Use the method of characteristics to solve nonlinear first order pde.
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.
$\endgroup$2 Answers
$\begingroup$$u_t+u_x^2=t$
$u_{tx}+2u_xu_{xx}=0$
Let $v=u_x$ ,
Then $v_t+2vv_x=0$ with $v(x,0)=0$
Follow the method in :
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{ds}=2v=2v_0$ , letting $x(0)=f(v_0)$ , we have $x=2v_0s+f(v_0)=2vt+f(v)$ , i.e. $v=F(x-2vt)$
$v(x,0)=0$ :
$F(x)=0$
$\therefore v=0$
$u_x=0$
$u(x,t)=g(t)$
$u_t=g_t(t)$
$\therefore g_t(t)=t$
$g(t)=\dfrac{t^2}{2}+C$
$\therefore u(x,t)=\dfrac{t^2}{2}+C$
$u(x,0)=0$ :
$C=0$
$\therefore u(x,t)=\dfrac{t^2}{2}$
$\endgroup$ $\begingroup$Another way (probably better) way is to use the method of characteristic for non-linear first order PDE. We have:
$$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 $u(x,t) = \frac{t^2}{2}$
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