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Consider the signaling problem $u_t + c(u)u_x= 0, t> 0, x> 0$

$u(x, 0) = u_0, x> 0,$

$u(0, t) = g(t), t> 0,$

where $c$ and $g$ are given functions and $u_0$ is a positive constant. If $c (u) > 0$, under what conditions on the signal $g$ will no shocks form? Determine the solution in this case in the domain $x > 0, t> 0$.

Here's what I have:

Characteristic lines are given by: $x = ut + x_o$

Solving for $du/dt$ gives: $u = k(x_o)$

The initial condition gives: $u(x_o, 0) = u_o = k(x_o) = k(ut-x)$

$u(x-ut, 0) = k(ut-x)$

The boundary condition gives: $u(0, t) = g(t)$

$u(0, \frac{x-x_o}{u}) = g(\frac{x-x_o}{u})$

$u(x,t) = g(\frac{x-x_o}{u})$

But from here I am quite confused... I don't see how I could solve for $u$ or determine conditions on $g$ where a shock will not develop?

EditPiAf
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Ben Gins
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1 Answers1

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$$u_t+C(u)u_x=0\quad\text{where }C(u)\text{ is a given function}$$ GENERAL SOLUTION :

The system of characteristic differential equations is : $$\frac{dt}{1}=\frac{dx}{C(u)}=\frac{du}{0}$$ A first equation of characteristic cuves comes from $du=0\quad\to\quad u=c_1$ .

A second equation of characteristic cuves comes from $\frac{dt}{1}=\frac{dx}{C(c_1)}\quad\to\quad x-C(c_1)t=c_2$

The general solution of the PDE is expressed on the form of implicit equation $\Phi\left(c_1,c_2\right)=0$ where $\Phi$ is any differentiable function of two variables. $$\Phi\left(u\:,\:x-C(u)t\right)=0$$ This is a manner to express any relationship between the two variables. This is equivalent to express the relationship by any function $F$ : $$x-C(u)t=F(u)$$ where $F$ is any derivable function.

In general, this implicit equation cannot be solved for $u$ on closed form.

DETERMINATION OF THE FUNCTION $F$ according to the condition $u(0,t)=g(t)$ :

$0-C\left(g(t)\right)t=F\left(g(t)\right)$

Let $g(t)=X \quad\to\quad t=g^{-1}(X)\quad$ where $g^{-1}$ is the inverse function of $g$.

$$-C(X)g^{-1}(X)=F(X)$$ Thus the function $F$ is now determined, given the functions $C$ and $g$.

PARTICULAR SOLUTION FITTED WITH THE GIVEN CONDITION :

With the particular function $F$ found above :

$x-C(u)t=F(u)$ with $F(u)=-C(u)g^{-1}(u)$

$$x-C(u)t=-C(u)g^{-1}(u)$$ $$x+C(u)\left(g^{-1}(u)-t \right)=0$$ $g^{-1}(u)=t-\frac{x}{C(u)}$ $$u=g\left(t-\frac{x}{C(u)}\right)$$ The result is on the form of implicit equation. Solving for $u$ in order to obtain an explicit form $u(x,t)$ is generally not possible, except in case of particular functions $C$ and $g$.

JJacquelin
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