APM346--2020S > Chapter 2

S2.2P Problem 2 (6)

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Anna Kiseleva:
I have 2 questions regarding the problem.

First and foremost, after solving the characteristics, I am getting u as a function of u. I know there is a way to deal with it via the use of Inverse Function theorem (?), but I am not sure how it's done.
I have arrived at:

u=xt-f(3(t-x-y) + xy + 2u)
First, am I getting the correct equation?
Second, how do I proceed from here?

I was also wondering what's the general rule of dealing with the IVP for 3 and higher dimensional equations? After plugging in t=0, or whatever the Cauchy problem condition is, I am getting f in terms of x and y and no straight-forward way of plugging t back in.

Matter-of-factly, my question isn't specific to equation (6).

Thanks in advance; I understand if I've made the question in any way unclear. :(

Yan Zhou:
I have attached my solution to equation (6). I hope it may be helpful. I am confused with the following IVP problem as well. Maybe someone else can help.

jfarrellhfx:
Hmm... Here's how I think I would start:
Begin with the equation for the integral curves:

$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{x}$$

I would try to solve for $u$ first:
$$du = x \ dt$$

But the integral is in terms of the variable $t$, but you have an $x$ in there... is there a way to get $x$ in terms of $t$?

Victor Ivrii:
Please, use MathJax for proper displaying equations. Also you need either repeat a problem here, or to provide a clickable link, like this

So, we have equation
\begin{equation}
u_t+3u_x-2u_y=x
\label{eqn-1}
\end{equation}
with the IVP
\begin{equation}
u |_{t=0}=0.
\label{eqn-2}
\end{equation}
Writing characteristics
\begin{equation}
\frac{dt}{1}=\frac{dx}{3}=\frac{dy}{-2}=\frac{du}{x}.
\label{eqn-3}
\end{equation}
Solving the first equality: $x-3t=c_1$, second $y+2t =c_2$ and the last one $u-\frac{x^2}{6}=C$, with $c_1, c_2, C$ constants along characteristics, which are marked by $c_1,c_2$. Then $C=\varphi(c_1,c_2)$ and finally
\begin{equation}
\boxed{u = \frac{x^2}{6} + \varphi (x-3t, y+2t)}
\label{eqn-4}
\end{equation}
is the general solution to (\ref{eqn-1}).

Plugging (\ref{eqn-4}) into (\ref{eqn-2}) we get $\frac{x^2}{6} + \varphi (x, y) =0\implies \varphi(x,y)= -\frac{x^2}{6}$ and plugging into (\ref{eqn-4}) we get
\begin{equation}
\boxed{u = \frac{x^2}{6} - \frac{(x-3t)^2}{6} = xt - \frac{3}{2}t^2.}
\label{eqn-5}
\end{equation}