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APM346-2015F => APM346--Home Assignments => HA4 => Topic started by: Emily Deibert on October 09, 2015, 05:07:44 PM

Title: HA4-P1
Post by: Emily Deibert on October 09, 2015, 05:07:44 PM

Problem 1: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.1 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.1)

(a) Here we have a Dirichlet boundary condition. In the domain ${t>0, x \geqslant ct}$, the solution is given by the D'Alembert formula:
\begin{equation}
u(x,t) = \frac{1}{2}[g(x+ct) + g(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} h(y)dy
\end{equation}
So plugging in the initial conditions as specified in the problem, we arrive at the solution:
\begin{equation}
u(x,t) = \phi(x+ct)
\end{equation}
In the domain ${0<x<ct}$, the solution is given by:
\begin{equation}
u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_{0}^{x+ct} h(y)dy + p\bigg(t - \frac{x}{c}\bigg) - \frac{1}{2}g(ct-x) - \frac{1}{2c} \int_0^{ct-x} h(y)dy
\end{equation}
So plugging in the initial and boundary conditions as specified in the problem, we arrive at the solution:
\begin{equation}
u(x,t) = \phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x)
\end{equation}
So the solution to (a) is
\begin{equation}
u(x,t) =
\begin{cases}
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) + \chi\bigg(t-\frac{x}{c}\bigg) - \phi(ct-x) & \text{for } \{0<x<ct\}
\end{cases}
\end{equation}


(b) In this case we have a Neumann boundary condition. The solution for the domain ${t>0, x \geqslant ct}$ is the same as in part (a), namely:
\begin{equation}
u(x,t) = \phi(x+ct)
\end{equation}
For the domain ${0<x<ct}$, the solution is given by:
\begin{equation}
u(x,t) = \frac{1}{2}g(x+ct) + \frac{1}{2c} \int_0^{x+ct} h(y)dy - c \int_0^{t-\frac{x}{c}} q(t')dt' + \frac{1}{2}g(ct-x) + \frac{1}{2c} \int_0^{ct-x} h(y)dy
\end{equation}
Plugging in our initial and boundary conditions as specified in the problem, we arrive at the solution:
\begin{equation}
u(x,t) = \phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x)
\end{equation}
where
\begin{equation}
\int\chi(t')dt' = X(t')
\end{equation}
So the solution to (b) is
\begin{equation}
u(x,t) =
\begin{cases}
\phi(x+ct) & \text{for } \{t>0, x \geqslant ct\}\\
\phi (x+ct) - \phi (0) - cX\bigg(t-\frac{x}{c}\bigg) + cX(0) + \phi(ct-x) & \text{for } \{0<x<ct\}
\end{cases}
\end{equation}

Title: Re: HA4-P1
Post by: Rong Wei on October 14, 2015, 12:51:25 AM

My solution is Ï•(x+ct)−1/c∫^ct−x₀X(t−xc)+Ï•(ct−x)for for {0<x<ct},  Because the boundary condition is U_x|x=0=χ(t), so we should integral χ(t). I plug in the derivative of U respect to x, and I got this answer, may be I'm wrong? I'm not sure,  :-[

Title: Re: HA4-P1
Post by: Rong Wei on October 14, 2015, 06:17:43 PM

Now I understand its the part we discussed should be c integrate X（x） from t - x/c to 0