Question 3.4: Find a weak solution to the following boundary value problem...
Find a weak solution to the following boundary value problem:
\frac{d^{2}u }{dx^{2} }=6x-\sin (x),0\leq x\leq 1subjected to :
\left.\begin{matrix} u(x)\mid _{x=1}=\sin (1) \\\frac{du}{dx}\mid _{x=0}=0 \end{matrix} \right\} (3.28)The weighted residual method is used in the steps as follows.
Step 1. Equation (3.28) has one differential equation (N_{D}=1 ) and two boundary conditions (N_{BC}=2 ) . Therefore, the number of the polynomial terms (N) must be larger than N_{D}+ N_{BC}=3 . Let N be 4, and the approximate solution is assumed as
\tilde{u} (x)=c_{0}+ c_{1}x+ c_{2}x^{2}+ c_{3}x^{3} (3.29)
Equation (3.29) has unknown constants (c_{0},c_{1},c_{2},c_{3} ); thus, four equations about (c_{0},c_{1},c_{2},c_{3} ) are derived from the conditions of the weak solution.
Step 2. Equation (3.28) involves in two boundary conditions, substituting Eq. (3.29) into Eq. (3.28) gives
Using Eq. (3.30) in Eq. (3.29) gets
\tilde{u}(x)=(\sin (1) – c_{2} -c_{3}) + c_{2}x^{2}+ c_{3}x^{3} (3.31)
Equation (3.31) is used to obtain the first and second derivatives of \tilde{u}(x) as
\left.\begin{matrix}\frac{d\tilde{u}(x) }{dx}=2c_{2}x+ 3c_{3}x^{2} \\\\ \frac{d^{2} \tilde{u}(x) }{dx^{2} }=2c_{2}+ 6c_{3}x \end{matrix} \right\} (3.32)Step 3. Equation (3.31) include two unknown constants (c_{2}, c_{3}). Therefore, two different test functions have to be selected to define another two equations by using the condition of a weak-solution (Eq. 3.23). Using Eq. (3.32) into Eq. (3.28) yields the expression of the residual as
\bar{\Re }(x)=\int\limits_{0}^{1}{ν(x) · ((2c_{2}+ 6c_{3}x)-6x+ \sin (x))dx}=0 (3.33)Next, different methods are used in selecting test functions ν(x) to evaluate residuals in Eq. (3.33).