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| | \end{equation} | | \end{equation} |
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| − | ===Solutions===
| |
| | | | |
| − | 1. Using the boundary conditions, we get a solution that looks like
| + | [[Tutorial Week 11 Solutions]] |
| − | $$u(x,t) =\sum\limits_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{2}\right)e^{-n^2\pi^2t}$$['''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 $n=1,2,\dots$.
| + | |
Latest revision as of 23:09, 30 December 2020
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}
Tutorial Week 11 Solutions
