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Consider \begin{cases} u_t- 2 u_x = u^2 \\ u(x,0) = g(x) \end{cases}

Determine the region above the $x$-axis for which there is a classical solution.

Answer: First, we need to solve the IVP. Using the method of characteristics, we get the following IVPS:

  1. $\begin{cases} \dfrac{\,dt}{\,ds}=1\\ t(r,0)=0 \end{cases}$
  2. $\begin{cases} \dfrac{\,dx}{\,ds}=-2\\ x(r,0)=r \end{cases}$
  3. $\begin{cases} \dfrac{\,du}{\,ds}=u^2\\ u(r,0)=g(r) \end{cases}$

Solutions of the IVPS:

  1. $\begin{aligned}[t] \dfrac{\,dt}{\,ds}=1 \\ \,dt=\,ds \\ \int \,dt=\int \,ds \\ t = s + \mathcal{C} \end{aligned}$

    where $C$ is a constant. Using $t(r,0)=0$, we find that $\mathcal{C}=0$.

    Hence $t=s$.

  2. $\begin{aligned}[t] \dfrac{\,dx}{\,ds}=-2 \\ \,dx=-2\,ds \\ \int \,dx=-2\int \,ds \\ x = -2s + \mathcal{C} \end{aligned}$

    where $C$ is a constant. Using $x(r,0)=r$, we find that $\mathcal{C}=r$.

    Hence $x=-2s+r$.

  3. $\begin{aligned}[t] \dfrac{\,du}{\,ds}=u^2 \\ \dfrac{1}{u^2}\,du=\,ds \\ \int u^{-2}\,du=\int \,ds \\ -\dfrac{1}{u}=-u^{-1} = s + \mathcal{C} \end{aligned}$

    where $C$ is a constant. Using $u(r,0)=g(r)$, we find that $\mathcal{C}=-1/g(r)$.

    Hence

    $\begin{aligned}[t] -\dfrac{1}{u} = s -\dfrac{1}{g(r)} \\ \dfrac{1}{u} = \dfrac{1}{g(r)}-s \\ \dfrac{1}{u} = \dfrac{1-s(g(r))}{g(r)} \\ u = \dfrac{g(r)}{1-s(g(r))} \end{aligned}$

Note that $s=t$ (by the $1^\text{st}$ IVP) and $r=x+2t$ (by the $2^\text{nd}$ IVP). Hence $$u(x,t)=\dfrac{g(x+2t)}{1-t(g(x+2t))}$$

Next, we need to determine the region above the $x$-axis for which this solution is a classical solution. How would I determine this region?

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Next, we need to determine the region above the $x$-axis for which this solution is a classical solution

Recall that a classical solution is a solution which is differentiable as many times as needed if you want to plug the function into the PDE (for example, if the PDE contains the term $u_{xxxx}$, then the fourth derivate $u_{xxxx}$ must exist in order for $u$ to be a classical solution).

In our case, we need to show that the first derivatives $u_t$ and $u_x$ exists. \begin{equation*} \begin{aligned} u_x(x,t) & =\dfrac{[1-tg(x+2t)]g_x(x+2t)-g(x+2t)[-tg_x(x+2t)]}{(1-tg(x+2t))^2} =\dfrac{g_x(x+2t)}{(1-tg(x+2t))^2} \\ u_t(x,t) & =\dfrac{[1-tg(x+2t)]2g_t(x+2t)-g(x+2t)[-2tg_t(x+2t)]}{(1-tg(x+2t))^2} =\dfrac{2g_t(x+2t)}{(1-tg(x+2t))^2} \end{aligned} \end{equation*}

Recall that $g(x)\in C^1(\mathbb{R})$ (i.e. $g_x$ and $g_t$ exists). Note that $u$, $u_x$ and $u_t$ exists only if $$1-tg(x+2t)\neq 0 \iff 1\neq tg(x+2t).$$

If $t=0$, then $1=\neq 0$. Hence our solution $$u(x,0)=g(x),$$ which we know from the original problem.

If $t\neq 0$, then $g(x+2t)\neq 1/t$. Note that $g$ takes a value of $x$ and $t$ but only returns $1/t$. Hence the region above the $x$-axis for which $$u(x,t)=\dfrac{g(x+2t)}{1-t(g(x+2t))}$$ is a classical solution is \begin{cases} \mathbb{R} & \text{ when } t=0\\ \mathbb{R}\backslash \{1/t: \forall t>0\} & \text{ when } t>0 \end{cases}

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