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.

  • $\begingroup$ How to verify the condition for of differentability on characteristics base curve? $\endgroup$ – Mittal G Jan 27 at 15:22
  • $\begingroup$ Why doing that ? Just check the above result $u(x,t)$ in putting it into the PDE. $\endgroup$ – JJacquelin Jan 27 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.