Posted Saturday October 28 2023.
Suppose you want to solve a linear partial differential equation of order m in n variables: for example when n = 3 and m = 2 you might be considering the wave equation in 2 dimensions:
$$
\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0
$$
This equation can be written as L[u] = 0, where L is the linear differential operator L[u] = ∂t2[u] − ∂x2[u] − ∂y2[u]. (Yes, L is a polynomial over ∂t, ∂x, ∂y, and yes, it is order 2 because this polynomial has degree 2.) In general, given some open set U ⊆ ℝn and linear differential operator L : Cm(U, ℝ) → C0(U, ℝ), you want to find solutions u ∈ Cm(U, ℝ) such that L[u] = 0. Because L is linear, the set of all solutions u ∈ Cm(U, ℝ) such that L[u] = 0 is itself a vector space called the kernel (or null space) or L. In notation, if S is the set of solutions to the equation L[u] = 0, then S = kernel(L) = {u ∈ Cm(U, ℝ) ∣ L[u] = 0}. It’s not hard to check these solutions form a vector space: try it! (solution: because L is linear, for any u, u′ ∈ S, L[u + u′] = L[u] + L[u′] = 0 + 0 = 0 ⟹ u + u′ ∈ S and for any u ∈ S and c ∈ ℝ, L[c ⋅ u] = c ⋅ L[u] = c ⋅ 0 = 0 ⟹ c ⋅ u ∈ S. And don’t forget: 0 ∈ S because L[0] = 0.)
To pose an initial value problem (IVP) you need one of the dimensions to be time - say it’s the first dimension. Now let D = {(0, x) ∣ x ∈ ℝn − 1} be the subset of the domain where t = 0. Given some initial data f : D → ℝ, the initial condition becomes u|D = f, so the IVP is
L[u] = 0, u|D = f.
We can write the solution space of this problem as S = kernel(L) ∩ R(D, f), where R(D, f) = {u ∣ u|D = f} (R is for restriction). S is not a vector space, because if u, u′ ∈ S are solutions, then (u + u′)|D = u|D + u′|D = f + f = 2f, which is not equal to f unless f = 0. What we have instead is that S is an affine space over kernel(L) ∩ R(D, 0). This follows from the more general argument that R(D, f) is an affine space over R(D, 0) - try proving it! (solution: if u ∈ R(D, f) and v ∈ R(D, 0) then (u + v)|D = u|D + v|D = f + 0 = f.)
An initial boundary value problem (IBVP) in n variables might take the form
L[u] = 0, u|t = 0 = f, u|x = 0 = g
where (x, t) ∈ ℝn − 1 × (0, ∞), f : ℝn − 1 → ℝ, and g : (0, ∞) → ℝ. (For an example take Shearer & Levy 4.2.2.) So, we have some initial condition f along with some forcing term g which is applied at x = 0 for all time t. We want to find a solution u ∈ kernel(L) ∩ R(D1, f) ∩ R(D2, g), where D1 = ℝn − 1 and D2 = (0, ∞).
Sometimes, as in the case of the wave equation, splitting the problem into two other IBVPs leads to a solution. We consider the IBVPs
L[u] = 0, u|t = 0 = 0, u|x = 0 = g
and
L[u] = 0, u|t = 0 = f, u|x = 0 = 0
In the first, the initial condition is replaced with zero, and in the second the boundary condition is replaced with zero.
We can show if u1 is a solution to the first and u2 is a solution to the second, it follows that u1 + u2 is a solution to the original IBVP, i.e. it belongs to the solution space S = kernel(L) ∩ R(D1, f) ∩ R(D2, g) where D1 = ℝn − 1 and D2 = (0, ∞). If u1 ∈ kernel(L) ∩ R(D1, f) ∩ R(D2, 0) and u2 ∈ kernel(L) ∩ R(D1, 0) ∩ R(D2, g), then
u1 + u2 ∈ kernel(L) since u1, u2 ∈ kernel(L) and it is a vector space,
u1 + u2 ∈ R(D1, f) since u1 ∈ R(D1, f), u2 ∈ R(D1, 0), and R(D1, f) is an affine space over R(D1, 0),
u1 + u2 ∈ R(D2, g) since u1 ∈ R(D2, 0), u2 ∈ R(D2, g), and R(D2, g) is an affine space over R(D2, 0),
Thus u1 + u2 ∈ S, so it solves the IBVP.
Appendix: seems like we are using the following theorem: if {(Di, fi)}, i ∈ I, is a (finite) set of pairs where each Di ⊆ U and f : Di → ℝ, and {ui}i ∈ I ⊆ C(U, ℝ) satisfies ui ∈ R(Di, fi) ∩ ⋂j ≠ iR(Dj, 0) for all i, then ∑iui ∈ ⋂iR(Di, fi).
But it’s just a special case of the more general fact about affine spaces: if {Xi}, i ∈ I is a (presumably finite) collection of subspaces of some vector space X, and {Ai} is a collection of affine spaces, each Ai affine over Xi, and if {ai} is a collection such that ai ∈ Ai and ai ∈ Xj for all j ≠ i, and a = ∑iai, then a ∈ Ai for all i.