Science:Math Exam Resources/Courses/MATH307/December 2012/Question 02 (a)

MATH307 December 2012

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Question 02 (a)
The boundary value problem ${\begin{aligned}&f''(x)+xf(x)=1,\quad 0<x<1\\&f(0)=1,\quad f'(1)=1\end{aligned}}$ can be approximated by an $(N+1)\times (N+1)$ system of linear equations of the form $(L+(\Delta x)^{2}Q)\mathbf {F} =\mathbf {b}$ (a) Write down L, Q, b and $\Delta x$ when N = 4.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. With N = 4, we will consider f(x) at 5 evenly spaced values of x between 0 to 1, where $x_{n}={\frac {n}{N}},\ 0<n<N$. This gives us 4 segments joining the points. For $f(x_{n})$ where $0<n<N$, we can approximate Failed to parse (syntax error): {\displaystyle f′′(x_{n}) = \frac{(f(x_{n+1}) - f(x_{n})) - (f(x_{n}) - f(x_{n-1}))}{(\Delta x)^{2}}} for $0<n<N$ This is equivalent to ${\frac {1}{(\Delta x)^{2}}}{\begin{bmatrix}1&&-2&&1&&0&&0\\0&&1&&-2&&1&&0\\0&&0&&1&&-2&&1\end{bmatrix}}{\begin{bmatrix}f(x_{0})\\f(x_{1})\\f(x_{2})\\f(x_{3})\\f(x_{4})\end{bmatrix}}$ Since we know that $f(0)=1$ and $f'(1)=1$, we can add a row to this matrix to include these boundary conditions to get L $L={\frac {1}{(\Delta x)^{2}}}{\begin{bmatrix}(\Delta x)^{2}&&0&&0&&0&&0\\1&&-2&&1&&0&&0\\0&&1&&-2&&1&&0\\0&&0&&1&&-2&&1\\0&&0&&0&&-\Delta x&&\Delta x\end{bmatrix}}$ To find Q, we will consider the second half of the ODE: $xf(x)$. We have $f''(x_{n})$ for $0<n<N$, so we will find $x(f(x_{n})$ for these points. We can ignore n = 0 and n = N because here, we are considering their boundary conditions instead. Since in the equation, Q is multiplied by $(\Delta x)^{2}$, we divide by $(\Delta x)^{2}$. $Q={\frac {1}{(\Delta x)^{2}}}{\begin{bmatrix}0&0&0&0&0\\0&\Delta x&0&0&0\\0&0&2\Delta x&0&0\\0&0&0&3\Delta x&0\\0&0&0&0&0\end{bmatrix}}={\frac {1}{(\Delta x)}}{\begin{bmatrix}0&0&0&0&0\\0&1&0&0&0\\0&0&2&0&0\\0&0&0&3&0\\0&0&0&0&0\end{bmatrix}}$ We will combine L and Q to determine b $(L+(\Delta x)^{2}Q){\vec {F}}={\vec {b}}$ We have arranged rows 1 to 3 of the matrices to correspond to the left hand side of the ODE when plugged into this equation, so the corresponding rows of ${\vec {b}}$ will have entries equal to 1. Row 0 and 4 will have entries equal to their corresponding boundary conditions, so both are equal to one as well. ${\vec {b}}={\begin{bmatrix}1\\1\\1\\1\\1\end{bmatrix}}$ $\Delta x$ is the spacing between the points. Since the points go from $0<x<1$, are evenly spaced and we have 5 of them, $\Delta x={\frac {1}{4}}=0.25$ We can plug $\Delta x=0.25$ into our expressions for Q and L above to get $L=16{\begin{bmatrix}{\frac {1}{16}}&&0&&0&&0&&0\\1&&-2&&1&&0&&0\\0&&1&&-2&&1&&0\\0&&0&&1&&-2&&1\\0&&0&&0&&{\frac {-1}{4}}&&{\frac {1}{4}}\end{bmatrix}}$ Q = $4{\begin{bmatrix}0&0&0&0&0\\0&1&0&0&0\\0&0&2&0&0\\0&0&0&3&0\\0&0&0&0&0\end{bmatrix}}$