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Let $(x,y) \in \mathbb{T} \times [0,1]$, ie the periodic strip ($\mathbb{T}$ is the circle). Consider the Dirichlet Problem: $$\Delta u(x,y)=f(x,y)$$ $$u(x,0) = a(x)\qquad u(x,1) = b(x)$$

By expanding all functions as Fourier series in $x$ like $u(x,y) = \sum \hat{u}_k(y)e^{ikx}$ we get a series of ODEs. The $k=0$ case can be solved by straight integration but I omit this part. The $k\neq0$ case is solved by variation of parameters, writing $\hat{u}_k(y) = h_k(y)+g_k(y)$ as the sum of homoegenous and general parts we get $$h_k(y) = \frac{\hat{b}_k\sinh(ky) - \hat{a}_k\sinh[k(y-1)]}{\sinh(k)}$$ $$g_k(y) =\int_0^y \frac{\sinh[k(y-t)]}{k} \hat{f}_k(t)dt - \frac{\sinh(ky)}{\sinh(k)}\int_0^1 \frac{\sinh[k(1-t)]}{k} \hat{f}_k(t)dt$$

My issue is with the convergence of the solution $u$. For example, suppose I would like to show that the solution is at the very least in $L^2$, ie $$\|u\|_{L^2}^2 = \sum_k \|\hat{u}_k(\cdot)\|_{L^2[0,1]}^2 < \infty$$

For the homogeneous part, integrating $\sinh^2(ky)$ I get $$\|h_k(y)\|_{L^2[0,1]}^2 \approx \frac{|\hat{a}_k|^2 + |\hat{b}_k|^2}{k}$$ which is what I expected. On the other hand the general part I get: $$\|g_k(y)\|_{L^2[0,1]}^2 \approx \frac{\sinh^2(k)}{k^4}\|\hat{f}_k(y)\|_{L^2[0,1]}^2$$ and here there must be something wrong because the $\sinh^2(k)$ blows up.

Of course I haven't specified in what space lie $a(x), b(x)$ and $f(x,y)$ nevertheless there seems to be something wrong with this general term. I would need some very strong conditions on $f$ to dominate the $\sinh^2(k)$ growth in the numerator.

What gives?

1 Answers 1

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Let's isolate the issue: you solved the ODE $g_k''-k^2g_k=\hat f_k$ with zero boundary conditions on the interval $[0,1]$, and got $g_k$ that is much larger than $\hat f_k$ when $k$ is larger. This cannot happen, since $g_k$ is just the convolution of $\hat f_k$ with Green's function; so you made a mistake somewhere in the process of solving for $g_k$. I'll use the Green function method: $$g_k(y) = \int_0^1 \hat f_k(z) G(y,z)\,dz$$ where Green's function $G(y,z)$ must satisfy $G_{yy}-k^2G = \delta_z$ with zero boundary conditions. So, $$ G(y,z) = \begin{cases} A\sinh(ky),\quad &y\le z, \\ B\sinh(k(1-y)),\quad &y\ge z \\ \end{cases} $$ The function must be continuous at $y=z$ and its derivative must jump by $1$ there. Hence $$ \begin{cases} A\sinh(kz) - B\sinh(k(1-z)) &= 0 \\ Ak \cosh(kz) + Bk\cosh(k(1-z)) &= -1 \\ \end{cases} $$ Solving this system yields $$ \begin{cases} A &= \dfrac{\left(- e^{2 k} + e^{2 k z}\right) e^{- k z}}{k \left(e^{2 k} - 1\right)} \\ B& = \dfrac{e^{ k(1- z)}(1- e^{2 k z}) }{k \left(e^{2 k} - 1\right)} \end{cases} $$ The upshot is that $G$ is of size $O(1/k)$. Indeed, the part $A\sinh(ky)$, $y\le z$, is controlled by $Ae^{kz}$ which is $$\frac{\left(- e^{2 k} + e^{2 k z}\right)}{k \left(e^{2 k} - 1\right)}$$ Exponentials cancel out when $k$ is positive and large; and are small when $k$ is negative and large. Similarly, $$Be^{k(1-z)} = \frac{e^{ 2k(1- z)}(1- e^{2 k z}) }{k \left(e^{2 k} - 1\right)} $$ where again, exponentials are large but balanced when $k$ is positive, and small when $k$ is negative.

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    Thank you very much. Actually my expression for $g_k$ was good, with some algebra one can show it is equivalent to your green function. I went wrong though with my estimate of its order. I think it is easier the way you did it, directly on the Green's function - I fuddled it trying on my solution. – 2017-01-08
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    I wonder what is the simplest way to estimate $\partial_y^s g_k(y)$, if there is some way we can do this directly from the green's function? Otherwise it seems to be a very cumbersome computation, and I end up at the end with a similar problem I had in the original post. – 2017-01-09
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    What I mean is it seems I have to differentiate the expression $\int_0^y G(y,z)\hat{f}_k(z)dz + \int_y^1 G(y,z) \hat{f}_k(z) dz$ – 2017-01-09
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    The anisotropy introduced by separating the variables is awkward, if the goal is to study the regularity of solution. I think it would be better either to keep problem in its natural 2D setting (and use planar Green's function, with a logarithmic singularity), or to go Fourier series all the way, expanding $f(x,y)=\sum c_{m,n} \exp(2\pi i(mx+ny))$. In the latter case the contribution of $f$ to the solution is the same series with coefficients $c_{m,n}/(m^2+n^2)$, so you can see that the second-order derivatives of this function are in $L^2$. And the boundary terms contribute something harmonic – 2017-01-09
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    I'm trying to solve this Dirichlet problem for $u \in H^s(\mathbb{T})\otimes H^r[0,1]$ with $f$ in a space $A$ such that $\Delta: H^s(\mathbb{T})\otimes H^r[0,1]\rightarrow A$ is an isomorphism. Like you said the computation is awkward, and I'm getting some strange results. Amonst others I figured the boundary functions $a(x),b(x)$ should be in $H^{s-1/2}(\mathbb{T})$ but instead I'm getting that they should be in $H^{s+r-1/2}$ which I don't think is right. The problem is very anisotropic, with the definiton of $A$ being awkward as well, ultimately I had to define it as a subsapce of $H^{-2}$. – 2017-01-09
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    Yes, looks strange. I suggest posting a new question that focuses on regularity on the scale of Sobolev spaces, instead of explicit computations like this one. – 2017-01-09
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    One thing still bothering me about this problem is the $L^2(0,1)$ norm of $g_k(y)$. If we take $f \in H^{s-2}$ then we expect $g \in H^s$. In other words, we should have $(k^2)^\sigma \|\partial_y^{s-\sigma} g_k(y) \|_{L^2(0,1)}^2 \approx \|f\|_{H^{s-2}}^2$ for any $\sigma \leq s$. But I only get $\|g_k(y)\|_{L^2(0,1)}^2 \approx k^{-3} \|f_k(y)\|_{L^2(0,1)}^2$ and thus I am missing a factor of $k$ in the denominator. I should have $k^{-4}$. – 2017-01-30