User:Egm6936.f10/Illustrative examples

 change the title of the subpage. call it "Stochastic 1-D heat problem", which is more meaningful. Egm6321.f12 (talk) 18:24, 3 September 2012 (UTC)

In this section, we will employ a stochastic heat conduction example to illustrate some concepts used in gPC to handle stochastic problems. This example mainly comes from D. Xiu et al. 2002. For any other information, one should resort to the original article.

 refer to my PEA2 S12 lecture notes for the computational formulation. also document all difficulties you encountered, and errors that you made. for example, the meaning of $$\displaystyle \sigma$$ in the formulation below (it is not the standard deviation, but must be multiplied by a constant to get the standard deviation). Egm6321.f12 (talk) 18:24, 3 September 2012 (UTC)

=One-dimensional model=

The statement of the problem is as follows:
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$$\displaystyle \frac{d}{dx}[\kappa(x;\omega) \frac{du}{dx}(x;\omega)]=0, x \in [0,1] $$ (ab1) The boundary conditions are
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$$\displaystyle u(0;\omega)=0,\;\;\;u(1;\omega)=1 $$ (ab2) and the random diffusivity is
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$$\displaystyle \kappa(x;\omega)=1+\epsilon(\omega)x $$ (ab3) where $$\displaystyle \omega$$ is the event, outcome, $$\displaystyle \epsilon(\omega)$$ is the associated random variable, and $$\displaystyle \kappa(x;\omega)>0$$. Before starting to solve this problem, we provide the exact solution of this problem for comparison with the gPC solution. The exact solution is For $$\displaystyle \epsilon(\omega)\neq 0:$$
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$$\displaystyle u^{e}(x,\omega)= \frac{log[1+\epsilon(\omega)x]}{log[1+\epsilon(\omega)]} $$ (ab4) For $$\displaystyle \epsilon(\omega)= 0:$$
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$$\displaystyle u^{e}(x,\omega)= x $$ (ab5)
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Legendre Chaos and Uniform distribution
In this subsection, we assume $$\displaystyle \epsilon(\omega) = \sigma \xi(\omega)$$, where $$\displaystyle \xi (\omega)\doteq \mathcal{U}\left ( -1,1 \right )$$ is a continuous uniform random variable with standard deviation $$\displaystyle \sigma $$. So, the associated polynomial chaos is the Legendre polynomial chaos.

From stochastic to deterministic
Applying Legendre Chaos expansion to $$\displaystyle \kappa (x;\omega)$$ and $$\displaystyle u (x;\omega)$$, we have


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$$\displaystyle \kappa (x;\omega)=\sum_{i=0}^{N}\kappa_{i}(x)P_i(\xi(\omega)),\;\;\;\;\;\;\;\;u (x;\omega)=\sum_{j=0}^{N}u_{j}(x)P_j(\xi(\omega)) $$ (ab6) Generally, the expansion is infinite. But if $$\displaystyle \kappa$$ and $$\displaystyle u$$ are polynomials, the expansions are always finite and exact due to the property of polynomial basis. In above expansions, instead of extending to a infinite, the summation is truncated at $$\displaystyle N$$ for both expansions. And it is not necessary that all expansions should have the same $$\displaystyle N$$. $$\displaystyle N$$ also is the highest order of the employed polynomials. The random parameter $$\displaystyle \omega$$ is absorbed into the Legendre basis $$\displaystyle P(\xi(\omega))$$. For simplicity, the notation $$\displaystyle \xi$$ and $$\displaystyle P(\xi)$$ will be used in following content. The coefficients $$\displaystyle \kappa_{i} $$ and $$\displaystyle u_{j} $$ are deterministic. Plugging the Legendre expansion (ab6) into governing equation (ab1), we obtain
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$$\displaystyle \frac{d}{dx}[\sum_{i=0}^{N}\kappa_{i}(x)P_i(\xi) \frac{d}{dx}\sum_{j=0}^{N}u_{j}(x)P_j(\xi)]=0 $$ (ab7) Upon simplification, equation (ab7) can be rewritten as
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$$\displaystyle \sum_{i=0}^{N}\sum_{j=0}^{N}\frac{d}{dx}[\kappa_{i}(x)\frac{d}{dx}u_{j}(x)]P_i(\xi)P_j(\xi)=0 $$ (ab8) A Galerkin projection of the above equation onto each Legendre polynomial basis $$\displaystyle \{P_{k}\}$$ is then conducted to ensure that the error is orthogonal to the Legendre polynomial space. By projecting and employing the orthogonality property of Legendre basis, we obtain for each $$\displaystyle k=0,1,...,N $$
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$$\displaystyle \sum_{i=0}^{N}\sum_{j=0}^{N}\frac{d}{dx}[\kappa_{i}(x)\frac{d}{dx}u_{j}(x)]c_{ijk}=0 $$ (ab9a)
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Or written in the matrix form as:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} \kappa_{0}\\ \kappa_{1} \\ ...\\ \kappa_{N} \end{matrix} \right \}^{T} \begin{bmatrix} c_{00k} & c_{01k} & ... & c_{0Nk}\\c_{10k} & c_{11k} & ... &c_{1Nk}\\... & ... & ... &...\\c_{N0k} & c_{N1k} & ... &c_{NNk} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\...\\u_{N}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab9b)
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In above equation (ab9), $$\displaystyle c_{ijk} = \langle P_{i}P_{j}P_{k}\rangle$$, which is the weighted inner product, can be independently evaluated from the Legendre polynomial basis. The relation between number of final equations $$\displaystyle P+1 $$, the dimensionality of the random inputs $$\displaystyle M $$ and the highest order of the polynomials $$\displaystyle N $$ is $$\displaystyle (P+1) = \frac{(M+N)!}{(M)!(N)!} $$. In this case, the dimensionality of the random inputs $$\displaystyle M = 1 $$. Hence, equation (ab9) is a set of $$\displaystyle (N+1)$$ coupled PDEs. Next, I will write out the equations for different $$\displaystyle N (N = 1, 2, ..., 7)$$ to show how these equations will look like.

