Differentiability of P.D.E With given initial conditions on Its Characteristic base curve.

Question

I am facing a major problem in

$$u_t +u u_x=1,\;\;\;\; \text{with initial condition}\;\; u \left(\frac{t^2}{4},t\right)=\frac{t}{2}$$

this P.D.E., here $x \in \mathbb{R} $ and $t>0$

Solution I tried - The auxiliry equation of this equation is $$\frac{dx}{u}=\frac{dt}{1}=\frac{du}{1}$$

considering the first and last terms and then 2nd and last we get two equations $$\frac{u^2}{2}-x=C_1\label{1}\tag{1}$$ and $$u-t=C_2.\label{2}\tag{2}$$

Now considering $$t=z, x=\frac{z^2}{4} \;\; \text{we get} \;\; u=\frac{z}{2} $$

after doing all the calculations in end I got $$u=\pm \sqrt{x-\frac{t^2}{4}}+\frac{t}{2}$$ The question they are asking is

Does The solution is differentiable On the Characteristic Base Curve?

And another important question for me is what does equation \eqref{1} and \eqref{2} represents?
I can solve such kind of questions but can't visualize them; can some one please help me to understand this question and provide me geometrical interpretation of such questions?

Thank You.

Answer 1

From this post:

Using the method of characteristics, one writes

• $\frac{\text d t}{\text d s}=1$. Letting $t(0) = t_0$ gives $t = s + t_0$.
• $\frac{\text d u}{\text d s}=1$. Letting $u(0) = \frac{1}{2} t_0$ gives $u = s + \frac{1}{2} t_0$.
• $\frac{\text d x}{\text d s}=u$. Letting $x(0) = \frac{1}{4} {t_0}^2$ gives $x = \frac{1}{2}s^2 + \frac{1}{2} t_0 s + \frac{1}{4} {t_0}^2$.

which yields the same solution $u(x,t)$ as proposed in OP.

The characteristic curves are parametrized by $s$. Differentiating the solution along a characteristic curve amounts to differentiate $u$ w.r.t. $s$ for a fixed $t_0$. You may be able to finish from here.