# Analytical solution of 2D Transient Heat conduction with Neumann condition

I want to know the analytical solution of a transient heat equation in a 2D square with inhomogeneous Neumann Boundary.

The transient heat conduction equation in a 2D square cavity :

$$\frac{dT}{dt}=\nabla^2T$$

and the boundary are:

$$\cases{T(0,y)=T_1\\T(L,y)=T_0\\\frac{\partial T(x,0)}{\partial y}=a\\\frac{\partial T(x,L)}{\partial y}=a}$$

The initial Temperature inside is $T_0$

If there is no analytical solution, then what is the solution of steady heat conduction (with $\nabla^2T=0$)?

Really appreciate anyone who can help!

• $T$ depends on 3 variables: $T(t,x,y)$. That may be the cause of the confusion. Jun 21 '18 at 23:13
• +1 for cute fox. No seriously the problem is interesting too. Jun 23 '18 at 9:09

This is very involved problem. To solve, you'll need to break the solution into successively smaller pieces. First, separate the steady-state and transient solutions, then split up the boundary conditions in order to use separation of variables.

To start off, I'm going to label

$$T(x,y,t) = u(x,y,t) + v(x,y)$$

where $v(x,y)$ is the time-independent, steady-state solution, and $u(x,y,t)$ is the decaying, transient solution.

The steady-state solution should satisfy $\nabla^2v = 0$ and all the boundary conditions as listed. Since all boundaries are inhomogeneous, we need so split it up further

$$v(x,y) = v_1(x,y) + v_2(x,y)$$

such that

\begin{matrix} \begin{cases} \nabla^2 v_1 = 0 \\ \\ v_1(0,y) = T_1,\ v_1(L,y) = T_0 \\ \\ \dfrac{\partial}{\partial y}v_1(x,0) = \dfrac{\partial}{\partial y}v_1(x,L) = 0 \end{cases} &&& \begin{cases} \nabla^2v_2 = 0 \\ \\ v_2(0,y) = v_2(L,y) = 0 \\ \\ \dfrac{\partial}{\partial y}v_2(x,0) = \dfrac{\partial}{\partial y}v_2(x,L) = a \end{cases} \end{matrix}

The homogeneous boundaries allow us to use separation of variables to solve each individual problem

• The first one is easier since you can intuitively guess that it's constant in $y$ and linear in $x$. This turns out to be

$$v_1(x,y) = T_0\frac{x}{L} + T_1\frac{L-x}{L} = \frac{(T_0-T_1)x}{L} + T_1$$

• For the second problem, the homogeneous boundary condition on $x$ returns a series solution of the form

$$v_2(x,y) = \sum_{n=1}^\infty \sin\left(\frac{n\pi x}{L}\right)\left[A_n\cosh\left(\frac{n\pi y}{L}\right) + B_n \cosh\left(\frac{n\pi(L-y)}{L}\right)\right]$$

Then, applying the remaining boundary conditions on $y$ will give

$$A_n = -B_n = \begin{cases} \dfrac{4aL}{n^2\pi^2\sinh(n\pi)}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases}$$

For a more thorough explanation of why I'm using hyperbolic functions instead of exponentials, check out my answer for this similar problem. Short answer: It makes the math easier

The remaining transient solution is homogeneous on all boundaries and has initial conditions that cancel out the steady-state:

\begin{cases} \dfrac{\partial u}{\partial t} = \nabla^2 u \\ \\ u(0,y,t) = u(L,y,t) = \dfrac{\partial}{\partial y}u(x,0,t) = \dfrac{\partial}{\partial y}u(x,L,t) = 0 \\ \\ u(x,y,0) = T_0 - v(x,y) \end{cases}

Applying separation of variables once more and matching the homogeneous boundaries, we obtain

$$u(x,y,t) = \sum_{n,m} c_{n,m} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{L}\right)\exp \left[-(n^2+m^2)\frac{\pi^2}{L^2}t\right]$$

The initial condition for this last piece will have you solve a double Fourier series in $x$, $y$. You can use linearity to simplify calculations a bit, i.e.

\begin{align} u_1(x,y,0) &= T_0 - T_1 - \frac{T_0-T_1}{L}x = \sum_{n,m} c_{1(n,m)} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{L}\right) \\ u_2(x,y,0) &= -\sum_{n=2k+1} \frac{4aL}{n^2\pi^2\cosh(n\pi)} \sin\left(\frac{n\pi x}{L}\right)\left[\cosh\left(\frac{n\pi y}{L}\right) - \cosh\left(\frac{n\pi(L-y)}{L}\right)\right] \\ &\qquad = \sum_{n,m} c_{2(n,m)} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi y}{L}\right) \end{align}

The first boundary function is constant in $y$ and the second is already a partial Fourier series, so they simplify to \begin{align} T_0 - T_1 - \frac{T_0-T_1}{L}x &= \sum_{n=1}^\infty c_{1(n,0)} \sin\left(\frac{n\pi x}{L}\right) \\ -\frac{4aL}{n^2\pi^2\sinh(n\pi)} \left[\cosh\left(\frac{n\pi y}{L}\right) - \cosh\left(\frac{n\pi(L-y)}{L}\right)\right] &= c_{2(n,0)} + \sum_{m=1}^\infty c_{2(n,m)}\cos\left(\frac{m\pi y}{L}\right) \end{align}

The last round of integration finishes the job

• Thanks a lot for your help!! I can understand the steady solution process, but the transient results like the last two equations are hard for me. Do I have to solve the last two equations to obtain 'c_1(n,0) and c_2(n,m) '? and does the c(n,m) in u(x,y,t) equals to c1(n,m)+c2(n,m)? Jun 25 '18 at 22:28
• Yes, $c_1+c_2=c$ for all indices $n,m$. You'll need to obtain the Fourier series of those LHS functions and then match them up. The integrals are a bit cumbersome, but they are doable. Jun 26 '18 at 4:45
• You can also obtain the entire $u(x,y,0)$ Fourier series as a double integral, but I find splitting it up this way makes it a bit easier. Since the cosine series of a constant is that same constant, you can conclude that the coefficient of the linear $x$ piece, which I've marked $c_{1(n,m)}$, is zero for $m\ne 0$. Jun 26 '18 at 4:53

Consider the method of separation of variables. Small googling provides a solution to 2D heat equation. Perhaps you will have to adjust the solution a little so that it works for your exact boundary conditions.

http://ramanujan.math.trinity.edu/rdaileda/teach/s12/m3357/lectures/lecture_3_6_short.pdf

In the end, you will likely arrive at an infinite sum of terms for your solution.

• Really appreciate your answer! I know I may solve it with separation of variables, but the thing is I'm not familiar with it, and I have read some similar cases to figure out the process of separation of variables, but I'm still unable to solve this issues. That is the reason why I post the question. Jun 21 '18 at 20:41
• This should be a comment, not an answer Jun 23 '18 at 8:26