# How to tell if PDE boundary conditions are homogeneous or not?

Say we have Laplace's equation: $$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0$$

These are the boundary conditions.

Now, to solve this by separation of variables, we need the boundary conditions to be homogeneous.

We can apparently (according to my mark-scheme) do so by defining a new variable $$u = T-T_0$$.

Then, the boundary conditions become $$u=0$$ for $$x=0, x=1, y=1$$ and $$T_0 \sin (2\pi x)$$ for $$y = 0$$.

Now obviously, the first three of those are homogeneous as they are equal to 0, but the fourth BC is what I'm not so sure about. Is $$u(x,0) = T_0 \sin (2\pi x)$$ a homogeneous boundary condition? My understanding of a PDE having homogeneous BCs was that the function (in this case $$u$$) is zero at the boundaries, which isn't completely fulfilled by the fourth boundary condition.

However, the mark-scheme proceeds from here to solve by separation of variables, which leads me to wonder if I'm misunderstanding what a homogeneous boundary condition is?

It is enough for three of the four boundary conditions to be "homogeneous" (i.e. equal to $$0$$) so that separation of variables is analytically tractable for the 2D Laplace equation on a square domain. Cf. for example https://tutorial.math.lamar.edu/classes/de/laplaceseqn.aspx for more details.
• @Vj123 For the method described in the link to work, three of the four functions need to be $0$, but the fourth one can (at least in principle) be any continuous function with convergent (in a suitable sense, sorry for being imprecise) Fourier series, for which it is enough to be twice continuously differentiable. So it doesn't have to be linear. (I am sorry for not being able to provide more details, I need to prepare a talk that I am giving tomorrow.) Apr 7, 2022 at 22:02