Toronto Math Forum
APM346-2018S => APM346--Tests => Quiz-1 => Topic started by: Victor Ivrii on January 25, 2018, 08:26:36 AM
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Draw characteristics and find the general solution of the following equation
$$
u_t + (t^2+1)u_x =0.
$$
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$$ dt/1 = dx/(t^2 +1) $$
$$ (t^2+1)dy = dx $$
$$ t^3 + t = x + C $$
$$ C = t^3 +t - x $$
$$ U = \phi ( t^3 +t - x) $$
I can't sketch here, but they should roughly look like cubics. They'll be steeper near the origin because of +t
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My plot for the characteristics $C = t^3 +t - x$ is attached
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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 $\{t=0\}$
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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*}