Difference between revisions of "Tutorial Week 11"
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\right. | \right. | ||
\end{equation} | \end{equation} | ||
| + | |||
| + | ===Solutions=== | ||
| + | |||
| + | 1. Using the boundary conditions, we get a solution that looks like | ||
| + | $$u(x,t) =\sum\limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t}$$['''1 pt''' for form]. | ||
| + | Using the initial condition we can find $B_n$ | ||
| + | $$B_n = \frac{1}{2}\int_{-2}^{2}x\sin\left(\dfrac{n\pi x}{2}\right)dx=\int_{0}^{2}x\sin\left(\dfrac{n\pi x}{2}\right)dx$$['''1 pt''' for coefficient integral]. | ||
| + | Solving, we get | ||
| + | \begin{equation} | ||
| + | B_n = (-1)^{n+1}\dfrac{4}{n \pi} | ||
| + | \end{equation}['''1 pt''' for coefficient value]. | ||
| + | |||
| + | 2. $u(x,t)$ has steady state solution $s(x)=1+2x$ ['''1 pt''']. | ||
| + | Defining $w(x,t)=u(x,t)-s(x)$: | ||
| + | \begin{equation} | ||
| + | \left\{ | ||
| + | \begin{array}{lr} | ||
| + | w_{t} = 4w_{xx}\\ | ||
| + | w(0,t) = 0\\ | ||
| + | w(2,t) = 0\\ | ||
| + | w(x,0)=-x, | ||
| + | \end{array} | ||
| + | \right. | ||
| + | \end{equation} ['''1 pt''' for initial condition]. | ||
| + | which from question 1 has solution: | ||
| + | $$w(x,t) =\sum\limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t},$$ | ||
| + | where | ||
| + | \begin{equation} | ||
| + | B_n =- (-1)^{n+1}\dfrac{4}{n \pi}=(-1)^{n}\dfrac{4}{n \pi} | ||
| + | \end{equation} ['''1 pt''' for correctly calculating $w$]. | ||
| + | Thus, | ||
| + | $$u(x,t) =1+2x+\sum\limits_{n=1}^\infty (-1)^{n}\dfrac{4}{n \pi}\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t}$$ ['''1 pt''' for putting it all together]. | ||
| + | |||
| + | |||
| + | 3. To satisfy the boundary condition $u(0,t)=0$, the trig function is of the form $\sin(\omega x)$ ['''1 pt''']. To satisfy the boundary condition $u_{x}(2,t) = 0$, $\cos(2\omega)=0$. Thus, $2\omega=\frac{\pi}{2}+(n-1)\pi$ for $n=1,2,\dots$ ['''1 pt''']. Equivalently, $\omega=\frac{(2n-1)\pi}{4}$ for $n=1,2,\dots$. | ||
Revision as of 14:12, 13 April 2017
Worksheet Questions
You can print the PDF.
1. Solve the following 1D heat equation defined on the interval $0 \leq x \leq 2$, and having the initial condition $u(x,0) = x$ for $0\leq x \leq 2$. \begin{equation} \left\{ \begin{array}{lr} u_{t} = 4u_{xx}\\ u(0,t) = 0\\ u(2,t) = 0 \end{array} \right. \end{equation} 2. Solve the following 1D heat equation defined on the interval $0 \leq x \leq 2$, and having the initial condition $u(x,0) = x+1$ for $0\leq x \leq 2$. \begin{equation} \left\{ \begin{array}{lr} u_{t} = 4u_{xx}\\ u(0,t) = 1\\ u(2,t) = 5 \end{array} \right. \end{equation} 3. What family of trig function should you use in order to solve the following 1D heat equation defined on the interval $0 \leq x \leq 2$, and having the initial condition $u(x,0) = x$ for $0\leq x \leq 2$? Specify the function, $\sin(\omega x)$ or $\cos(\omega x)$, and the spatial frequencies $\omega$ as a function of an integer $n$. \begin{equation} \left\{ \begin{array}{lr} u_{t} = 4u_{xx}\\ u(0,t) = 0\\ u_{x}(2,t) = 0 \end{array} \right. \end{equation}
Solutions
1. Using the boundary conditions, we get a solution that looks like $$u(x,t) =\sum\limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t}$$[1 pt for form]. Using the initial condition we can find $B_n$ $$B_n = \frac{1}{2}\int_{-2}^{2}x\sin\left(\dfrac{n\pi x}{2}\right)dx=\int_{0}^{2}x\sin\left(\dfrac{n\pi x}{2}\right)dx$$[1 pt for coefficient integral]. Solving, we get \begin{equation} B_n = (-1)^{n+1}\dfrac{4}{n \pi} \end{equation}[1 pt for coefficient value].
2. $u(x,t)$ has steady state solution $s(x)=1+2x$ [1 pt]. Defining $w(x,t)=u(x,t)-s(x)$: \begin{equation} \left\{ \begin{array}{lr} w_{t} = 4w_{xx}\\ w(0,t) = 0\\ w(2,t) = 0\\ w(x,0)=-x, \end{array} \right. \end{equation} [1 pt for initial condition]. which from question 1 has solution: $$w(x,t) =\sum\limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t},$$ where \begin{equation} B_n =- (-1)^{n+1}\dfrac{4}{n \pi}=(-1)^{n}\dfrac{4}{n \pi} \end{equation} [1 pt for correctly calculating $w$]. Thus, $$u(x,t) =1+2x+\sum\limits_{n=1}^\infty (-1)^{n}\dfrac{4}{n \pi}\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t}$$ [1 pt for putting it all together].
3. To satisfy the boundary condition $u(0,t)=0$, the trig function is of the form $\sin(\omega x)$ [1 pt]. To satisfy the boundary condition $u_{x}(2,t) = 0$, $\cos(2\omega)=0$. Thus, $2\omega=\frac{\pi}{2}+(n-1)\pi$ for $n=1,2,\dots$ [1 pt]. Equivalently, $\omega=\frac{(2n-1)\pi}{4}$ for $n=1,2,\dots$.
