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$\newcommand{\erf}{\operatorname{erf}}$
Find solution $u(x,t)$ to
\begin{align}
&u_t=u_{xx} && -\infty<x<\infty, \ t>0,\label{eq-5-1}\\[4pt]
&u|_{t=0}=\left\{\begin{aligned}
&x\qquad && |x|<1,\\[2pt]
&0 && |x|>1
\end{aligned}\right.,\\[4pt]
&\max |u|<\infty.
\end{align}
Calculate integral.
Hint:
For $u_t=ku_{xx}$ use $\displaystyle{G(x,t)=\frac{1}{\sqrt{4\pi kt}}\exp (- (x-y)^2/4kt)}$.
To calculate integral make change of variables and use $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-z^2}\,dz}$.
Find solution $u(x,t)$ to
\begin{align}
&u_t=u_{xx} && -\infty<x<\infty, \ t>0,\label{eq-5-1}\\[4pt]
&u|_{t=0}=\left\{\begin{aligned}
&x\qquad && |x|<1,\\[2pt]
&0 && |x|>1
\end{aligned}\right.,\\[4pt]
&\max |u|<\infty.
\end{align}
Calculate integral.
Hint:
For $u_t=ku_{xx}$ use $\displaystyle{G(x,t)=\frac{1}{\sqrt{4\pi kt}}\exp (- (x-y)^2/4kt)}$.
To calculate integral make change of variables and use $\displaystyle{\erf(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-z^2}\,dz}$.
458
Consider the PDE with boundary conditions:
\begin{align}
&u_{tt}-c^2u_{xx} + \omega^2 u =0,\qquad&&0<x<L,\label{eq-4-1}\\[2pt]
&(u_x -\alpha u)|_{x=0}=0,\label{eq-4-2}\\[2pt]
&(u_x +\beta u)|_{x=L}=0\label{eq-4-3}
\end{align}
where $c>0$ and $\omega>0$ areconstant. Prove that the energy $E(t)$ defined as
\begin{equation}
E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + c^2u_{x}^2 + \omega^2 u^2)\,dx +\frac{c^2\alpha}{2}u(0,t)^2+
\frac{c^2\beta}{2}u(L,t)^2
\end{equation}
does not depend on $t$.
\begin{align}
&u_{tt}-c^2u_{xx} + \omega^2 u =0,\qquad&&0<x<L,\label{eq-4-1}\\[2pt]
&(u_x -\alpha u)|_{x=0}=0,\label{eq-4-2}\\[2pt]
&(u_x +\beta u)|_{x=L}=0\label{eq-4-3}
\end{align}
where $c>0$ and $\omega>0$ areconstant. Prove that the energy $E(t)$ defined as
\begin{equation}
E(t)= \frac{1}{2}\int_0^L \bigl( u_t^2 + c^2u_{x}^2 + \omega^2 u^2)\,dx +\frac{c^2\alpha}{2}u(0,t)^2+
\frac{c^2\beta}{2}u(L,t)^2
\end{equation}
does not depend on $t$.
461
Consider the first order equation:
\begin{equation}
(t^2+1)u_t + u_x = u.
\label{eq-1-1}
\end{equation}
(a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
(b) Write the general solution.
(c) Solve equation (\ref{eq-1-1}) with the initial condition $u(x,0)= \cos(x)$. Explain why the solution is fully determined by the initial condition.
\begin{equation}
(t^2+1)u_t + u_x = u.
\label{eq-1-1}
\end{equation}
(a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
(b) Write the general solution.
(c) Solve equation (\ref{eq-1-1}) with the initial condition $u(x,0)= \cos(x)$. Explain why the solution is fully determined by the initial condition.
464
Problem 9: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.9
Reflects new numeration of the problems
Reflects new numeration of the problems
465
Problem 8: http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.6.P.html#problem-2.6.8
Reflects new numeration of the problems
Reflects new numeration of the problems