N = 1
When $$\displaystyle N = 1 :$$ The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi $$ (ab10) Therefore, we can immediately compute the coefficients $$\displaystyle c$$ as follows:
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$$\displaystyle c_{000}=\mathbb E\langle P_0 P_0 P_0 \rangle = \int_{-1}^{+1} (1 \cdot 1 \cdot 1) \frac{1}{2}\, d \xi = 1 $$ (ab11)
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$$\displaystyle c_{100}=c_{010}=c_{001}=\mathbb E\langle P_1 P_0 P_0 \rangle = \int_{-1}^{+1} (\xi \cdot 1 \cdot 1) \frac{1}{2} \, d \xi = 0 $$ (ab12)
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$$\displaystyle c_{110}=c_{101}=c_{011}=\mathbb E\langle P_1 P_1 P_0 \rangle = \int_{-1}^{+1} (\xi \cdot \xi \cdot 1) \frac{1}{2}\, d \xi = \frac{1}{3} $$ (ab13)
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$$\displaystyle c_{111}=\mathbb E \langle P_1 P_1 P_1 \rangle = \int_{-1}^{+1} (\xi \cdot \xi \cdot \xi) \frac{1}{2}\, d \xi = 0 $$ (ab14) And the two equations are:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} \kappa_{0}\\ \kappa_{1} \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0\\0 & \frac{1}{3} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab15)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} \kappa_{0}\\ \kappa_{1} \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3}\\\frac{1}{3} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab16) Since the conductivity is already given as $$\displaystyle \kappa(x;\omega)=1+\sigma \xi(\omega)x $$, we can directly have the components of $$\displaystyle \kappa_{i}(x), i = 0,1$$ as
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$$\displaystyle \kappa_{0}(x)=1;\;\;\;\;\; \kappa_{1}(x)=\sigma x $$ (ab17) Substituting above quantities into equation (ab15) and (ab16), these two equations can be further modified as
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0\\0 & \frac{1}{3} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab18)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3}\\\frac{1}{3} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab19)
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N = 2
When $$\displaystyle N = 2 :$$ The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi;\;\;\;\;\;P_{2} = \frac{1}{2}(3\xi^2 -1) $$ (ab20) Except above calculated values for the coefficient $$\displaystyle c$$, we have additionals as follows:
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$$\displaystyle c_{200}=c_{020}=c_{002}=\mathbb E \langle P_2 P_0 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{2}(3\xi^2 -1) \cdot 1 \cdot 1) \frac{1}{2} \, d \xi = 0 $$ (ab21)
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$$\displaystyle c_{012}=c_{021}=c_{102}=c_{120}=c_{201}=c_{210}=\mathbb E\langle P_0 P_1 P_2 \rangle = \int_{-1}^{+1} (1 \cdot \xi \cdot \frac{1}{2}(3\xi^2 -1)) \frac{1}{2} \, d \xi = 0 $$ (ab22)
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$$\displaystyle c_{112}=c_{121}=c_{211}=\mathbb E \langle P_1 P_1 P_2 \rangle = \int_{-1}^{+1} (\xi \cdot \xi \cdot \frac{1}{2}(3\xi^2 -1)) \frac{1}{2}\, d \xi = \frac{2}{15} $$ (ab23)
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$$\displaystyle c_{221}=c_{212}=c_{122}=\mathbb E\langle P_2 P_2 P_1 \rangle = \int_{-1}^{+1} (\frac{1}{2}(3\xi^2 -1) \cdot \frac{1}{2}(3\xi^2 -1) \cdot \xi ) \frac{1}{2} \, d \xi = 0 $$ (ab24)
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$$\displaystyle c_{220}=c_{202}=c_{022}=\mathbb E \langle P_2 P_2 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{2}(3\xi^2 -1) \cdot \frac{1}{2}(3\xi^2 -1) \cdot 1) \frac{1}{2} \, d \xi = \frac{1}{5} $$ (ab25)
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$$\displaystyle c_{222}=\mathbb E\langle P_2 P_2 P_2 \rangle = \int_{-1}^{+1} (\frac{1}{2}(3\xi^2 -1) \cdot \frac{1}{2}(3\xi^2 -1) \cdot \frac{1}{2}(3\xi^2 -1)) \frac{1}{2} \, d \xi = \frac{2}{35} $$ (ab26) In this case, we have $$\displaystyle \kappa_{2}(x) = 0$$.
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Substituting above known quantities into equation (ab9), the three equations can be simplified as follows:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0 & 0\\0 & \frac{1}{3} &0\\0 & 0 &\frac{1}{5} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab27)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3} & 0\\\frac{1}{3} & 0 &\frac{2}{15}\\0 & \frac{2}{15} &0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab28)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & \frac{1}{5}\\0 & \frac{2}{15} &0\\\frac{1}{5} & 0 &\frac{2}{35} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab29)
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N = 3
When $$\displaystyle N = 3 :$$

The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi;\;\;\;\;\;P_{2} = \frac{1}{2}(3\xi^2 -1);\;\;\;\;\;P_{3} = \frac{1}{2}(5\xi^3 -3\xi) $$ (ab30) In this case, we have $$\displaystyle \kappa_{3}(x) = 0$$.
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Finally, the four equations can be simplified as follows:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\0 \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0 & 0 &0 \\ 0 & \frac{1}{3} &0 &0 \\ 0 & 0 & \frac{1}{5} &0 \\ 0 & 0 & 0 & \frac{1}{7} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'}\\u_{3}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab31)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & 0 &\frac{2}{15} & 0 \\ 0 & \frac{2}{15} &0 & \frac{3}{35} \\ 0 & 0 & \frac{3}{35} &0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'}\\u_{3}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab32)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & \frac{1}{5} &0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} \\ \frac{1}{5} & 0 &\frac{2}{35} &0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab33)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & \frac{1}{7} \\ 0 & 0 & \frac{3}{35} & 0 \\ 0 & \frac{3}{35} & 0 &\frac{4}{105} \\ \frac{1}{7} & 0 & \frac{4}{105} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab34)
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N = 4
When $$\displaystyle N = 4 :$$

The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi;\;\;\;\;\;P_{2} = \frac{1}{2}(3\xi^2 -1);\;\;\;\;\;P_{3} = \frac{1}{2}(5\xi^3 -3\xi);\;\;\;\;\;P_{4} = \frac{1}{8}(35\xi^4 -30\xi^2 +3) $$ (ab35)
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In this case, we have $$\displaystyle \kappa_{4}(x) = 0$$.

