2D Heat Equation Solutions

[Metronome]

When solving a [imath]2[/imath]D Heat Equation, suppose I separate the solution into time and space, i.e., [imath]f_1(T(t)) = f_2(Z(x,\ y)) = \lambda[/imath], and then separate space into its dimensions, i.e., [imath]f_3(X(x)) = f_4(Y(y),\ \lambda) = r[/imath]. The problem of this sort I worked seems to have two nontrivial paths, one in the case that [imath]\lambda = r \neq 0[/imath] and another in the case that [imath]\lambda \neq r, \lambda \neq 0, r \neq 0[/imath]. Usually in other problems I have encountered only one nontrivial path.

After I have solved for [imath]u[/imath] in each of the paths, the former being a Fourier Series solution and the latter being a double Fourier Series solution, am I supposed to combine the answers into a single particular solution to the problem somehow, or are these separate particular solutions which I would choose between based on some physical measurement to determine whether or not [imath]\lambda = r[/imath]?

 

[nasi112]

It is hard to follow you in this way. Since you are good in writing symbols, I suggest that you write the whole problem with its initial and boundary conditions. Then, I could see the problem scope and I might help you. But as I can remember the general solution for heat equation is the sum of all other solutions and you can always check if your solution is correct by plugging the initial and boundary conditions.

 

[Metronome]

It is hard to follow you in this way. Since you are good in writing symbols, I suggest that you write the whole problem with its initial and boundary conditions. Then, I could see the problem scope and I might help you. But as I can remember the general solution for heat equation is the sum of all other solutions and you can always check if your solution is correct by plugging the initial and boundary conditions.

Okay, to simplify the post, I will pose a Laplace Equation with similar behavior instead. Say I have [imath]\triangle v = 0, v_x(0, y) = 0, v_x(1, y) = 0, v(x, 0) = 7, v(x, 1) = 7[/imath]. Separating variables yields [imath]X_{xx} = \lambda X, Y_{yy} = -\lambda Y[/imath]. If [imath]\lambda = 0[/imath], I get just the boundary value problem solution [imath]v = 7[/imath]. If [imath]\lambda \neq 0[/imath], I get [imath]v = cos(n\pi x)\frac{(14sinh(3n\pi) + 7e^{-3n\pi} - 7)e^{-n\pi y} + (7 - 7e^{-3n\pi})e^{n\pi y}}{2sinh(3n\pi)}[/imath]. My question is whether these two solutions should be combined, or selected between based on a physical observation of [imath]\lambda[/imath].

I had similar branching paths appear in the Heat Equation [imath]k\triangle u = u_t, u_x(t, 0, y) = 0, u_x(t, 1, y) = 0, u(t, x, 0) = 0, u(t, x, 1) = 0[/imath], except that the two branches involved a single and a double Fourier summation. I no longer have my full solution to that problem.

 

[nasi112]

My question is whether these two solutions should be combined

Yes the two solutions could be combined (Not necessary if you will start with n = 0).

You found \(\displaystyle v_1\) and \(\displaystyle v_2\).

Let us say \(\displaystyle v_1 = 7\).

\(\displaystyle v(x,y) = v_1 + v_2 = 7 + v_2\)

The \(\displaystyle 7\) came from the eigenvalue \(\displaystyle \lambda_0 = 0\), it corresponds to the index \(\displaystyle n = 0\).

You have two options to write the general solution.

\(\displaystyle v(x,y) = v_2\), and start your summation with \(\displaystyle n = 0\).
Or
\(\displaystyle v(x,y) = 7 + v_2\), and start your summation with \(\displaystyle n = 1\). (This is a safer approach.)

You have to be very careful about your boundary conditions.
\(\displaystyle v(x,0) = 7\). This means \(\displaystyle v_2 = 0\) (When first index \(\displaystyle n = 1\)). If you don't get \(\displaystyle v_2 = 0\), your solution is wrong and you failed to fulfill the boundary conditions.

I had similar branching paths appear in the Heat Equation [imath]k\triangle u = u_t, u_x(t, 0, y) = 0, u_x(t, 1, y) = 0, u(t, x, 0) = 0, u(t, x, 1) = 0[/imath], except that the two branches involved a single and a double Fourier summation. I no longer have my full solution to that problem.

The general solution for the two dimensional heat equation (homogeneous) will look similar to this (a little change will happen depending on your initial and boundary conditions):

\(\displaystyle u(x,y,t) = \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} A_{nm} \sin(\lambda_n x) \cos(\lambda_m y)e^{-\lambda_{nm} kt}\)

Did you arrive to this step? Why you no longer have your full solution? Where are you stuck at?