We want to solve \begin{align} (t+u)u_x +tu_t &= x-t \\ u(1,x) &= 1+x \end{align} Normally, I'd rewrite the equation as $\frac{t+u}{t}u_x + u_t = \frac{x-t}{t}$, then as per the method of characteristics, set $\frac{dx}{dt} = \frac{t+u}{t}$. Setting $v(t) = u(t, x(t)),$ we see $\dot{v}(t) = x(t)-t$. Now here, I'd normally solve for $v$, plug it back into the characteristic equation and proceed, but I'm not sure how to proceed. Might be something basic from ODE that I'm forgetting. Thanks.


$$(t+u)u_x+tu_t=x-t$$ Characteristic differential equations : $\quad \frac{dx}{t+u}=\frac{dt}{t}=\frac{du}{x-t}$

A first family of characteristic curves comes from :

$\frac{dx}{t+u}=\frac{du}{x-t}=\frac{dx+du}{(t+u)+(x-t)}=\frac{dx+du}{x+u}=\frac{d(x+u)}{(x+u)}=\frac{dt}{t} \quad\to\quad \frac{x+u}{t}=c_1$

A second family of characteristic curves comes from :

$\frac{dx}{t+u}=\frac{dt}{t}=\frac{dx-dt}{(t+u)-t}=\frac{dx-dt}{u}=\frac{du}{x-t} \quad\to\quad u^2-(x-t)^2=c_2$

With any independent $c_1,c_2$ , all above is valid only on the characteristic curves. Outside, $c_1$ and $c_2$ are not independent. The relationship can be expressed on the form of an implicit equation $\Phi(c_1,c_2)=0$ : $$\Phi\left(\frac{x+u}{t}\:,\:u^2-(x-t)^2\right)=0$$ or alternatively $c_1=F(c_2)$ where $F$ is any differentiable function : $$\frac{x+u}{t}=F\left(u^2-(x-t)^2\right)$$ This is an implicit form for the general solution of the PDE.

Then, with the condition : $u(1,x)=1+x$ $$\frac{x+(1+x)}{1}=F\left((1+x)^2-(x-1)^2\right) \quad\to\quad F(4x)=2x+1$$ This determines the function $F(X)=\frac{X}{2}+1$

With $X=u^2-(x-t)^2$ put into the above general solution: $$\frac{x+u}{t}=\frac{u^2-(x-t)^2}{2}+1 $$ This quadratic equation can be solved for $u$. One root $u=t-x$ doesn't satisfy the condition. The other root is the solution of the PDE consistent with the condition $u(1,x)=1+x$ : $$u(x,t)=x-t+\frac{2}{t}$$

  • $\begingroup$ Thanks for the answer; I hadn't learned the method of characteristics this way (I've only dealt with ones that could be solved by my method in the description). Your method seems more general/standard though. Is the general rule always to compare the $dx$ and $du$ terms then the $dx$ and $dt$ terms, or was that just some intuition? $\endgroup$ – Curious Dec 30 '16 at 2:11
  • $\begingroup$ In fact, this way to apply the method of characteristics isn't the most general, but it is more straightforward in the simple cases of first order PDEs. math.ualberta.ca/~xinweiyu/436.A1.12f/… $\endgroup$ – JJacquelin Dec 30 '16 at 6:58

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