Finally, the five equations can be simplified as follows:


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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{3} &0 & 0 & 0 \\ 0 & 0 & \frac{1}{5} &0 &0 \\ 0 & 0 & 0 &\frac{1}{7} &0 \\ 0 & 0 & 0 & 0 & \frac{1}{9} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab36)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\0  \\0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{15} & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} \\ 0 & 0 & 0 & \frac{4}{63} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab37)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1 \\ \sigma x \\ 0 \\ 0  \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & \frac{1}{5} & 0 & 0 \\ 0 & \frac{2}{15} &0 & \frac{3}{35} & 0 \\ \frac{1}{5} & 0 & \frac{2}{35} & 0 & \frac{2}{35} \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab38)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\0  \\0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & \frac{1}{7} & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 \\ \frac{1}{7} & 0 & \frac{4}{105} & 0 & \frac{2}{77} \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab39)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0  \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & \frac{1}{9} \\ 0 & 0 & 0 & \frac{4}{63} & 0 \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 \\ \frac{1}{9} & 0 & \frac{20}{693} & 0 & \frac{18}{1001} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab40)
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N = 5
When $$\displaystyle N = 5 :$$

The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi;\;\;\;\;\;P_{2} = \frac{1}{2}(3\xi^2 -1);\;\;\;\;\;P_{3} = \frac{1}{2}(5\xi^3 -3\xi);\;\;\;\;\;P_{4} = \frac{1}{8}(35\xi^4 -30\xi^2 +3);\;\;\;\;\;P_{5} = \frac{1}{8}(63\xi^5 -70\xi^3 +15\xi) $$ (ab41)
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In this case, we have $$\displaystyle \kappa_{5}(x) = 0$$.

Finally, the six equations can be simplified as follows:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0  \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{5} & 0  & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{7} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{9} & 0  \\ 0 & 0 & 0  & 0 & 0 & \frac{1}{11} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab42)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3} & 0 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{15} & 0 & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} & 0 \\ 0 & 0 & 0 & \frac{4}{63} & 0 & \frac{5}{99} \\ 0 & 0 & 0 & 0 & \frac{5}{99} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab43)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & \frac{1}{5} & 0 & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 & 0 \\ \frac{1}{5} & 0 & \frac{2}{35} & 0 & \frac{2}{35} & 0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 & \frac{10}{231} \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} & 0 \\ 0 & 0 & 0 & \frac{10}{231} & 0 & \frac{10}{429} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab44)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & \frac{1}{7} & 0 & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} & 0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 & \frac{10}{231} \\ \frac{1}{7} & 0 & \frac{4}{105} & 0 & \frac{2}{77} & 0 \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 & \frac{20}{1001} \\ 0 & 0 & \frac{10}{231} & 0 & \frac{20}{1001} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab45)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & \frac{1}{9} & 0 \\ 0 & 0 & 0 & \frac{4}{63} &0 & \frac{5}{99} \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} & 0 \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 & \frac{20}{1001} \\ \frac{1}{9} & 0 & \frac{20}{693} & 0 & \frac{18}{1001} & 0 \\ 0& \frac{5}{99} & 0 & \frac{20}{1001} & 0 & \frac{2}{143} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab46)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & \frac{1}{11} \\ 0 & 0 & 0 & 0 & \frac{5}{99} & 0 \\ 0 & 0 & 0 & \frac{10}{231} & 0 & \frac{10}{429} \\ 0 & 0 & \frac{10}{231} & 0 & \frac{20}{1001} & 0 \\ 0 & \frac{5}{99} & 0 & \frac{20}{1001} & 0 & \frac{2}{143} \\ \frac{1}{11} & 0 & \frac{10}{429} & 0 & \frac{2}{143} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab47)
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N = 6
When $$\displaystyle N = 6 :$$ The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi;\;\;\;\;\;P_{2} = \frac{1}{2}(3\xi^2 -1);\;\;\;\;\;P_{3} = \frac{1}{2}(5\xi^3 -3\xi);\;\;\;\;\;P_{4} = \frac{1}{8}(35\xi^4 -30\xi^2 +3);\;\;\;\;\;P_{5} = \frac{1}{8}(63\xi^5 -70\xi^3 +15\xi) $$ $$\displaystyle P_{6} = \frac{1}{16}(231\xi^6-315\xi^4+105\xi^2-5) $$ (ab48)
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In this case, we have $$\displaystyle \kappa_{6}(x) = 0$$.

