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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?

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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.

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  • $\begingroup$ 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

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