$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ###[Problems](id:sect-2.1.P) > 1. [Problem 1](#problem-2.1.P.1) > 2. [Problem 2](#problem-2.1.P.2) > 3. [Problem 3](#problem-2.1.P.3) > 4. [Problem 4](#problem-2.1.P.4) > 5. [Problem 5](#problem-2.1.P.5) > 6. [Problem 6](#problem-2.1.P.6) **[Problem 1.](id:problem-2.1.P.1)** a. Draw characteristics and find the general solution to each of the following equations \begin{gather} 2 u\_t +3u\_x=0; \\\\ u\_t+ t u\_x=0; \\\\ u\_t +xu\_x=0; \\\\ u\_t+x^2u\_x=0 \\\\ u\_t+x^3u\_x=0. \end{gather} b. Consider IVP problem $u|\_{t=0}=f (x)$ as $-\infty\< x \< \infty$ ; does solution always exists? If not, what conditions should satisfy $f(x)$? c. Where this solution is uniquely determined? d. Consider this equation in $\\{t\>0, x\>0\\}$ with the initial condition $u|\_{t=0}=f(x)$ as $x\>0$; where this solution defined? Is it defined everywhere in $\\{t\>0, x\>0\\}$ or do we need to impose condition at $x=0$? In the latter case impose condition $u|\_{x=0}=g(t)$ ($t\>0$) and solve this IVBP; e. Consider this equation in $\\{t\> 0, x\<0\\}$ with the initial condition $u|\_{t=0}=f(x)$ as $x\<0$; where this solution defined? Is it defined everywhere in $\\{t\>0, x\< 0\\}$ or do we need to impose condition at $x=0$? In the latter case impose condition $u|\_{x=0}=g(t)$ ($t\>0$) and solve this IVBP; f. Consider problems (d) as $t\<0$; g. Consider problems (e) as $t\<0$; **[Problem 2.](id:problem-2.1.P.2)** a. Find the general solution to each of the following equations \begin{gather} xu\_x+ yu\_y=0,\\\\ xu\_x-yu\_y=0 \end{gather} in $\\{(x,y)\ne (0,0)\\}$; when this solution is continuous at $(0,0)$? Explain the difference between these two cases; b. Find the general solution to each of the following equations \begin{gather} yu\_x+ xu\_y=0,\\\\ yu\_x-xu\_y=0 \end{gather} in $\\{(x,y)\ne (0,0)\\}$; when this solution is continuous at $(0,0)$? Explain the difference between these two cases; **[Problem 3.](id:problem-2.1.P.3)** In the same way consider equations \begin{gather} (x^2+1)yu\_x+(y^2+1)xu\_y=0; \\\\ (x^2+1)yu\_x-(y^2+1)xu\_y=0. \end{gather} **[Problem 4.](id:problem-2.1.P.4)** Find the solution of \begin{equation} \left\\{\begin{aligned} &u\_x+3u\_y=xy,\\\\ &u|\_{x=0}=0. \end{aligned} \right. \end{equation} **[Problem 5.](id:problem-2.1.P.5)** Find the general solutions to each of \begin{gather} yu\_x-xu\_y=x; \\\\ yu\_x-xu\_y=x^2; \\\\ yu\_x+xu\_y=x; \\\\ yu\_x+xu\_y=x^2; \end{gather} In one instance solution does not exist. Explain why. **[Problem 6.](id:problem-2.1.P.6)** Solve IVP \begin{align} &u\_t+uu\_x=0,\qquad t\> 0; \\\\ &u|_{t=0}=f(x) \end{align} and describe domain in $(x,t)$ where this solution is properly defined with one of the following initial data \begin{align} f(x)&=\hphantom{-}\tanh (x); \\\\[5pt] f(x)&=-\tanh (x);\\\\ f(x)&=\left\\{\begin{aligned} -1& && x<-a,\\\\ x/a& && -a\le x \le a,\\\\ 1& && x>a; \end{aligned}\right. \\\\ f(x)&=\left\\{\begin{aligned} 1& && x<-a,\\\\ -x/a& && -a\le x \le a,\\\\ -1& && x>a; \end{aligned}\right. \\\\ f(x)&=\left\\{\begin{aligned} -1& && x<0,\\\\ 1& && x> 0;\\\\ \end{aligned}\right. \\\\ f(x)&=\hphantom{-}\sin(x). \\\\ f(x)&=\left\\{\begin{aligned} &\sin (x) && |x|\<\pi,\\\\ &0 && |x|\>\pi, \end{aligned}\right.\\\\ f(x)&=\left\\{\begin{aligned} &-\sin (x) && |x|\<\pi,\\\\ &0 && |x|\>\pi, \end{aligned}\right. \end{align} Here $a>0$ is a parameter. ______ [$\Uparrow$](../contents.html)  [$\uparrow$](./S2.1.html)  [$\Rightarrow$](./S2.2.html)