Finally, the six equations can be simplified as follows:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0  \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{5} & 0  & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{7} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{9} & 0 & 0 \\ 0 & 0 & 0  & 0 & 0 & \frac{1}{11} & 0 \\ 0 & 0 & 0 & 0  & 0 & 0 & \frac{1}{13} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab49)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{15} & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} & 0 & 0 \\ 0 & 0 & 0 & \frac{4}{63} & 0 & \frac{5}{99} & 0 \\ 0 & 0 & 0 & 0 & \frac{5}{99} & 0 & \frac{6}{143} \\ 0 & 0 & 0 & 0 & 0 & \frac{6}{143} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab50)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & \frac{1}{5} & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 & 0  & 0 \\ \frac{1}{5} & 0 & \frac{2}{35} & 0 & \frac{2}{35} & 0  & 0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 & \frac{10}{231}  & 0 \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} & 0  & \frac{5}{143} \\ 0 & 0 & 0 & \frac{10}{231} & 0 & \frac{10}{429} & 0 \\ 0 & 0 & 0 & 0 & \frac{5}{143} & 0 & \frac{14}{715} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab51)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & \frac{1}{7} & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} & 0  & 0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 & \frac{10}{231}  & 0 \\ \frac{1}{7} & 0 & \frac{4}{105} & 0 & \frac{2}{77} & 0  & \frac{100}{3003} \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 & \frac{20}{1001}  & 0 \\ 0 & 0 & \frac{10}{231} & 0 & \frac{20}{1001} & 0 & \frac{7}{429} \\ 0 & 0 & 0 & \frac{100}{3003} & 0 & \frac{7}{429} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab52)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & \frac{1}{9} & 0 & 0 \\ 0 & 0 & 0 & \frac{4}{63} &0 & \frac{5}{99}  & 0 \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} & 0  & \frac{5}{143} \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 & \frac{20}{1001}  & 0 \\ \frac{1}{9} & 0 & \frac{20}{693} & 0 & \frac{18}{1001} & 0  & \frac{20}{1287} \\ 0& \frac{5}{99} & 0 & \frac{20}{1001} & 0 & \frac{2}{143} & 0 \\ 0 & 0 & \frac{5}{143} & 0 & \frac{20}{1287} & 0 & \frac{28}{2431} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab53)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & \frac{1}{11} & 0 \\ 0 & 0 & 0 & 0 & \frac{5}{99} & 0 & \frac{6}{143} \\ 0 & 0 & 0 & \frac{10}{231} & 0 & \frac{10}{429} & 0 \\ 0 & 0 & \frac{10}{231} & 0 & \frac{20}{1001} & 0 & \frac{7}{429} \\ 0 & \frac{5}{99} & 0 & \frac{20}{1001} & 0 & \frac{2}{143} & 0 \\ \frac{1}{11} & 0 & \frac{10}{429} & 0 & \frac{2}{143} & 0 & \frac{80}{7293} \\ 0 & \frac{6}{143} & 0 & \frac{7}{429} & 0 & \frac{80}{7293} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab54)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{13} \\ 0 & 0 & 0 & 0 & 0 & \frac{6}{143} & 0 \\ 0 & 0 & 0 & 0 & \frac{5}{143} & 0 & \frac{14}{715}  \\ 0 & 0 & 0 & \frac{100}{3003} & 0 & \frac{7}{429} & 0 \\ 0 & 0 & \frac{5}{143} & 0 & \frac{20}{1287} & 0 & \frac{28}{2431} \\ 0 & \frac{6}{143} & 0 & \frac{7}{429} & 0 & \frac{80}{7293} & 0 \\ \frac{1}{13} & 0 & \frac{14}{715} & 0 & \frac{28}{2431} & 0 & \frac{91}{10508} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab55)
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N = 7
When $$\displaystyle N = 7 :$$ The Legendre Chaos bases are
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$$\displaystyle P_{0} = 1;\;\;\;\;\;P_{1} = \xi;\;\;\;\;\;P_{2} = \frac{1}{2}(3\xi^2 -1);\;\;\;\;\;P_{3} = \frac{1}{2}(5\xi^3 -3\xi);\;\;\;\;\;P_{4} = \frac{1}{8}(35\xi^4 -30\xi^2 +3);\;\;\;\;\;P_{5} = \frac{1}{8}(63\xi^5 -70\xi^3 +15\xi) $$ $$\displaystyle P_{6} = \frac{1}{16}(231\xi^6-315\xi^4+105\xi^2-5);\;\;\;\;\;P_{7} = \frac{1}{16}(429\xi^7-693\xi^5+315\xi^3-35\xi)$$ (ab56)
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In this case, we have $$\displaystyle \kappa_{7}(x) = 0$$.

Finally, the six equations can be simplified as follows:
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0  \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{5} & 0  & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{7} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{9} & 0 & 0 & 0 \\ 0 & 0 & 0  & 0 & 0 & \frac{1}{11} & 0 & 0 \\ 0 & 0 & 0 & 0  & 0 & 0 & \frac{1}{13} & 0 \\ 0 & 0 & 0 & 0  & 0 & 0 & 0 & \frac{1}{15} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab57)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{15} & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{4}{63} & 0 & \frac{5}{99} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{5}{99} & 0 & \frac{6}{143} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{6}{143} & 0 & \frac{7}{195} \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{7}{195} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab58)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & \frac{1}{5} & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{2}{15} & 0 & \frac{3}{35} & 0 & 0 & 0 & 0 \\ \frac{1}{5} & 0 & \frac{2}{35} & 0 & \frac{2}{35} & 0  & 0 & 0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 & \frac{10}{231} & 0 & 0 \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} & 0  & \frac{5}{143} & 0 \\ 0 & 0 & 0 & \frac{10}{231} & 0 & \frac{10}{429} & 0 & \frac{21}{715} \\ 0 & 0 & 0 & 0 & \frac{5}{143} & 0 & \frac{14}{715} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{21}{715} & 0 & \frac{56}{3315} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab59)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & \frac{1}{7} & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{3}{35} & 0 & \frac{4}{63} & 0 & 0 & 0 \\ 0 & \frac{3}{35} & 0 & \frac{4}{105} & 0 & \frac{10}{231}  & 0 & 0 \\ \frac{1}{7} & 0 & \frac{4}{105} & 0 & \frac{2}{77} & 0  & \frac{100}{3003} & 0 \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 & \frac{20}{1001} & 0 & \frac{35}{1287} \\ 0 & 0 & \frac{10}{231} & 0 & \frac{20}{1001} & 0 & \frac{7}{429} & 0 \\ 0 & 0 & 0 & \frac{100}{3003} & 0 & \frac{7}{429} & 0 & \frac{205}{14832} \\ 0 & 0 & 0 & 0 & \frac{35}{1287} & 0 & \frac{205}{14832} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab60)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & \frac{1}{9} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{4}{63} &0 & \frac{5}{99} & 0 & 0 \\ 0 & 0 & \frac{2}{35} & 0 & \frac{20}{693} & 0 & \frac{5}{143} & 0 \\ 0 & \frac{4}{63} & 0 & \frac{2}{77} & 0 & \frac{20}{1001} & 0 & \frac{35}{1287} \\ \frac{1}{9} & 0 & \frac{20}{693} & 0 & \frac{18}{1001} & 0  & \frac{20}{1287} & 0 \\ 0& \frac{5}{99} & 0 & \frac{20}{1001} & 0 & \frac{2}{143} & 0  & \frac{201}{15706} \\ 0 & 0 & \frac{5}{143} & 0 & \frac{20}{1287} & 0 & \frac{28}{2431} & 0 \\ 0 & 0 & 0 & \frac{35}{1287} & 0 & \frac{201}{15706} & 0 & \frac{29}{2953} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab61)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & \frac{1}{11} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{5}{99} & 0 & \frac{6}{143} & 0 \\ 0 & 0 & 0 & \frac{10}{231} & 0 & \frac{10}{429} & 0 & \frac{21}{715} \\ 0 & 0 & \frac{10}{231} & 0 & \frac{20}{1001} & 0 & \frac{7}{429} & 0 \\ 0 & \frac{5}{99} & 0 & \frac{20}{1001} & 0 & \frac{2}{143} & 0 & \frac{201}{15706} \\ \frac{1}{11} & 0 & \frac{10}{429} & 0 & \frac{2}{143} & 0 & \frac{80}{7293} & 0 \\ 0 & \frac{6}{143} & 0 & \frac{7}{429} & 0 & \frac{80}{7293} & 0 & \frac{229}{25184} \\ 0 & 0 & \frac{21}{715} & 0 & \frac{201}{15706} & 0 & \frac{229}{25184} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab62)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{13} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{6}{143} & 0 & \frac{7}{195} \\ 0 & 0 & 0 & 0 & \frac{5}{143} & 0 & \frac{14}{715} & 0 \\ 0 & 0 & 0 & \frac{100}{3003} & 0 & \frac{7}{429} & 0 & \frac{205}{14832} \\ 0 & 0 & \frac{5}{143} & 0 & \frac{20}{1287} & 0 & \frac{28}{2431} & 0 \\ 0 & \frac{6}{143} & 0 & \frac{7}{429} & 0 & \frac{80}{7293} & 0 & \frac{229}{25184} \\ \frac{1}{13} & 0 & \frac{14}{715} & 0 & \frac{28}{2431} & 0 & \frac{91}{10508} & 0 \\ 0 & \frac{7}{195} & 0 & \frac{205}{14832} & 0 & \frac{229}{25184} & 0 & \frac{97}{13441} \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab63)
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$$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left [ \left \{ \begin{matrix} 1\\ \sigma x \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix} \right \}^{T} \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{15} \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{7}{195} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{21}{715} & 0 & \frac{56}{3315} \\ 0 & 0 & 0 & 0 & \frac{35}{1287} & 0 & \frac{205}{14832} & 0 \\ 0 & 0 & 0 & \frac{35}{1287} & 0 & \frac{201}{15706} & 0 & \frac{29}{2953} \\ 0 & 0  & \frac{21}{715} & 0 & \frac{201}{15706} & 0 & \frac{229}{25184} & 0 \\ 0 & \frac{7}{195} & 0 & \frac{205}{14832} & 0 & \frac{229}{25184} & 0 & \frac{97}{13441} \\ \frac{1}{15} & 0 & \frac{56}{3315} & 0 & \frac{29}{2953} & 0 & \frac{97}{13441} & 0 \end{bmatrix} \left \{ \begin{matrix} u_{0}^{'}\\u_{1}^{'} \\u_{2}^{'} \\u_{3}^{'} \\u_{4}^{'} \\u_{5}^{'} \\u_{6}^{'} \\u_{7}^{'} \end{matrix} \right \} \right ] = 0 $$ (ab64)
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These equations are deterministic and can be solved by any conventional method, e.g., finite difference method, finite element method, etc.

