*Question 2a* The given equations are clearly homogenous and linear. The discriminant of $u_{tt}-4u_{xx}=0$ is given by $0^2 -(1)(-4)=4>0$ hence this equation is hyperbolic. The discriminant of $u_{t}=8u_{xx}$ is $0$ hence this equation is parabolic. The discriminant of $u_{xx}+u_{yy}=0$ is given by $0^2-(1)(1)=-1<0$ hence this equation is elliptic. *Question 2b* WTS $u(x,t)=\frac{1}{\sqrt{ 4\pi kt }}e^{-x^{2}/4kt}$ is the solution to the heat equation $u_{t}=ku_{xx}$ where $x\in \mathbb{R},t>0.$ Using the product rule, it is derivatives are given $\begin{align} u_{t}&=\frac{x^{2}}{4kt^2\sqrt{ 4\pi kt }} e^{-x^{2}/4kt} - \frac{1}{2t\sqrt{ 4\pi kt }}e^{-x^{2}/4kt} \\ &= e^{-x^{2}/4kt} \left( \frac{x^{2}-2kt}{4kt^2 \sqrt{ 4\pi kt }} \right) \\ u_{x} &= -\frac{x}{2kt\sqrt{ 4\pi kt }}e^{-x^{2}/4kt} \\ u_{xx} & = \frac{x^2}{4k^{2}t^{2} \sqrt{ 4\pi k t }} e^{-x^{2}/4kt} - \frac{1}{2kt\sqrt{ 4\pi kt }}e^{-x^2/4kt} \\ &= e^{-x^{2}/4kt} \left( \frac{x^{2}-2kt}{4k^{2}t^{2}\sqrt{ 4\pi kt } } \right) \end{align}$so indeed $u(x,t)$ satisfies $u_{t}=ku_{xx}.$ $\square$ Setting $k=0.5$ gives $u(x,t)=\frac{1}{\sqrt{ 2\pi t }}e^{-x^2/2t}.$ Below is a sketch of $u$ at different times ![[Pasted image 20241012142314.png|600]] As $t\to0,$ $u(0,t)\to \infty$ and for all $x\neq0,$ $u(x,t)\to0.$ $\square$ *Question 2c* Assume $u^*$ is a solution to $(2).$ Then $u(x)=u^*(x)+A$is also a solution for all $A\in \mathbb{R},$ since $u_{x}(x)=u^{*}_{x}(x) \implies u_{x}(0)=0=u_{x}(1)$ and $u_{xx}(x)=u^{*}_{xx}(x)=f(x).$ $\square$ Integrating $(2a)$ with respect to $x$ between $0$ and $1$ and using the Fundamental Theorem of Calculus (FTC), we get $\int_{0}^{1} f(\tilde{x})\; d\tilde{x}=\int_{0}^{1} u_{xx}(\tilde{x}) \; d \tilde{x} = u_{x}(1)-u_{x}(0)=0. $ On the other hand, assume $\int_{0}^{1} f(x) dx = 0.$ For $x\in [0,1],$ define $u(x)=\int_{0}^{x}\int_{0}^{x}f(\tilde{x})\;d\tilde{x}\; d\tilde{x}$Then by FTC, $u_{xx}(x)=f(x)$ and $u_{x}(x)=\int_{0}^{x} f(\tilde{x})\;d\tilde{x}$ which gives $u_{x}(0)=\int_{0}^{0}f(\tilde{x})\; d\tilde{x}=0= \int_{0}^{1}f(\tilde{x})\; d\tilde{x}=u_{x}(1)$Therefore $u(x)$ is a solution. That is, $\int_{0}^{1}f(x)\;dx=0$ is a necessary condition for the existence of a solution for $(2)$. $\square$ *Question 2d* Solving its characteristic equation $\xi'(t)=\frac{1+t^{2}}{2},\quad \xi(0)=x_{0}$we get$\xi(t)=\frac{t}{2}+\frac{t^{3}}{6}+x_{0}.$ Now $x=\xi(t)= \frac{t}{2}+\frac{t^{3}}{6}+x_{0} \iff x_{0} = x-\frac{t^{2}}{2} -\frac{t^{3}}{6}$so that the solution is $u(x,t)=\Phi(x_{0})=1-2x+t+\frac{t^{3}}{3}.$ *Question 2e* Let $u=ve^{\alpha x+\beta y}.$ Compute its partial derivatives $\begin{align} u_{x}&=e^{\alpha x+\beta y}(v_{x} + \alpha v) &u_{y}&=e^{\alpha x+\beta y}(v_{y}+\beta v) \\ u_{xx} &= e^{\alpha x+ \beta y}(v_{xx}+2\alpha v_{x}+\alpha^2v) &u_{yy} &= e^{\alpha x+\beta y} (v_{yy}+2\beta v_{y}+\beta^{2}v ) \end{align}$Substituting these in the elliptic equation, we get $\begin{align} 0 = e^{\alpha x +\beta y}(v_{xx}&+2\alpha v_{x}+\alpha^{2}v &+ 3v_{yy}+6\beta v_{y}+3\beta^{2}v \\ &-2v_{x}-2\alpha v &24v_{y}+24\beta v \\ & &+5v ) \end{align}$Since $e^{\alpha x+\beta y}>0,$ we get$0=v_{xx}+3v_{yy}+(2\alpha-2)v_{x}+(6\beta+24)v_{y}+(\alpha^{2}-2\alpha+3\beta^{2}+24\beta+5)v$Setting $\alpha=1,\beta=-4,$ we get $0=v_{xx}+3v_{yy}-44v$ Now by repeatedly using the chain rule, $\frac{\partial^2 v}{\partial y'^{2}}=\left( \frac{dy}{dy'} \right)^2 \frac{\partial^{2} v}{\partial y^{2}}$hence setting $y'=\gamma y$ where $\gamma=\frac{1}{\sqrt{ 3 }}$ gives $v_{xx}+v_{y'y'}-44v=0.$