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APM3462015F => APM346Home Assignments => HA5 => Topic started by: Rong Wei on October 16, 2015, 08:30:07 PM

http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter3/S3.2.P.html

I'm not sure :[

(a) Here is my typed solution:
We have the Dirichlet problem \begin{equation}
\begin{cases}
u_t = ku_{xx} & x>0, t>0 \\
u_{t=0} = g(x) \\
u_{x=0} = 0
\end{cases}
\label{eq:problem}
\end{equation}
By method of continuation, we extend the initial function to the full line (recalling the odd extension for Dirichlet problems): \begin{equation}
g_{extended}(x) = \begin{cases} g(x) & x>0 \\
g(x) & x<0 \\
0 & x=0
\label{eq:gext}
\end{cases}
\end{equation}
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. \begin{equation}
w(x,t) = \int_{\infty}^{\infty} G(x,y,t)g_{extended}(y)dy
\end{equation}
So using the definition of $g_{extended}$ as given in \ref{eq:gext}, we can rewrite this as: \begin{equation}
w(x,t) = \int_{0}^{\infty} G(x,y,t)g(y)dy  \int_{\infty}^{0} G(x,y,t)g(y)dy
\end{equation}
In the second equation above, let's make a change of variables from y to y. Then this yields: \begin{equation}
w(x,t) = \int_0^{\infty}[G(x,y,t)  G(x, y, t)]g(y)dy \end{equation}
This is our solution. Written in terms of the definition of G, we arrive at the final solution to \ref{eq:problem}: \begin{equation}
u(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{(xy)^2/(4kt)}  e^{(x+y)^2/(4kt)}]g(y)dy \end{equation}
(b) Part (b) is very similar, only here we have a Neumann problem: \begin{equation}
\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}
\end{equation}
Recall that for Neumann problems, the continuation is even. So we write the extension of the initial function to the entire line: \begin{equation}
g_{extended}(x) = \begin{cases} g(x) & x \geq 0 \\
g(x) & x \leq 0 \end{cases}
\label{eq:geven}
\end{equation}
Proceeding similarly as in part (a), the solution for the whole line will be: \begin{equation}
w(x,t) = \int_{\infty}^{\infty} G(x,y,t)g_{extended}(y)dy
\end{equation}
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}: \begin{equation}
w(x,t) = \int_{0}^{\infty} G(x,y,t)g(y)dy + \int_{\infty}^{0} G(x,y,t)g(y)dy
\end{equation}
Let's again make the change of variables, from y to y. We arrive at: \begin{equation}
w(x,t) = \int_{0}^{\infty} [G(x,y,t) + G(x,y,t)]g(y)dy
\end{equation}
Using the definition of G as before, we arrive at the final solution to \ref{eq:neumann}: \begin{equation}
u(x,t) = \frac{1}{2\sqrt{\pi{}kt}}\int_0^{\infty}[e^{(xy)^2/(4kt)} + e^{(x+y)^2/(4kt)}]g(y)dy \end{equation}

@ 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.

yes! emily, I believe so

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