Existence/uniqueness and solution of quasilinear PDE

Question

So I have this question in a textbook that I have been trying to solve for review but I can't $$u_t+uu_x=1$$ $$u(x,t)=t\text{ when } 2x-t^2=0$$ So the book says that the solution does not exist at all for this problem. I will show you my solution. So first of all I parametrize by $t(0)=s,x(0)=s^2/2, u(0)=s$. With these you get the characteristic equations \begin{align} &\frac{dx}{dr} = u \nonumber \\ &\frac{dt}{dr} = 1 \nonumber \\ &\frac{du}{dr} =1 \nonumber \end{align} Using the initial conditions, we get \begin{align} &x = \frac{r^2}{2}+sr+ \frac{s^2}{2} \nonumber \\ & t = r + s \nonumber \\ &u = r + s\nonumber \end{align} Testing the non-characteristic condition we get $$\frac{\partial (x,t)}{\partial (r,s)}=\begin{vmatrix} r+s & r+s\\ 1 & 1 \notag \end{vmatrix} = 0$$ So the non-characteristic condition does not hold, and according to the problem I am supposed to check for existence and uniqueness. The non-characteristic condition implies that the solution either doesn't exist or isn't unique. But I get the solution, which just shows that it isn't unique. Did I do something wrong? Or am I interpreting the solution wrong? Since the textbook says it doesn't exist. $$ u(x,t)=t=\pm\sqrt{2x}$$

Answer 1

$$\begin{cases} u_t+uu_x=1\\ u(x,t)=t\text{ when } 2x-t^2=0 \end{cases}$$ A solution is : $$u(x,t)=\frac{x}{t}+\frac{t}{2}$$

Proof :

$\begin{cases} u_x=\frac{1}{t}\\ u_t=-\frac{x}{t^2}+\frac{1}{2}\end{cases} \quad\implies\quad u_t+uu_x=-\frac{x}{t^2}+\frac{1}{2}+\left(\frac{x}{t}+\frac{t}{2} \right)\frac{1}{t} =1$

Thus, the PDE $\quad u_t+uu_x=1\quad $ is satisfied.

And when $2x-t^2=0 \quad\to\quad x=\frac{t^2}{2}\quad;\quad u=\frac{x}{t}+\frac{t}{2} = \frac{\frac{t^2}{2}}{t}+\frac{t}{2} =t$

Thus the condition $\quad u(x,t)=t\quad$ when $\quad 2x-t^2=0\quad$ is satisfied.

So, with the actual wording of the problem, there is a solution. This is apparently in contradiction with the textbook which says that the solution does not exist.

Probably, there is something important missing in the copy of the wording. For example it might be specified that $u(x,t)$ must exist at $t=0$. If this forgotten (or implicit) condition was added, $u(x,t)=\frac{x}{t}+\frac{t}{2}$ would no longer be a correct solution due to the term with $\frac{1}{t}$ .

IN ADDITION :

$u_t+uu_x=1\quad\to\quad$ Set of ODEs for the characteristic curves : $$\frac{dx}{u}=\frac{dt}{1}=\frac{du}{1}$$ First family of characteristic curves, from $\quad \frac{du}{1}=\frac{dt}{1}$ $$u-t=c_1$$

Second family of characteristic curves, from $\quad \frac{dx}{u}=\frac{dx}{c_1+t}=\frac{dt}{1}\quad\to\quad c_1t+\frac{t^2}{2}-x=c_2$ $$(u-t)t+\frac{t^2}{2}-x=c_2$$ The general solution of the PDE can be expressed on various equivalent forms, for example : $$(u-t)t+\frac{t^2}{2}-x=F(u-t)$$ In this implicit equation, $F(X)$ is an arbitrary function. $$(u-t)t+\frac{1}{2}(t^2-2x)=F(u-t)$$

With condition, when $\quad 2x-t^2=0 \quad,\quad u(x,t)=t $ : $$(u-t)t+\frac{1}{2}(t^2-2x)=F(u-t)=(t-t)t+\frac{1}{2}(0)=F(t-t)$$ $$F(0)=0$$ Thus, the specified condition restricts the infinite set of functions $F(X)$ to a smaller set of functions $F(X)$ with $F(0)=0$ , still infinite.

Among them, for example $F(X)=X \quad\to\quad (u-t)t+\frac{1}{2}(t^2-2x)=(u-t)$

or another example : $F(X)=X^2 \quad\to\quad (u-t)t+\frac{1}{2}(t^2-2x)=(u-t)^2$

For each of these example, solving the implicit equation for $u$ (when possible) leads to a solution of the problem.

The simplest example is with $F(X)=0$ which leads to a particular solution : $$(u-t)t+\frac{1}{2}(t^2-2x)=0 \quad\to\quad u(x,t)=\frac{t}{2}+\frac{x}{t}$$

This is the particular solution given as an example at the beginning. Of course, they are an infinity of solutions provided by the above implicit equation.