Boundary conditions
The way to determinize the boundary conditions is the same as for the governing equation.

N = 1
When $$\displaystyle N = 1 :$$


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$$\displaystyle u (0;\omega)=\sum_{j=0}^{1}u_{j}(0)P_j(\xi(\omega))=0,\;\;\;\;\;\;\; u (1;\omega)=\sum_{j=0}^{1}u_{j}(1)P_j(\xi(\omega))=1 $$ (ab65)
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The stochastic Galerkin's method is employed here to get rid of the uncertainties in equation(ab65).


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$$\displaystyle \sum_{j=0}^{1}u_{j}(0)a_{ij}=0,\;\;\;\;\;\;\; \sum_{j=0}^{1}u_{j}(1)a_{ij}=p_i $$ (ab66)
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where $$\displaystyle a_{ij}= \mathbb E \langle P_i P_j \rangle$$ is the weighted inner product, and $$\displaystyle p_i = \mathbb E \langle P_i 1 \rangle = \mathbb E \langle P_i P_0 \rangle $$. The value of $$\displaystyle a_{ij}$$ and $$\displaystyle p_i$$ can be evaluated using the orthogonal property of Legendre polynomials.


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$$\displaystyle p_0 = a_{00}=\mathbb E \langle P_0 P_0 \rangle = \int_{-1}^{+1} (1 \cdot 1 )\frac{1}{2} \, d \xi = 1 $$ (ab67)
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$$\displaystyle p_1 = a_{10}=a_{01}=\mathbb E \langle P_1 P_0 \rangle = \int_{-1}^{+1} (\xi \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab68)
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$$\displaystyle a_{11}=\mathbb E \langle P_1 P_1 \rangle = \int_{-1}^{+1} (\xi \cdot \xi )\frac{1}{2} \, d \xi = \frac{1}{3} $$ (ab69)
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Plugging above coefficients into equation(ab66), the boundary conditions are:


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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{01}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{01}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab70)
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$$\displaystyle u_{0}(0)a_{01}+u_{1}(0)a_{11}=0,\;\;\;\;\;\;\; u_{0}(1)a_{01}+u_{1}(1)a_{11}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab71)
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N = 2
When $$\displaystyle N = 2 :$$

The same procedure is used for $$\displaystyle N = 2 $$.


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$$\displaystyle p_2 = a_{02}= a_{20}=\mathbb E \langle P_2 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{2}(3\xi^2 -1) \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab72)
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$$\displaystyle a_{12}=a_{21}=\mathbb E \langle P_1 P_2 \rangle = \int_{-1}^{+1} ( \xi \cdot \frac{1}{2}(3\xi^2 -1) )\frac{1}{2} \, d \xi = 0 $$ (ab73)
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$$\displaystyle a_{22}=\mathbb E \langle P_2 P_2 \rangle = \int_{-1}^{+1} (\frac{1}{2}(3\xi^2 -1) \cdot \frac{1}{2}(3\xi^2 -1) )\frac{1}{2}\, d \xi = \frac{2}{5} $$ (ab74)
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Finally, we have the boundary conditions as


