Toronto Math Forum
APM346-2015F => APM346--Tests => Test 1 => Topic started by: Victor Ivrii on October 21, 2015, 08:47:17 PM
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Consider the first order equation:
\begin{equation}
(t^2+1)u_t + u_x = u.
\label{eq-1-1}
\end{equation}
(a) Find the characteristic curves and sketch them in the $(x,t)$ plane.
(b) Write the general solution.
(c) Solve equation (\ref{eq-1-1}) with the initial condition $u(x,0)= \cos(x)$. Explain why the solution is fully determined by the initial condition.
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For part a) b) I get, $$\frac{dt}{t^2+1} = \frac{dx}{1}\\Let\ t = \tan{\theta}\\So\ \theta = \arctan{\ t}\\So\ 1 + t^2 = 1 + \tan^2{\theta} = \sec^2{\theta}\\and\ \frac{dt}{d\theta} = \sec^2{\theta}\\So\ \int \frac{1}{1+t^2} dt = \int \frac{sec^2{\theta}}{\sec^2{\theta}} d\theta\\= \int 1 d\theta\\= \theta\\= \arctan{t}\\For \int 1 dx = x\\ Hence\ x = \arctan{t}\\the\ characteristic\ curves\ are\ t = \tan{x}\\then\ we\ have\ c = x - \arctan{t}\\ u = \phi{(c)} = \phi{(x - \arctan{t})}$$
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We have $$\frac{dt}{t^{2}+1}=\frac{dx}{1}=\frac{du}{u}$$
Xi Yue is right until the last line. Instead we need to proceed as follows: $$\frac{du}{u}=dx\Rightarrow \ln(u)=x+d\Rightarrow u=De^{x}=\phi(x-\arctan(t))e^{x}$$
So for part c): $$u(x,0)=\phi(x)e^x=\cos(x)\Rightarrow \phi(x)=\cos(x)e^{-x}$$ since $\arctan(0)=0$. Therefore, the solution to the IVP is:
$$u(x,t)=\cos(x-\arctan(t))e^{\arctan(t)-x}e^{x}=\cos(x-\arctan(t))e^{\arctan(t)}$$
The solution is fully determined because the original equation was first order in time, with no restrictions on x. Therefore one initial condition uniquely defines a solution.
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One can look at the contour plot on this page http://www.wolframalpha.com/input/?i=x-arctan%28t%29 to see the characteristic curves
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I think the characteristic curve is like this? As c is a random real constant.
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:)