Polynomial chaos stochastic collocation and Galerkin UQ — Examples

Take the standard Cauchy law, whose density is $\rho (x)=1\mathrm{/}(\pi (1+x^2))$ on $\mathbb{R}$. Its second moment integral is $\int x^{2}d\rho (x)={\pi }^{-1}{\int }_{\mathbb{R}}{x}^2\mathrm{/}(1+x^2)dx\mathrm{.}$ For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $x^2\mathrm{/}(1+x^2)\ge 1\mathrm{/}2$, so the integral over $[-R,-1]\cup [1,R]$ is at least $(R-1)\mathrm{/}\pi$.

Letting $R\to \mathrm{\infty }$ shows the second moment is infinite. Hence $x\mathrm{\notin }{L}^2(\rho )$, the degree-one monomial Gram entry is not finite, and [F1] fails; the finite Gram--Schmidt construction cannot supply a degree-one chaos basis for this law.

Let $\rho =({\delta }_{-1}+{\delta }_1)\mathrm{/}2$ and take $p=2$. All moments are finite. The coefficient vector of $q(x)=x^2-1$ in the ordered monomials $(1,x,x^2)$ is $(-1,0,1)\mathrm{\ne }0$. Yet $q(-1)=q(1)=0$, so $\int q(x{)}^{2}d\rho (x)=0$.

Therefore the degree-two Gram quadratic form vanishes on a nonzero vector and is singular despite finiteness of every moment. The stated sufficiency claim is false.

At cutoff $p_1=0$, the only moment-Gram matrix is $[1]$, so the uniform probability law is admissible by [F1]. Its support is $\mathrm{\Gamma }=[-1,1]$, and for every $y\in \mathrm{\Gamma }$, $1\le 2+y\le 3$. Thus the diffusion definition applies with $a_{\ast }=1$ and $a^{\ast }=3$, and $F=0$ is a bounded affine forcing.

The pointwise weak equation is ${\int }_0^1(2+y)u^{\mathrm{\prime }}(x)v^{\mathrm{\prime }}(x)dx=0$ for every $v\in H_0^1(0,1)$. Taking $v=u$ and using $2+y\ge 1$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{H_0^1}^2\le 0$, so $u=0$ for every $y$. This solution is a strongly measurable square-integrable zero function and satisfies the Bochner equation; uniqueness also follows from [F3].

The first two moments are $\mathbb{E}[Y]=0$ and $\mathbb{E}[Y^2]=1$, so the degree-one Gram matrix is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, positive definite. Thus [F1] applies. Gram--Schmidt gives ${\psi }_0=1$ and ${\psi }_1=Y$, since $Y$ already has mean zero and unit norm and its leading coefficient is positive.

Substitution gives $k(Y)=\kappa {\psi }_0+\kappa \eta {\psi }_1$ at both support points, so the degree-one projection residual is identically zero. Its coefficients are $c_0=\mathbb{E}[k(Y)]=\kappa$ and $c_1=\mathbb{E}[k(Y)Y]=\kappa \eta$. By [F3], the mean is $c_0$ and variance is $c_1^2={\kappa }^2{\eta }^2$.

The condition $\mathrm{\mid }\eta \mathrm{\mid }<1$ implies $1+\eta Y\ge 1-\mathrm{\mid }\eta \mathrm{\mid }>0$ at both states, so the reaction rate is positive on its whole support. Every function on this support also has the exact form $g(Y)=\frac{g(1)+g(-1)}{2}+\frac{g(1)-g(-1)}{2}Y$; hence this two-dimensional finite basis spans $L^2(\rho )$ for this particular law.

For one uniform marginal, the moment Gram matrix through degree two is $\left(\begin{array}{ccc}1& 0& 1\mathrm{/}3\\ 0& 1\mathrm{/}3& 0\\ 1\mathrm{/}3& 0& 1\mathrm{/}5\end{array}\right)$. Its quadratic form at $(c_0,c_1,c_2)$ is $(c_0+c_2\mathrm{/}3{)}^2+c_1^2\mathrm{/}3+4c_2^2\mathrm{/}45$, which is positive for every nonzero vector; all required moments are finite on $[-1,1]$. Thus the cutoff is admissible. In three coordinates, the total-degree set has one zero multi-index, three degree-one indices, three indices $2e_j$, and three indices $e_i+e_j$ with $i<j$. These are disjoint and exhaust the nonnegative triples with sum at most two, so its cardinality is $1+3+3+3=10$.

The full degree-two tensor box has $3$ choices in each of three coordinates, hence $3^3=27$ multi-indices. The three-node Gaussian rules are defined for the positive uniform measures on $[-1,1]$; their tensor rule with $m_1=m_2=m_3=3$ has $3^3=27$ distinct nodes by [F3]. Thus in this example the total-degree basis uses ten modes compared with twenty-seven tensor modes and node evaluations.