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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{10}+u_{2}(0)a_{20}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{10}+u_{2}(1)a_{20}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab75)
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$$\displaystyle u_{0}(0)a_{01}+u_{1}(0)a_{11}+u_{2}(0)a_{21}=0,\;\;\;\;\;\;\; u_{0}(1)a_{01}+u_{1}(1)a_{11}+u_{2}(1)a_{21}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab76)
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$$\displaystyle u_{0}(0)a_{02}+u_{1}(0)a_{12}+u_{2}(0)a_{22}=0,\;\;\;\;\;\;\; u_{0}(1)a_{02}+u_{1}(1)a_{12}+u_{2}(1)a_{22}=p_2 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{5}u_{2}(0)=0,\;\;\;\;\;\;\; \frac{1}{5}u_{2}(1)=0 $$ (ab77)
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N = 3
When $$\displaystyle N = 3 :$$
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$$\displaystyle p_3 =\mathbb E \langle P_3 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{2}(5\xi^3 -3x) \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab78) The essential boundary conditions are as following
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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{01}+u_{2}(0)a_{02}+u_{3}(0)a_{03}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{01}+u_{2}(1)a_{02}+u_{3}(1)a_{03}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab79)
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$$\displaystyle u_{0}(0)a_{10}+u_{1}(0)a_{11}+u_{2}(0)a_{12}+u_{3}(0)a_{13}=0,\;\;\;\;\;\;\; u_{0}(1)a_{10}+u_{1}(1)a_{11}+u_{2}(1)a_{12}+u_{3}(1)a_{13}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab80)
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$$\displaystyle u_{0}(0)a_{20}+u_{1}(0)a_{21}+u_{2}(0)a_{22}+u_{3}(0)a_{23}=0,\;\;\;\;\;\;\; u_{0}(1)a_{20}+u_{1}(1)a_{21}+u_{2}(1)a_{22}+u_{3}(1)a_{23}=p_2 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{5}u_{2}(0)=0,\;\;\;\;\;\;\; \frac{1}{5}u_{2}(1)=0 $$ (ab81)
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$$\displaystyle u_{0}(0)a_{30}+u_{1}(0)a_{31}+u_{2}(0)a_{32}+u_{3}(0)a_{33}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}=p_3 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{7}u_{3}(0)=0,\;\;\;\;\;\;\; \frac{1}{7}u_{3}(1)=0 $$ (ab82)
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N = 4
When $$\displaystyle N = 4 :$$
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$$\displaystyle p_4 =\mathbb E \langle P_4 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{8}(35\xi^4 - 30\xi^2 +3) \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab83) The essential boundary conditions are as following
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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{01}+u_{2}(0)a_{02}+u_{3}(0)a_{03}+u_{4}(0)a_{04}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{01}+u_{2}(1)a_{02}+u_{3}(1)a_{03}+u_{4}(1)a_{04}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab84)
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$$\displaystyle u_{0}(0)a_{10}+u_{1}(0)a_{11}+u_{2}(0)a_{12}+u_{3}(0)a_{13}+u_{4}(0)a_{14}=0,\;\;\;\;\;\;\; u_{0}(1)a_{10}+u_{1}(1)a_{11}+u_{2}(1)a_{12}+u_{3}(1)a_{13}+u_{4}(1)a_{14}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab85)
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$$\displaystyle u_{0}(0)a_{20}+u_{1}(0)a_{21}+u_{2}(0)a_{22}+u_{3}(0)a_{23}+u_{4}(0)a_{24}=0,\;\;\;\;\;\;\; u_{0}(1)a_{20}+u_{1}(1)a_{21}+u_{2}(1)a_{22}+u_{3}(1)a_{23}+u_{4}(1)a_{24}=p_2 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{5}u_{2}(0)=0,\;\;\;\;\;\;\; \frac{1}{5}u_{2}(1)=0 $$ (ab86)
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$$\displaystyle u_{0}(0)a_{30}+u_{1}(0)a_{31}+u_{2}(0)a_{32}+u_{3}(0)a_{33}+u_{4}(0)a_{34}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{34}=p_3 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{7}u_{3}(0)=0,\;\;\;\;\;\;\; \frac{1}{7}u_{3}(1)=0 $$ (ab87)
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$$\displaystyle u_{0}(0)a_{40}+u_{1}(0)a_{41}+u_{2}(0)a_{42}+u_{3}(0)a_{43}+u_{4}(0)a_{44}=0,\;\;\;\;\;\;\; u_{0}(1)a_{40}+u_{1}(1)a_{41}+u_{2}(1)a_{42}+u_{3}(1)a_{43}+u_{4}(1)a_{44}=p_4 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{9}u_{4}(0)=0,\;\;\;\;\;\;\; \frac{1}{9}u_{4}(1)=0 $$ (ab88)
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N = 5
When $$\displaystyle N = 5 :$$
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$$\displaystyle p_5 =\mathbb E \langle P_5 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{8}(63\xi^5 - 70\xi^3 +15\xi) \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab89) The essential boundary conditions are as following
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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{01}+u_{2}(0)a_{02}+u_{3}(0)a_{03}+u_{4}(0)a_{04}+u_{5}(0)a_{05}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{01}+u_{2}(1)a_{02}+u_{3}(1)a_{03}+u_{4}(1)a_{04}+u_{5}(1)a_{05}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab90)
