Convolution preserve the boundary condition

Question

Here, I want to know if convolution will preserve the Neumann condition or not. Suppose $K$ is a continuous function on some interval $[-L,L]$, and $u$ is some 'good enouth' function on $[0,L]$ that satisfies the Neumann boundary condition, i.e., $u'(0) = u'(L) = 0$. Define function $H$ by convolution,

$$ H[u] (x) = \int_0^L K(x-y) u(y) dy $$

Does $H[u]$ satisfy Neumann BC?, i.e., Does $H_x [u] (0) = H_x[u] (L) = 0$ ?

I suppose $u$ to be good enough since I actually don't know which specific condition should be imposed on $u$.

My attempt:
If we assume that the following holds: $H$ is self-adjoint and compact, i.e., $K$ is symmetric and the Hilbert-Schmidt condition $\int_0^L \int_0^L K(x-y)^2 dy dx < \infty$ holds. Suppose there exists a sequence of eigenvalue $\lambda_n$ and eigenfunction $\phi_n(x)$, we can try to expand $K$ and $u$ as

$$ K(x-y) = \sum_{n=0}^{\infty} \lambda_n \phi_n(x) \phi_n(y), \quad u(x) = \sum_{n=0}^{\infty} a_n \phi_n(x) $$ then by substituting them into the convolution, we get \begin{align} H[u]_x = H \left[ \sum_{n=0}^\infty a_n \phi_n \right]_x = \sum_{n=0}^\infty a_n \lambda_n \phi_n' \end{align} I have also tried some example kernel $K(x-y)$. For example, one can take the kernel to be the Green function of $-\frac{\partial^2}{\partial x^2} + I$ with Neumann BC, and for this kernel, the eigenfunction is $\cos(n \pi x/L)$. So all the $\phi'(0) = \phi'(L) = 0$, hence the convolution satisfies the Neumann BC.

My question:

• For general kernel, even if I expand them as series, I still could not find the connection between Neumann BC of $u$ and Neumann BC of $H[u]$.
• Is there a relatively simpler way without using series expansion to prove this? Like integration by parts (I tried IBP but failed) or other tricks?
• If we have to use the property of Hilbert-Schmidt theory, can we differentiate the series term by term here? Will this introduce new convergence problems so that $u$ requires stronger conditions?

Answer 1

I will set $L=1$ with no loss of generality. You are assuming a filter that is invariant by translation, so generically, it will not "know" about the boundary condition so your answer is no. Take for example for $h<1$: $$ K = \frac1{2h}1_{(-h,h)} $$ Say you start with $u=1$, then: $$ H[u](x) = \begin{cases}\frac{h+x}{2h} & x\leq h \\ 1 & h\leq x\leq 1-h \\ \frac{1+h-x}{2h} & 1-h\leq x\end{cases} $$ and for continuous $u$ in general, $H[u]$ is differentiable at $0$ and $1$ with: $$ H[u]'(0) = \frac{u(h)}{2h} \quad H[u]'(1) = -\frac{u(1-h)}{2h} $$ so Neuman's boundary conditions are compromised.

The easiest way to amend this is to look for more general kernels: $K(x,y)$ which are not translation invariant. A simple generic class that you can construct can be motivated by going in Fourier space. You can decompose you $u$ as: $$ u(x) = \sum_{n=0}^\infty u_n2\cos(\pi nx) \\ u_n = \int_0^1u(x)\cos(\pi nx)dx $$ so a natural kernel is: $$ \begin{align} H[u](x) &= \sum_{n=0}^\infty K_nu_n\cos(\pi nx) \\ &= \int_0^1dyu(y)\sum_{n=0}^\infty K_n2\cos(\pi nx)\cos(\pi ny) \\ &= \int_0^1dyu(y)\frac{k(x-y)+k(x+y)}2\\ k(\xi) &:= \sum_{n=0}^\infty k_n2\cos(\pi nx) \end{align} $$ In general, a kernel of the form: $$ K(x,y) = \frac{k(x-y)+k(x+y)}2 $$ with $k$ supported on $(-1,2)$ satisfying: $$ \forall x\in[0,1], k(x) = k(-x) = k(2-x) $$ will preserve the Neumann boundary conditions for smooth enough $k$ (in Fourier space, $K_n$ decays fast enough).

You could have also motivated it from the method of images. Indeed, if you want $u$ to satisfy Neumann's boundary conditions, you can equivalently extend it on $\mathbb R$ by parity (now extended to $(-1,1)$) and $2$ periodicity. From this perspective, $H[u]$ will typically not be even and 2-periodic, which is why Neuman's boundary conditions are generically violated. The quick fix is to instead define: $$ H[u](x) = \sum_{n\in\mathbb Z}(K*u)(2n+x)+(K*u)(2n-x) $$ which by construction is even and 2-periodic so the BC's are preserved.

This second method is equivalent to the first as: $$ k(\xi) = \sum_{n\in\mathbb Z}K(2n+\xi)+K(2n-\xi) $$ which is related to the Fourier approach by the Poisson summation formula.