I find this problem challenging:

Use the method of characteristics to solve $u_t+u_x^2=t$ with $u(x,0)=0$.

I know I'm supposed to let $p=u_x$ and $q=u_t$. Then I get $F(x,t,u,p,q)=p^2+q-t=0$. But what to do from there eludes me.

Any help/hints to the solution process would be greatly appreciated.




Let $v=u_x$ ,

Then $v_t+2vv_x=0$ with $v(x,0)=0$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dx}{ds}=2v=2v_0$ , letting $x(0)=f(v_0)$ , we have $x=2v_0s+f(v_0)=2vt+f(v)$ , i.e. $v=F(x-2vt)$

$v(x,0)=0$ :


$\therefore v=0$




$\therefore g_t(t)=t$


$\therefore u(x,t)=\dfrac{t^2}{2}+C$

$u(x,0)=0$ :


$\therefore u(x,t)=\dfrac{t^2}{2}$


Another way (probably better) way is to use the method of characteristic for non-linear first order PDE. We have:

$$ds = \frac{dx}{F_p} = \frac{dt}{F_t} = \frac{du}{pF_p + pF_p} = \frac{-dp}{F_x + pF_u} = \frac{-dq}{F_t + qF_u}$$

where $p = u_x$ and $q= u_y$

For our equation we obtain the following relations:

$$ds = \frac{dx}{2p} = \frac{dt}{1} = \frac{du}{2p^2+q} = \frac{-dp}{0} = \frac{dq}{1}$$

From this we obtain that $p = C_1$ is a constant and also $dq = dt \implies q=t+C_2$. Plugging into the PDE we obtain the relation $\boxed{C_2 = -C_1^2}$. Now as $q$ depends only on $t$ and $p$ only on $x$ we can integrate the following

$$du = pdx + qdt = C_1dx + (t-C_2^2)dt $$

$$\implies u = C_1x + \frac{t^2}{2} - C_1^2t + D$$

This gives you the complete integral of the PDE. Plugging in the initial values we have that $0 = C_1x + D \implies C_1 = D =0$, which yields the aprticular solution $u(x,t) = \frac{t^2}{2}$

  • $\begingroup$ Where can I find a derivation of the first equation? I want some intuitive explanation for it. $\endgroup$ – Prince Kumar Aug 22 '18 at 7:08
  • $\begingroup$ @PrinceKumar At the moment I don't have any resource at hands, so unfortunately I can't explicitly direct you to anything. However I think a quick Google search should yield some nice results. If I'm not mistaken this method is known as method of charactersitics $\endgroup$ – Stefan4024 Aug 22 '18 at 13:59
  • $\begingroup$ Yes. But I get many articles describing this for the case of 1st Order Linear PDE or at most Quasilinear, but not a general non-linear case. That's why I wanted to know any textbook sources as standard textbooks are much better at explaining such complex topics in simple manner. $\endgroup$ – Prince Kumar Aug 22 '18 at 19:24

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.