MHB Solving the Shock Wave Problem for $u_t + uu_x = 0$

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The discussion focuses on solving the equation $u_t + uu_x = 0$ with specific initial conditions, allowing for the formation of a shock wave. The solution indicates that the shock occurs at time $t=1$, where the function $u$ becomes discontinuous. The entropy condition is confirmed to be satisfied since $u^-$ (1) is greater than $u^+$ (0) at the shock. Participants clarify the definition of the entropy condition and verify its application in this context. The conversation concludes with agreement on the findings regarding the shock wave and entropy condition.
evinda
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Hello! (Wave)

I want to solve the equation $u_t+uu_x=0$ with the initial condition $u(x,0)=1$ for $x \leq 0$, $1-x$ for $0 \leq x \leq 1$ and $0$ for $x \geq 1$. I want to solve it for all $t \geq 0$, allowing for a shock wave. I also want to find exactly where the shock is and show that it satisfies the entropy condition.

I have tried the following.

We get that $u(x(t),t)=c$.

The characteristic line that passes through the points $(x,t)$ and $(x_0,0)$ has slope

$\frac{x-x_0}{t-0}=\frac{dx}{dt}=u(x,t)=u(x_0,0)=\left\{\begin{matrix}
1, & \text{ if } x_0 \leq 0,\\
1-x_0, & \text{ if } 0 \leq x_0 \leq 1,\\
0, & \text{ if } x_0 \geq 1.
\end{matrix}\right.$$\Rightarrow x-x_0=\left\{\begin{matrix}
t, & \text{ if } x_0 \leq 0,\\
t(1-x_0), & \text{ if } 0 \leq x_0 \leq 1,\\
0, & \text{ if } x_0 \geq 1.
\end{matrix}\right.$

$\Rightarrow u(x,t)=\left\{\begin{matrix}
1, & \text{ if } x \leq t,\\
\frac{1-x}{1-t}, & \text{ if } t \leq x \leq 1,\\
0, & \text{ if } x \geq 1.
\end{matrix}\right.$Do we have a shock at the time when $u$ is not continuous? But how can we find such a time?
 
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evinda said:
Hello! (Wave)

I want to solve the equation $u_t+uu_x=0$ with the initial condition $u(x,0)=1$ for $x \leq 0$, $1-x$ for $0 \leq x \leq 1$ and $0$ for $x \geq 1$. I want to solve it for all $t \geq 0$, allowing for a shock wave. I also want to find exactly where the shock is and show that it satisfies the entropy condition.

I have tried the following.

We get that $u(x(t),t)=c$.

The characteristic line that passes through the points $(x,t)$ and $(x_0,0)$ has slope

$\frac{x-x_0}{t-0}=\frac{dx}{dt}=u(x,t)=u(x_0,0)=\left\{\begin{matrix}
1, & \text{ if } x_0 \leq 0,\\
1-x_0, & \text{ if } 0 \leq x_0 \leq 1,\\
0, & \text{ if } x_0 \geq 1.
\end{matrix}\right.$$\Rightarrow x-x_0=\left\{\begin{matrix}
t, & \text{ if } x_0 \leq 0,\\
t(1-x_0), & \text{ if } 0 \leq x_0 \leq 1,\\
0, & \text{ if } x_0 \geq 1.
\end{matrix}\right.$

$\Rightarrow u(x,t)=\left\{\begin{matrix}
1, & \text{ if } x \leq t,\\
\frac{1-x}{1-t}, & \text{ if } t \leq x \leq 1,\\
0, & \text{ if } x \geq 1.
\end{matrix}\right.$Do we have a shock at the time when $u$ is not continuous? But how can we find such a time?

Hey evinda!

What will happen to $u$ when $t$ reaches $1$? (Wondering)

Btw, your solution doesn't cover $t>1$ does it?
The second condition breaks down, and the first and third condition will overlap. (Worried)
 
The shock wave happens when $t=1$.

Which is the entropy condition? (Thinking)
 
evinda said:
The shock wave happens when $t=1$.

Which is the entropy condition?

I don't know what an entropy condition is. (Crying)

Do you have a definition? (Wondering)
 
I like Serena said:
I don't know what an entropy condition is. (Crying)

Do you have a definition? (Wondering)

I found the definition at page $12$: https://web.stanford.edu/class/math220a/handouts/conservation.pdf.

In our case, ee have $f'(u)=u$.
At $t=1$ we have $u^-=1$ and $u^+=0$, and so we have $u^- > u^+$, which means that the entropy condition is satisfied.

Right? (Thinking)
 
I like Serena said:
Yes, I believe so. (Nod)

Nice, thank you! (Smile)
 

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