Sketching the solution to the IVP (Characteristic curves)

[Leb]

Homework Statement

Sketch the solution to the IVP
$u_{t}+uu_{x}=0 \\ u(x,0) = e^{-x^{2}}$

Homework Equations

Monge's equations $\frac{dt}{d\tau} = 1 \\ \frac{dx}{d\tau}=u \\ \frac{du}{d\tau}=0$

The Attempt at a Solution

I think I do not really need the Monge's equation, but I have no idea, how to sketch the wave profile. Could someone please explain it to me step by step ?
P.S.
I also got stuck on solving the Monge's equations. Particularly in the part where $dx=u(x,t)d\tau$ I tried taking $t_{0} = 0$ which gave me $t=\tau$ and also taking $u(x,t)=u_{0}=F({\sigma})=const.$ which makes it independent of t. Pluging it back in $dx=u_{0}dt --> x = u_{0}t + x_{0}$ where $x_{0} =x'*u_{0}=const.$ Then I blindly assume $x_(0) = \sigma$ which then gives me that the Characteristic Curves should be of the form $x=u_{0}t+\sigma$. Now, since we are given that at t=0 $u(x,0)=e^{-x^{2}}$ and $u(x,0)=u_{0}(\sigma)=e^{-\sigma^{2}}$. So from this, I naively deduce, that each characteristic passes through $(\sigma,0)$ with gradient $e^{\sigma^{2}}$. Can someone point out which assumptions/steps are wrong ? Thanks.

 

[mighty2000]

Hello,
To sketch the solution to the IVP, we can use the method of characteristics. This involves finding the characteristic curves which are given by the equations:
\frac{dt}{d\tau} = 1 \\ \frac{dx}{d\tau}=u \\ \frac{du}{d\tau}=0
From the initial condition, we know that at t=0, u(x,0)=e^{-x^{2}}. This means that at the point (x,0) on the x-axis, the value of u is e^{-x^{2}}. Therefore, the characteristic curve passing through this point has a slope of e^{-x^{2}}. This gives us the equation:
\frac{dx}{dt} = e^{-x^{2}}
This is a separable differential equation that can be solved using integration. After solving for x, we get the equation for the characteristic curve:
x = ln(t+C)
where C is a constant of integration.
To sketch the solution, we can plot several characteristic curves on the x-t plane, with each curve having a different value of C. This will give us a family of curves that represent the solution to the IVP. Additionally, we can plot the initial condition u(x,0)=e^{-x^{2}} on the x-t plane, which will be a horizontal line at y=e^{-x^{2}}. The solution to the IVP will be the region above this line bounded by the characteristic curves. This can be seen in the attached graph.

As for your attempt at solving the Monge's equations, it seems like you are on the right track. However, the assumption x_0=\sigma is not necessary and may lead to incorrect results. Instead, you can use the initial condition u(x,0)=e^{-x^{2}} to determine the value of C in the characteristic curve equation, giving you the equation x=ln(t+e^{-x^{2}}). This will give you the same family of curves as described above.