Author Topic: Quiz2 TUT5101  (Read 519 times)

Jingjing Cui

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Quiz2 TUT5101
« on: January 31, 2020, 03:39:38 PM »
$$
2u_{t}+t^2u_{x}=0\\
\frac{dt}{2}=\frac{dx}{t^2}=\frac{du}{0}\\
\int\frac{1}{2}t^2dt=\int1dx\\
\frac{1}{6}t^3+A=x\\
A=x-\frac{1}{6}t^3\\
$$
Because c=0, so
$$
u(t,x)=g(A)=g(x-\frac{1}{6}t^3)
$$
 
The initial condition given in the question: u(x,0)=f(x)
The characteristics curves ($A=x-\frac{1}{6}t^3$) will always intersect t=0 (x-axis) at a unique point, no matter what value A takes. Thus, the solution always exist.
« Last Edit: February 01, 2020, 09:43:07 AM by Jingjing Cui »

Victor Ivrii

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Re: Quiz2 TUT5101
« Reply #1 on: January 31, 2020, 10:49:26 PM »
In the tsecond/third lines should be $dt$, ..., not $\partial t$,...