### Author Topic: Q1-T5102-P2  (Read 3757 times)

#### Victor Ivrii

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##### Q1-T5102-P2
« on: January 25, 2018, 08:26:36 AM »
Draw characteristics and find the general solution of the following equation
$$u_t + (t^2+1)u_x =0.$$

#### Andrew Hardy

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##### Re: Q1-T5102-P2
« Reply #1 on: January 25, 2018, 09:04:06 AM »
$$dt/1 = dx/(t^2 +1)$$
$$(t^2+1)dy = dx$$
$$t^3 + t = x + C$$
$$C = t^3 +t - x$$
$$U = \phi ( t^3 +t - x)$$

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

#### Elliot Jarmain

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##### Re: Q1-T5102-P2
« Reply #2 on: January 25, 2018, 09:49:52 AM »
My plot for the characteristics $C = t^3 +t - x$ is attached

#### Victor Ivrii

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##### Re: Q1-T5102-P2
« Reply #3 on: January 25, 2018, 10:31:51 AM »
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 $\{t=0\}$
« Last Edit: January 25, 2018, 10:36:26 AM by Victor Ivrii »

#### Elliot Jarmain

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##### Re: Q1-T5102-P2
« Reply #4 on: January 25, 2018, 10:37:02 AM »
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*}