### Author Topic: HA5-P4  (Read 2354 times)

#### Catch Cheng

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« on: October 17, 2015, 04:52:25 PM »

#### Catch Cheng

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•  • Posts: 12
• Karma: 0 ##### Re: HA5-P4
« Reply #1 on: October 17, 2015, 05:08:15 PM »
Please correct me if something is wrong, thank you.
« Last Edit: October 17, 2015, 06:11:54 PM by Catch Cheng »

#### Catch Cheng

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« Reply #2 on: October 17, 2015, 06:13:06 PM »
This is 4(b).

#### Catch Cheng

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« Reply #3 on: October 17, 2015, 08:45:19 PM »
This is 4(c) and 4(d), please correct me if something is wrong. Thanks.
« Last Edit: October 17, 2015, 08:58:42 PM by Catch Cheng »

#### Yeming Wen

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« Reply #4 on: October 20, 2015, 09:32:17 PM »
I think 4(c) is asking us to find the solution for convection heat equation, which it is in 3(e), not the ordinary heat equation.

#### Zaihao Zhou

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« Reply #5 on: October 20, 2015, 11:13:33 PM »
Yeming is right. I will post the result here. Correct me if I'm wrong.

Since the question asks a Dirichlet condition for the convection, in problem 3(c) we have shown the transformation $u(x,t) = U(x-ct,t)$ is not appropriate since $u(0,t) =U(-ct,t) =0$ is not a boundary condition for $U$.

Thus we use transformation $u(x,t) = v(x,t)e^{\alpha x + \beta t}$. Note in this case we have
\begin{equation}
u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0
\end{equation}
This is a valid boundary condition for v.

Now we figure out how $g(x)$ changed here.
\begin{equation}
u(x,0) = v(x,0)e^{\alpha x} = g(x) = e^{-\epsilon |x|} \rightarrow v(x,0) = e^{-\epsilon |x| - \alpha x}
\end{equation}
Where $\alpha = \frac{c}{2k}$

Then general solution of Dirichlet problem gives
\begin{equation}
u(x,t) = \int_0^\infty (G(x,y,t) - G(x,-y,t))e^{-(\epsilon + \frac{c}{2k}) y}dy
\end{equation}
We can take $y$ out from $|y|$ since its all positive.

Do the same thing Catch has been doing. The final answer is
\begin{equation}
u(x,t) = \frac{e^{c^2-2cx}}{2}(1+erf(\frac{x-ct}{\sqrt{4kt}})) - \frac{e^{c^2+2cx}}{2}(1+erf(\frac{x+ct}{\sqrt{4kt}}))
\end{equation}

#### Victor Ivrii ##### Re: HA5-P4
« Reply #6 on: October 21, 2015, 05:49:37 AM »
$\newcommand{\erf}{\operatorname{erf}}$

Yes, you are right.

PS Except to produce nice looking output we need to use \erf rather than erf. But wait! such command is not defined and to remedy it we do once on the top of the page (not in each post)
Code: [Select]
$\newcommand{\erf}{\operatorname{erf}}$and then \erf(x) results not in $\Erf(x)$ but in $\erf(x)$