Errors in posing a boundary value problem

It is not enough to solve a boundary value problem — it must be solved correctly, and for that one must at least avoid errors at the problem formulation stage. We will consider a couple of typical inaccuracies.

Compatibility of initial and boundary conditions

The first typical error is to prescribe initial and boundary conditions that do not agree with each other at the junction. In a nonstationary problem the solution is sought for $t$ $\ge 0$ and $M$ $\in G$; on the boundary $S$ for $t$ $= 0$ the initial and boundary conditions describe the same point. If they give different values there, a discontinuity arises at that point.

Let us consider the Dirichlet problem with the initial condition $T(M, 0)$ $= T_0(M)$ and the boundary condition $T(M, t)$ $= \Phi(M, t)$, $M$ $\in S$. The zeroth-order compatibility condition requires the data to coincide

If condition (2.19$\displaystyle T_0(M)$ $\displaystyle = \Phi(M, 0), \quad M$ $\displaystyle \in S.$) is violated, the Fourier series of the solution approaches the boundary with different limits in $t$ and in $S$: the convergence is nonuniform, the Gibbs phenomenon appears, and no classical (smooth) solution exists.

But even the fulfillment of (2.19$\displaystyle T_0(M)$ $\displaystyle = \Phi(M, 0), \quad M$ $\displaystyle \in S.$) is not enough. The equation itself $\frac{\partial T}{\partial t}$ $= a^2 \cdot \Delta T$ relates the derivatives of the data: differentiating the boundary condition with respect to time and taking $t$ $= 0$, we obtain the first-order compatibility condition

Higher-order conditions are constructed similarly; for a solution of class $C^k$ compatibility conditions are required up to roughly order $k$. Violating them does not make the problem unsolvable, but it limits the smoothness of the solution — something easy to forget when formally writing down “arbitrary” $T_0(M)$ and $\Phi(M, t)$. For the Neumann problem the compatibility conditions are the same but are written for the normal derivative:

Solvability condition of the stationary Neumann problem

And here let us also discuss the solvability condition of the stationary Neumann problem. First, it should be noted that the solution of the Neumann problem is defined up to a constant, which can be fixed at any point of the geometry. But the main issue is that for the Neumann problem it is essential that the sum of the energy fluxes — incoming, outgoing, sources and sinks inside the geometry — equals zero. If this condition is not met, the solution will have an incorrect physical meaning. For example, what happens if a heat flux enters a rod while its other end is insulated? Then the temperature will grow indefinitely in the steady state, which essentially will not exist. In practice, of course, this cannot happen, because the rod is not one-dimensional and heat can leak through the lateral surface, and pure Neumann conditions do not exist anyway; nevertheless, the model must be correct. Let us consider an example of a stationary Neumann problem. The time derivative in equation (1.8$\displaystyle \frac{\partial T(M, t)}{\partial t}$ $\displaystyle = a^2 \cdot \Delta T(M, t)$ $\displaystyle + f(M, t),$) then vanishes, and it takes the form

The Neumann boundary conditions for it are given by the boundary condition (1.14$\displaystyle \lambda \cdot \frac{\partial T(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S,$).

Green's second formula is known

If we set $U(M)$ $= -1$, we obtain

Now let us substitute equation (2.21$\displaystyle a^2 \cdot \Delta T(M)$ $\displaystyle + f(M)$ $\displaystyle = 0.$) and the boundary conditions (1.14$\displaystyle \lambda \cdot \frac{\partial T(M, t)}{\partial \vec{n}}$ $\displaystyle = \Phi(M, t), \quad M$ $\displaystyle \in S,$)

that is, the amount of heat produced by the sources inside the geometry equals the amount of heat leaving through the boundary. If the condition is violated, the solution will have an incorrect physical meaning.