### Author Topic: HA5-P1  (Read 4304 times)

#### Rong Wei

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##### HA5-P1
« on: October 16, 2015, 08:30:07 PM »
« Last Edit: October 17, 2015, 04:19:04 PM by Victor Ivrii »

#### Rong Wei

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##### Re: HA4-P1
« Reply #1 on: October 16, 2015, 08:58:54 PM »
I'm not sure

#### Emily Deibert

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##### Re: HA5-P1
« Reply #2 on: October 17, 2015, 05:48:59 PM »
(a) Here is my typed solution:

We have the Dirichlet problem
\begin{cases}
u_t = ku_{xx} & x>0, t>0 \\
u|_{t=0} = g(x) \\
u|_{x=0} = 0
\end{cases}
\label{eq:problem}

By method of continuation, we extend the initial function to the full line (recalling the odd extension for Dirichlet problems):
g_{extended}(x) = \begin{cases} g(x) & x>0 \\
-g(-x) & x<0 \\
0 & x=0
\label{eq:gext}
\end{cases}

So for the entire line (i.e. $-\infty<x<\infty$), we arrive at a solution with the formula as given in the textbook; call it w. The solution to the given problem \ref{eq:problem} will just be this solution in the region of x>0.
w(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g_{extended}(y)dy

So using the definition of $g_{extended}$ as given in \ref{eq:gext}, we can rewrite this as:
w(x,t) = \int_{0}^{\infty} G(x,y,t)g(y)dy - \int_{-\infty}^{0} G(x,y,t)g(-y)dy

In the second equation above, let's make a change of variables from -y to y. Then this yields:
w(x,t) = \int_0^{\infty}[G(x,y,t) - G(x, -y, t)]g(y)dy
This is our solution. Written in terms of the definition of G, we arrive at the final solution to \ref{eq:problem}:
u(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/(4kt)} - e^{-(x+y)^2/(4kt)}]g(y)dy

(b) Part (b) is very similar, only here we have a Neumann problem:
\begin{cases}
u_t = ku_{xx} & x>0, t>0 \\
u|{t=0} = g(x) \\
u_x|{x=0} = 0
\end{cases}
\label{eq:neumann}

Recall that for Neumann problems, the continuation is even. So we write the extension of the initial function to the entire line:
g_{extended}(x) = \begin{cases} g(x) & x \geq 0 \\
g(-x) & x \leq 0 \end{cases}
\label{eq:geven}

Proceeding similarly as in part (a), the solution for the whole line will be:
w(x,t) = \int_{-\infty}^{\infty} G(x,y,t)g_{extended}(y)dy

The solution to \ref{eq:neumann} will just be this solution in the region x>0.
So let's write this in terms of the definition in \ref{eq:geven}:
w(x,t) = \int_{0}^{\infty} G(x,y,t)g(y)dy + \int_{-\infty}^{0} G(x,y,t)g(-y)dy

Let's again make the change of variables, from -y to y. We arrive at:
w(x,t) = \int_{0}^{\infty} [G(x,y,t) + G(x,-y,t)]g(y)dy

Using the definition of G as before, we arrive at the final solution to \ref{eq:neumann}:
u(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{-(x-y)^2/(4kt)} + e^{-(x+y)^2/(4kt)}]g(y)dy
« Last Edit: October 17, 2015, 06:00:01 PM by Emily Deibert »

#### Emily Deibert

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##### Re: HA5-P1
« Reply #3 on: October 17, 2015, 06:00:50 PM »
@ Rong Wei, I believe we both got the same solution, just that you wrote yours in terms of $G_0$ and I wrote mine in terms of the exponential definition of G.

#### Rong Wei

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##### Re: HA5-P1
« Reply #4 on: October 17, 2015, 06:27:23 PM »
yes! emily, I believe so

#### Victor Ivrii

Result is the same but the complete answer should not mention continuation but contain integral from $0$ to $\infty$ and $\bigl[G_0(x-y,t) \pm G_0(x+y,t)\bigr]= G(x,y,t)$ for Neumann/Dirichlet b.c. and $G(x,y,t)$ is a Green's function for IBVP here. See also problem on the interval