Terminology — Question 5

Question 5 – Solution

Strategy. Check whether the data line intersects all characteristic lines or follows just one of them.

Step 1: Recover the characteristic form. Along $x=t+s$, the chain rule gives $\frac{d}{dt}u(t+s,t)=u_x+u_t=0$. Thus $u(t+s,t)=u(s,0)$. Defining $F(s)=u(s,0)$ proves $\boxed{u(x,t)=F(x-t)}$ for every solution. Conversely this form has $u_t=-F'$ and $u_x=F'$, so it solves the PDE.

Step 2: Use a noncharacteristic data line. The condition becomes $F((a-1)t)=h(t)$. If $a\ne 1$, the argument ranges over all real numbers exactly once. Consequently $$\boxed{F(s)=h\!\left(\frac{s}{a-1}\right),\qquad u(x,t)=h\!\left(\frac{x-t}{a-1}\right).}$$ This is $C^1$, solves the PDE and takes the prescribed trace. The characteristic representation proves uniqueness in the whole stated $C^1$ class.

Step 3: Diagnose characteristic data. If $a=1$, the condition reduces to $F(0)=h(t)$ for every $t$. A solution exists exactly when $h$ is constant, say $H$. Then any $C^1$ function with $F(0)=H$ works. For example, $u=H$ and $u=H+x-t$ are distinct solutions. Thus compatible characteristic data leave infinitely many fields undetermined.

Step 4: Interpret the two sine traces. For $a=0$, $u(x,t)=\sin(t-x)$ is the unique solution. For $a=1$, the varying trace $\sin t$ contradicts constancy along that characteristic, so no solution exists. The diagram shows the unique intersection with $x=0$ and the characteristic line $x=t$; the latter supplies information on just one member of the family.