### Author Topic: HA5-P4  (Read 4544 times)

#### Catch Cheng

• Jr. Member
• Posts: 12
• Karma: 0
##### HA5-P4
« on: October 17, 2015, 04:52:25 PM »

#### Catch Cheng

• Jr. Member
• 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

• Jr. Member
• Posts: 12
• Karma: 0
##### Re: HA5-P4
« Reply #2 on: October 17, 2015, 06:13:06 PM »
This is 4(b).

#### Catch Cheng

• Jr. Member
• Posts: 12
• Karma: 0
##### Re: HA5-P4
« 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

• Full Member
• Posts: 19
• Karma: 6
##### Re: HA5-P4
« 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

• Full Member
• Posts: 29
• Karma: 0
##### Re: HA5-P4
« 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

u(0,t) = v(0,t)e^{\beta t} = 0 \rightarrow v(0,t) = 0

This is a valid boundary condition for v.

Now we figure out how $g(x)$ changed here.

u(x,0) = v(x,0)e^{\alpha x} = g(x) = e^{-\epsilon |x|} \rightarrow v(x,0) = e^{-\epsilon |x| - \alpha x}

Where $\alpha = \frac{c}{2k}$

Then general solution of Dirichlet problem gives

u(x,t) = \int_0^\infty (G(x,y,t) - G(x,-y,t))e^{-(\epsilon + \frac{c}{2k}) y}dy

We can take $y$ out from $|y|$ since its all positive.

Do the same thing Catch has been doing. The final answer is

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}}))

#### Victor Ivrii

$\newcommand{\erf}{\operatorname{erf}}$
$\newcommand{\erf}{\operatorname{erf}}$and then \erf(x) results not in $\Erf(x)$ but in $\erf(x)$