APM346-2018S > Quiz-1

Q1-T5102-P2

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Victor Ivrii:
Draw characteristics and find the general solution of the following equation
$$
u_t + (t^2+1)u_x =0.
$$

Andrew Hardy:
$$ dt/1 = dx/(t^2 +1) $$
$$ (t^2+1)dy = dx $$
$$ t^3 + t = x + C $$
$$ C = t^3 +t - x  $$
$$ U = \phi ( t^3 +t - x) $$

I can't sketch here, but they should roughly look like cubics. They'll be steeper near the origin because of +t

Elliot Jarmain:
My plot for the characteristics $C = t^3 +t - x$ is attached

Victor Ivrii:
And where did you guys learned to integrate like this?  >:(

Also, do not post graphics as pdf attachments. Convert to png or jpeg.

Further, the plot with this range is actually misleading because it obscures the fact that characteristics are never parallel to $\{t=0\}$

Elliot Jarmain:
Along the characteristic curves:
\begin{gather*}
   \frac{dt}{1} = \frac{dx}{t^2+1} \\
   \int{t^2+1 \, dt} = \int{dx}\\
   \frac{t^3}{3} + t + C = x
\end{gather*}
Also $f(x,t,u) = 0$ implies $u$ is constant along the characteristic curves,
therefore:
\begin{equation*}
   u = \phi(x-\frac{t^3}{3}-t)
\end{equation*}

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