text{Eigenvalues }{\lambda }_n\text{: solutions to equation }({\alpha }_1{\alpha }_2-{\beta }_1{\beta }_2{\lambda }^2)\sin (\lambda L)+({\alpha }_1{\beta }_2+{\alpha }_2{\beta }_1)\lambda \cos (\lambda L)=0\\ & X_n(x)={\beta }_1{\lambda }_n\cos ({\lambda }_{n}x)+{\alpha }_1\sin ({\lambda }_{n}x),n=0,1,2,\cdots \end{array}$

$\begin{array}{c}{Y}^{″}-\delta Y=0\\ {\alpha }_{3}Y(0)-{\beta }_3{Y}^{\prime }(0)=0\\ {\alpha }_{4}Y(M)+{\beta }_4{Y}^{\prime }(M)=0\end{array}\}\begin{array}{cc}& -\delta ={\hat{\lambda }}^2\\ & \text{Eigenvalues }{\hat{\lambda }}_m\text{: solutions to equation }({\alpha }_3{\alpha }_4-{\beta }_3{\beta }_4{\hat{\lambda }}^2)\sin (\hat{\lambda }M)+({\alpha }_3{\beta }_4+{\alpha }_4{\beta }_3)\hat{\lambda }\cos (\hat{\lambda }M)=0\\ & Y_m(x)={\beta }_3({\hat{\lambda }}_m)\cos ({\hat{\lambda }}_{m}y)+{\alpha }_3\sin ({\hat{\lambda }}_{m}y),m=0,1,2,\cdots \end{array}$

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

Step 3.2: Solve Non-homogeneous IBVP

Setup Problem

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

$v(x,y,t):=\sum _{m,n=0}^{\mathrm{\infty }}{T}_{mn}(t)X_n(x)Y_m(y)$

$q(x,y,t):=\sum _{m,n=0}^{\mathrm{\infty }}{Q}_{mn}(t)X_n(x)Y_m(y),Q_{mn}(t)=\frac{\underset{0}{\overset{L}{\int }}\underset{0}{\overset{M}{\int }}q(x,y,t)X_n(x)Y_m(y)dydx}{\underset{0}{\overset{L}{\int }}{X}_n^2(x)dx\underset{0}{\overset{M}{\int }}{Y}_m^2(y)dy}$

Determine Coefficients

We then substitute expansion into the PDE:

$\frac{\partial }{\partial t}[\sum T_{mn}(t)X_n(x)Y_m(y)]=k\{\frac{\partial }{\partial x^2}[\sum T_{mn}(t)X_n(x)Y_m(y)]+\frac{\partial }{\partial y^2}[\sum T_{mn}(t)X_n(x)Y_m(y)]\}+\sum Q_{mn}(t)X_n(x)Y_m(y)$

$\Rightarrow \sum T_{m^{\prime }n}(t)X_n(x)Y_m(y)=k\{\sum T_{mn}(t){X_n}^{″}(x)Y_m(y)+\sum T_{mn}(t)X_n(x){Y_m}^{″}(y)\}+\sum Q_{mn}(t)X_n(x)Y_m(y)$

$\Rightarrow \sum T_{m^{\prime }n}(t)X_n(x)Y_m(y)=\sum k\{T_{mn}(t)[-{\lambda }_n^2{X}_n(x)]Y_m(y)+\sum T_{mn}(t)X_n(x)[-{\hat{\lambda }}_m^2{Y}_m(y)]\}+\sum Q_{mn}(t)X_n(x)Y_m(y)$

$\Rightarrow \sum [T_{m^{\prime }n}(t)+k({\lambda }_n^2+{\hat{\lambda }}_m^2)]X_n(x)Y_m(y)=\sum Q_{mn}(t)X_n(x)Y_m(y)$

This implies that $X_n(x)\otimes Y_m(y)$ forms an orthogonal basis. This means that we can write the following:

$\Rightarrow T_{m^{\prime }n}(t)+k({\lambda }_n^2+{\hat{\lambda }}_m^2)=Q_{mn}(t)$

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

$\mu (t)=e^{\int k({\lambda }_n^2+{\hat{\lambda }}_m^2)dt}=e^{k({\lambda }_n^2+{\hat{\lambda }}_m^2)t}$

Solving for our coefficient we get:

$T_{mn}(t)=e^{-k({\lambda }_n^2+{\hat{\lambda }}_m^2)t}\underset{0}{\overset{t}{\int }}{e}^{k({\lambda }_n^2+{\hat{\lambda }}_m^2)s}{Q}_{mn}(s)ds+C_{mn}{e}^{-k({\lambda }_n^2+{\hat{\lambda }}_m^2)t}$

Satisfy Initial Condition

We apply the initial condition to our equation above:

$\begin{array}{cc}v(x,y,0)& =f(x,y)-s(x,y,0)\\ & =\sum T_{mn}(0)X_n(x)Y_m(y)\\ & =\sum C_{mn}{X}_n(x)Y_m(y)\end{array}$

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

$C_{mn}=\frac{\underset{0}{\overset{L}{\int }}\underset{0}{\overset{M}{\int }}[f(x,y)-s(x,y,0)]X_n(x)Y_m(y)dydx}{\underset{0}{\overset{L}{\int }}{X}_n^2(x)dx\underset{0}{\overset{M}{\int }}{Y}_m^2(y)dy}$

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

Step 4: Combine Solutions

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.