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$$\displaystyle u_{0}(0)a_{10}+u_{1}(0)a_{11}+u_{2}(0)a_{12}+u_{3}(0)a_{13}+u_{4}(0)a_{14}+u_{5}(0)a_{15}=0,\;\;\;\;\;\;\; u_{0}(1)a_{10}+u_{1}(1)a_{11}+u_{2}(1)a_{12}+u_{3}(1)a_{13}+u_{4}(1)a_{14}+u_{5}(1)a_{15}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab91)
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$$\displaystyle u_{0}(0)a_{20}+u_{1}(0)a_{21}+u_{2}(0)a_{22}+u_{3}(0)a_{23}+u_{4}(0)a_{24}+u_{5}(0)a_{25}=0,\;\;\;\;\;\;\; u_{0}(1)a_{20}+u_{1}(1)a_{21}+u_{2}(1)a_{22}+u_{3}(1)a_{23}+u_{4}(1)a_{24}+u_{5}(1)a_{25}=p_2 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{5}u_{2}(0)=0,\;\;\;\;\;\;\; \frac{1}{5}u_{2}(1)=0 $$ (ab92)
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$$\displaystyle u_{0}(0)a_{30}+u_{1}(0)a_{31}+u_{2}(0)a_{32}+u_{3}(0)a_{33}+u_{4}(0)a_{34}+u_{5}(0)a_{35}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{34}+u_{5}(1)a_{35}=p_3 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{7}u_{3}(0)=0,\;\;\;\;\;\;\; \frac{1}{7}u_{3}(1)=0 $$ (ab93)
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$$\displaystyle u_{0}(0)a_{40}+u_{1}(0)a_{41}+u_{2}(0)a_{42}+u_{3}(0)a_{43}+u_{4}(0)a_{44}+u_{5}(0)a_{45}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{44}+u_{5}(1)a_{45}=p_4 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{9}u_{4}(0)=0,\;\;\;\;\;\;\; \frac{1}{9}u_{4}(1)=0 $$ (ab94)
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$$\displaystyle u_{0}(0)a_{50}+u_{1}(0)a_{51}+u_{2}(0)a_{52}+u_{3}(0)a_{53}+u_{4}(0)a_{54}+u_{5}(0)a_{55}=0,\;\;\;\;\;\;\; u_{0}(1)a_{50}+u_{1}(1)a_{51}+u_{2}(1)a_{52}+u_{3}(1)a_{53}+u_{4}(1)a_{54}+u_{5}(1)a_{55}=p_5 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{11}u_{5}(0)=0,\;\;\;\;\;\;\; \frac{1}{11}u_{5}(1)=0 $$ (ab95)
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N = 6
When $$\displaystyle N = 6 :$$
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$$\displaystyle p_6 =\mathbb E \langle P_6 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{16}(231\xi^6 - 315\xi^4 +105\xi^2 -5) \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab96) The essential boundary conditions are as following
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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{01}+u_{2}(0)a_{02}+u_{3}(0)a_{03}+u_{4}(0)a_{04}+u_{5}(0)a_{05}+u_{6}(0)a_{06}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{01}+u_{2}(1)a_{02}+u_{3}(1)a_{03}+u_{4}(1)a_{04}+u_{5}(1)a_{05}+u_{6}(1)a_{06}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab97)
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$$\displaystyle u_{0}(0)a_{10}+u_{1}(0)a_{11}+u_{2}(0)a_{12}+u_{3}(0)a_{13}+u_{4}(0)a_{14}+u_{5}(0)a_{15}+u_{6}(0)a_{16}=0,\;\;\;\;\;\;\; u_{0}(1)a_{10}+u_{1}(1)a_{11}+u_{2}(1)a_{12}+u_{3}(1)a_{13}+u_{4}(1)a_{14}+u_{5}(1)a_{15}+u_{6}(1)a_{16}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab98)
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$$\displaystyle u_{0}(0)a_{20}+u_{1}(0)a_{21}+u_{2}(0)a_{22}+u_{3}(0)a_{23}+u_{4}(0)a_{24}+u_{5}(0)a_{25}+u_{6}(0)a_{26}=0,\;\;\;\;\;\;\; u_{0}(1)a_{20}+u_{1}(1)a_{21}+u_{2}(1)a_{22}+u_{3}(1)a_{23}+u_{4}(1)a_{24}+u_{5}(1)a_{25}+u_{6}(1)a_{26}=p_2 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{5}u_{2}(0)=0,\;\;\;\;\;\;\; \frac{1}{5}u_{2}(1)=0 $$ (ab99)
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$$\displaystyle u_{0}(0)a_{30}+u_{1}(0)a_{31}+u_{2}(0)a_{32}+u_{3}(0)a_{33}+u_{4}(0)a_{34}+u_{5}(0)a_{35}+u_{6}(0)a_{36}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{34}+u_{5}(1)a_{35}+u_{6}(1)a_{36}=p_3 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{7}u_{3}(0)=0,\;\;\;\;\;\;\; \frac{1}{7}u_{3}(1)=0 $$ (ab100)
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$$\displaystyle u_{0}(0)a_{40}+u_{1}(0)a_{41}+u_{2}(0)a_{42}+u_{3}(0)a_{43}+u_{4}(0)a_{44}+u_{5}(0)a_{45}+u_{6}(0)a_{46}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{44}+u_{5}(1)a_{45}+u_{6}(1)a_{46}=p_4 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{9}u_{4}(0)=0,\;\;\;\;\;\;\; \frac{1}{9}u_{4}(1)=0 $$ (ab101)
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$$\displaystyle u_{0}(0)a_{50}+u_{1}(0)a_{51}+u_{2}(0)a_{52}+u_{3}(0)a_{53}+u_{4}(0)a_{54}+u_{5}(0)a_{55}+u_{6}(0)a_{56}=0,\;\;\;\;\;\;\; u_{0}(1)a_{50}+u_{1}(1)a_{51}+u_{2}(1)a_{52}+u_{3}(1)a_{53}+u_{4}(1)a_{54}+u_{5}(1)a_{55}+u_{6}(1)a_{56}=p_5 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{11}u_{5}(0)=0,\;\;\;\;\;\;\; \frac{1}{11}u_{5}(1)=0 $$ (ab102)
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$$\displaystyle u_{0}(0)a_{60}+u_{1}(0)a_{61}+u_{2}(0)a_{62}+u_{3}(0)a_{63}+u_{4}(0)a_{64}+u_{5}(0)a_{65}+u_{6}(0)a_{66}=0,\;\;\;\;\;\;\; u_{0}(1)a_{60}+u_{1}(1)a_{61}+u_{2}(1)a_{62}+u_{3}(1)a_{63}+u_{4}(1)a_{64}+u_{5}(1)a_{65}+u_{6}(1)a_{66}=p_6 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{13}u_{6}(0)=0,\;\;\;\;\;\;\; \frac{1}{13}u_{6}(1)=0 $$ (ab103)
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N = 7
When $$\displaystyle N = 7 :$$
