# 2.8 Method of Weighted Residuals

## 2.8.2 The Collocation Method

One approach to determine the $$N$$ unknown values of $$a_ j$$ would be to enforce the governing PDE at $$N$$ spatial points in the domain (i.e., to make sure that the the approximate solution exactly satisfies the PDE at $$N$$ points). Note that in general, the exact solution will not be a linear combination of the $$\phi _ j(x)$$, so it will not be possible for our approximate solution to satisfy the PDE at every point in the domain. To see this, let's substitute our approximate solution $$\tilde{T}(x)= 100 + a_1 \phi _1(x) + a_2 \phi _2(x)$$ into Equation  2.149. First, let's derive an expression for $$\tilde{T}_{xx}$$:

 $$\displaystyle \tilde{T}_{xx}$$ $$\displaystyle =$$ $$\displaystyle \frac{\partial ^2}{\partial x^2}\left[a_1 \phi _1(x) + a_2 \phi _2(x)\right],$$ (2.162) $$\displaystyle (\phi _1)_{xx}$$ $$\displaystyle =$$ $$\displaystyle -2,$$ (2.163) $$\displaystyle (\phi _2)_{xx}$$ $$\displaystyle =$$ $$\displaystyle -6x,$$ (2.164) $$\displaystyle \Rightarrow \tilde{T}_{xx}$$ $$\displaystyle =$$ $$\displaystyle -2 a_1 - 6 a_2 x.$$ (2.165)

Next, we define a residual for Equation  2.149,

 $R(\tilde{T},x) \equiv \left(k \tilde{T}_ x\right)_ x + f.$ (2.166)

The residual tells us by how much the approximate solution does not satisfy the governing PDE. If the solution $$\tilde{T}$$ were exact, then $$R = 0$$ for all $$x$$. Now, substitution of our chosen $$\tilde{T}$$ into the residual (recall $$k=1$$ and $$f=50 e^ x$$ in this example) gives

 $R(\tilde{T},x) = -2 a_1 - 6 a_2 x + 50 e^ x. \label{equ:steady_ dif1d_ residual}$ (2.167)

Clearly, since $$a_1$$ and $$a_2$$ are constants (i.e., they do not depend on $$x$$), there is no way for this residual to be zero for all $$x$$.

By setting the residual to be zero at $$N$$ different points $$x$$, we will obtain $$N$$ equations that we can solve to determine the $$N$$ unknown coefficients. The question remains, where should the $$N$$ points be selected? The points at which the governing equation will be enforced are known as the collocation points. For our example with $$N=2$$ , we will choose the relatively simple approach of equally subdividing the domain with $$N=2$$ interior collocation points. For this domain from $$-1 \leq x \leq 1$$, the equi-distant collocation points would be at $$x = \pm 1/3$$. Thus, the two conditions for determining $$a_1$$ and $$a_2$$ are,

 $$\displaystyle R(\tilde{T},-1/3)$$ $$\displaystyle =$$ $$\displaystyle 0,$$ (2.168) $$\displaystyle R(\tilde{T},1/3)$$ $$\displaystyle =$$ $$\displaystyle 0.$$ (2.169)

From Equation  2.167 this gives,

 $$\displaystyle -2 a_1 + 2 a_2 + 50 e^{-1/3}$$ $$\displaystyle =$$ $$\displaystyle 0,$$ (2.170) $$\displaystyle -2 a_1 - 2 a_2 + 50 e^{1/3}$$ $$\displaystyle =$$ $$\displaystyle 0.$$ (2.171)

Re-arranging this into a matrix form gives,

 $\left(\begin{array}{rr} -2 & 2 \\ -2 & -2 \end{array}\right) \left(\begin{array}{c} a_1 \\ a_2 \end{array}\right) = \left(\begin{array}{l} -50 e^{-1/3} \\ -50 e^{1/3} \end{array}\right).$ (2.172)
 $\Rightarrow \left(\begin{array}{c} a_1 \\ a_2 \end{array}\right) = \left(\begin{array}{l} 25 \cosh (1/3) \\ 25 \sinh (1/3) \end{array}\right) = \left(\begin{array}{r} 26.402 \\ 8.489 \end{array}\right).$ (2.173)

The results using this collocation method are shown in the figures below which include plots of $$\tilde{T}(x)$$, the error (i.e., $$\tilde{T}(x)-T(x)$$), and the residual. Note that the residual is exactly zero at the collocation points (i.e., $$x = \pm 1/3$$), though the approximation is not exact at these points (i.e., $$\tilde{T} \neq T$$ at $$x = \pm 1/3$$). This is because the residual $$R(\tilde{T},x)$$ measures a different thing to the error $$\tilde{T}(x)-T(x)$$. Remember, the residual tells us by how much the PDE is not satisfied at a given point, so it relates to the balance between the terms in the PDE (in our case between the term $$\left(k \tilde{T}_ x\right)_ x$$ and the heat source term $$f(x)$$). Even if the PDE is satisfied at a particular point (i.e., the residual is zero at that point), the solution might not be exact at that point.

Figure 2.27: Comparison of $$T$$ (solid) and $$\tilde{T}$$ (dashed) for collocation method.
Figure 2.28: Error $$\tilde{T}-T$$ for collocation method.
Figure 2.29: Residual $$R(\tilde{T},x)$$ for collocation method.