Question about the differentiability of solution on base characteristics curve.

Let $$u(x, t)$$ be a function that satisfies the PDE: $$u_t+uu_x = 1, x \in \mathbb{R}, t > 0$$, and the initial condition $$u\big(\frac{t^2}{4}, t\big) = \frac{t}{2}$$. Then show that the IVP has solutions none of which is differentiable on the characteristics base curve.

My Attempt: By Lagrange's method, we have $$\frac{dt}{1} = \frac{dx}{u}= \frac{du}{1}$$. On solving above, we get $$u-t=a$$ and $$x-\frac{u^2}{2} = c_2$$ which further implies
$$u=t+f\big(x-\frac{u^2}{2}\big)$$. Now $$u\big(\frac{t^2}{4}, t\big) = \frac{t}{2}$$ implies that $$\frac{t}{2}= t+f\big(\frac{t^2}{8}\big) \implies f\big(\frac{t^2}{8}\big) = -\frac{t}{2} \implies f(t) = \mp\sqrt{2t}$$. This means $$u = t\mp\sqrt{2}\sqrt{x-\frac{u^2}{2}}$$. How to proceed further?

Your calculus is correct. $$u = t\mp\sqrt{2}\sqrt{x-\frac{u^2}{2}}.$$ Then one have to solve this equation for $$u$$. $$(u-t)^2=2x-u^2$$ This is a simple quadratic equation. The solution is : $$u(x,t)=\frac{t\pm \sqrt{4x-t^2}}{2}$$ This function satisfies the PDE and the specified condition.
• Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE. – JJacquelin Jan 27 at 16:15