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$$\displaystyle p_7 =\mathbb E \langle P_7 P_0 \rangle = \int_{-1}^{+1} (\frac{1}{16}(429\xi^7 - 693\xi^5 +315\xi^3 -35\xi) \cdot 1 )\frac{1}{2} \, d \xi = 0 $$ (ab96) The essential boundary conditions are as following
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$$\displaystyle u_{0}(0)a_{00}+u_{1}(0)a_{01}+u_{2}(0)a_{02}+u_{3}(0)a_{03}+u_{4}(0)a_{04}+u_{5}(0)a_{05}+u_{6}(0)a_{06}+u_{7}(0)a_{07}=0,\;\;\;\;\;\;\; u_{0}(1)a_{00}+u_{1}(1)a_{01}+u_{2}(1)a_{02}+u_{3}(1)a_{03}+u_{4}(1)a_{04}+u_{5}(1)a_{05}+u_{6}(1)a_{06}+u_{7}(1)a_{07}=p_0 $$ $$\displaystyle \Downarrow$$ $$\displaystyle u_{0}(0)=0,\;\;\;\;\;\;\; u_{0}(1)=1 $$ (ab97)
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$$\displaystyle u_{0}(0)a_{10}+u_{1}(0)a_{11}+u_{2}(0)a_{12}+u_{3}(0)a_{13}+u_{4}(0)a_{14}+u_{5}(0)a_{15}+u_{6}(0)a_{16}+u_{7}(0)a_{17}=0,\;\;\;\;\;\;\; u_{0}(1)a_{10}+u_{1}(1)a_{11}+u_{2}(1)a_{12}+u_{3}(1)a_{13}+u_{4}(1)a_{14}+u_{5}(1)a_{15}+u_{6}(1)a_{16}+u_{7}(1)a_{17}=p_1 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{3}u_{1}(0)=0,\;\;\;\;\;\;\; \frac{1}{3}u_{1}(1)=0 $$ (ab98)
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$$\displaystyle u_{0}(0)a_{20}+u_{1}(0)a_{21}+u_{2}(0)a_{22}+u_{3}(0)a_{23}+u_{4}(0)a_{24}+u_{5}(0)a_{25}+u_{6}(0)a_{26}+u_{7}(0)a_{27}=0,\;\;\;\;\;\;\; u_{0}(1)a_{20}+u_{1}(1)a_{21}+u_{2}(1)a_{22}+u_{3}(1)a_{23}+u_{4}(1)a_{24}+u_{5}(1)a_{25}+u_{6}(1)a_{26}+u_{7}(1)a_{27}=p_2 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{5}u_{2}(0)=0,\;\;\;\;\;\;\; \frac{1}{5}u_{2}(1)=0 $$ (ab99)
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$$\displaystyle u_{0}(0)a_{30}+u_{1}(0)a_{31}+u_{2}(0)a_{32}+u_{3}(0)a_{33}+u_{4}(0)a_{34}+u_{5}(0)a_{35}+u_{6}(0)a_{36}+u_{7}(0)a_{37}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{34}+u_{5}(1)a_{35}+u_{6}(1)a_{36}+u_{7}(1)a_{37}=p_3 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{7}u_{3}(0)=0,\;\;\;\;\;\;\; \frac{1}{7}u_{3}(1)=0 $$ (ab100)
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$$\displaystyle u_{0}(0)a_{40}+u_{1}(0)a_{41}+u_{2}(0)a_{42}+u_{3}(0)a_{43}+u_{4}(0)a_{44}+u_{5}(0)a_{45}+u_{6}(0)a_{46}+u_{7}(0)a_{47}=0,\;\;\;\;\;\;\; u_{0}(1)a_{30}+u_{1}(1)a_{31}+u_{2}(1)a_{32}+u_{3}(1)a_{33}+u_{4}(1)a_{44}+u_{5}(1)a_{45}+u_{6}(1)a_{46}+u_{7}(1)a_{47}=p_4 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{9}u_{4}(0)=0,\;\;\;\;\;\;\; \frac{1}{9}u_{4}(1)=0 $$ (ab101)
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$$\displaystyle u_{0}(0)a_{50}+u_{1}(0)a_{51}+u_{2}(0)a_{52}+u_{3}(0)a_{53}+u_{4}(0)a_{54}+u_{5}(0)a_{55}+u_{6}(0)a_{56}+u_{7}(0)a_{57}=0,\;\;\;\;\;\;\; u_{0}(1)a_{50}+u_{1}(1)a_{51}+u_{2}(1)a_{52}+u_{3}(1)a_{53}+u_{4}(1)a_{54}+u_{5}(1)a_{55}+u_{6}(1)a_{56}+u_{7}(1)a_{57}=p_5 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{11}u_{5}(0)=0,\;\;\;\;\;\;\; \frac{1}{11}u_{5}(1)=0 $$ (ab102)
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$$\displaystyle u_{0}(0)a_{50}+u_{1}(0)a_{51}+u_{2}(0)a_{52}+u_{3}(0)a_{53}+u_{4}(0)a_{54}+u_{5}(0)a_{55}+u_{6}(0)a_{66}+u_{7}(0)a_{67}=0,\;\;\;\;\;\;\; u_{0}(1)a_{50}+u_{1}(1)a_{51}+u_{2}(1)a_{52}+u_{3}(1)a_{53}+u_{4}(1)a_{54}+u_{5}(1)a_{55}+u_{6}(1)a_{66}+u_{7}(1)a_{67}=p_6 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{13}u_{6}(0)=0,\;\;\;\;\;\;\; \frac{1}{13}u_{6}(1)=0 $$ (ab103)
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$$\displaystyle u_{0}(0)a_{70}+u_{1}(0)a_{71}+u_{2}(0)a_{72}+u_{3}(0)a_{73}+u_{4}(0)a_{74}+u_{5}(0)a_{75}+u_{6}(0)a_{76}+u_{7}(0)a_{77}=0,\;\;\;\;\;\;\; u_{0}(1)a_{70}+u_{1}(1)a_{71}+u_{2}(1)a_{72}+u_{3}(1)a_{73}+u_{4}(1)a_{74}+u_{5}(1)a_{75}+u_{6}(1)a_{76}+u_{7}(1)a_{77}=p_7 $$ $$\displaystyle \Downarrow$$ $$\displaystyle \frac{1}{13}u_{6}(0)=0,\;\;\;\;\;\;\; \frac{1}{13}u_{6}(1)=0 $$ (ab103)
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Solution of deterministics
For simplicity, only the solution When N = 2 is provided.

$$\displaystyle \sigma = 0.1 $$

$$\displaystyle \sigma = 0.5 $$

$$\displaystyle \sigma = 0.9 $$

The mean-square error, $$\displaystyle e_2(x)$$, of the numerical solution from the gPC expansion $$\displaystyle u_p(x,\omega)$$ is computed $$\displaystyle e_2(x) = (\mathbb E[u_p(x,\omega)-u_e(x,\omega)]^2)^{(1/2)}$$ The $$\displaystyle L_\infty$$ norm of $$\displaystyle e_2(x)$$ is plotted with respect to $$\displaystyle p$$, the order of polynomial chaos expansion.



Hermite Chaos and Normal distribution
In this subsection, we assume $$\displaystyle \epsilon(\omega)$$ is Gaussian random variable, with pdf
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$$\displaystyle f(\epsilon)=\frac{1}{\sqrt{2\pi}}e^{-\epsilon^2/2},\;\;\;\epsilon \in (-\infty,\infty). $$ (ab30)
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Except the basis is different from Legendre polynomial, the procedure is exactly the same. Which chaos to be choose totally depends on the distribution of the random variable, $$\displaystyle \epsilon(\omega) $$.

=Two-dimensional model=

$$\displaystyle $$

=References=