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I think I'm starting to get the hang of this, but I'm still somewhat stuck.

The class I'm taking is supposed to have a brief introduction to PDE's, but it's being taught as if this all review, so some of this stuff is coming pretty slowly.

Using the characteristic curves, I'm going to parametrize $x$ by $t$, and so by the chain rule I'll have $$\frac{d}{dt}u(x(t),t) = u_x\frac{dx}{dt}+u_t = 0 = (1+x^2)u_x+u_t.$$

Comparing these we get that $\frac{dx}{dt} = x^2+1$, which, by separation of variables gives us $x(t) = tan(t+C)$.

But I'm kind of stuck here. There's no other information given in the problem, but I thought that we needed to be given some other curve, $\Gamma$ so that we could introduce some initial conditions.

Do I already have enough information to solve this PDE, or is there something else I'm missing?

Any thoughts would be greatly appreciated.

Thanks in advance.

Bears
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  • From your parameterisation, you get two integral curves from your equation after 'so by the chain rule' (which should be $du/d\color{red}t$). You have found one, the other is \begin{align} \frac{du}{dt} = 0 &\implies u(x(t),t) = \text{constant in time} \ &\implies u(x(t),t) = u(x(t'),t') \quad \forall t' \in [0,T] \end{align} – Matthew Cassell Sep 05 '20 at 00:57
  • Yeah, that was supposed to be a $t$. Thanks for pointing that out.

    Your answer here makes sense. Thanks for your input.

    So, How would I put this all together to actually solve the PDE? Am I just a few steps away from the actual solution, or do I have more work to do?

     – Bears Sep 05 '20 at 01:05
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    The $t$s obscure things, in my opinion. From implicit function theorem you get the same DE $$\frac{dy}{dx} = \frac{1}{x^2+1} \implies y - \arctan x = C$$ In other words, the solution to the PDE is some one variable function $f$ given by $$u(x,y)=f(y-\arctan x)$$ which is as general as it would get. – Ninad Munshi Sep 05 '20 at 01:29
  • Okay. This is all coming together then.

    So, just to make sure I understand correctly, $f$ can be any (presumably differentiable) single variable function and that would solve the PDE? If I understand that right, that's a surprisingly loose constraint.

     – Bears Sep 05 '20 at 02:24
  • Yes that is correct. What kind of class is this, out of curiosity? Your username reminds me of my old school. – Ninad Munshi Sep 05 '20 at 03:21
  • User name unrelated; I just like bears, haha. It's an applied math class. It's basically functional analysis for differential and integral equations. Most of what I've done so far is in the realm of pure math, so I'm a bit out of my wheelhouse here. But I (at least formally) meet all the pre-reqs, so I'm hoping I'm not in over my head. – Bears Sep 05 '20 at 13:37

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