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We have the following setup: $$u_t+uu_x = -Du \ u(x,0)=sin x.$$ The question is to find the time $T_s$ of a first shock formation. So basically, I need to solve the equation using method of characteristics and go from there.

But when I solve the corresponding quasilinear equation,I am getting the implicit solution $$u(x,t) = dfrac{sin(x-tu)}{1+Dt}$$ which does not seem right.

Can anybody provide a solution using the method of characterisics so I can pinpoint where my mistake is?

First, it should be clearly stated what $D$ is here. Many times it represents a differential operator, but here I assume it is meant to be an arbitrary constant. I'll rewrite the equation with $q = u_x, p = u_t, z = u,$ so that $F(p,q,z) = 0,$
$$p - zq + Dz = 0.$$
Next, apply the method of characteristics, where only the $z, x$ characteristics are needed since its not fully nonlinear.

begin{align}
frac{dz}{dt} &= q partial_q F + p partial_p F = p -zq = -Dz; ;; z(x_0,0) = sin(x_0)\
frac{dx}{dt} &= partial_q F = -z; ;; x(t = 0) = x_0.
end{align}
Solving these gives us our solution,
begin{align}
z(x_0,t) &= sin(x_0)e^{-Dt}\
x &= frac{sin{(x_0)}(e^{-Dt} - 1) + Dx_0}{D}
end{align}
Now if we could solve the latter equation for $x_0 = x_0(x,t),$ we could plug this back into the equation for $z(x_0,t)$ to get $u(x,t)$ explicitly.

If I wanted to show a finite time blow up, I'd formally take the $x$ derivative of $z(x_0,t),$
begin{align}
frac{dz}{dx} &= cos{(x_0)}e^{-Dt}frac{dx_0}{dx} = cos{(x_0)}e^{-Dt} frac{1}{frac{dx}{dx_0}}\
&= frac{Dcos{(x_0)}e^{-Dt}}{cos{(x_0)}(e^{-Dt} - 1) + D}\
&= frac{De^{-Dt}}{e^{-Dt} + frac{D}{cos{(x_0)}}-1}\
end{align}
Perhaps I have errored, but this gradient only blows up if $e^{-Dt} = frac{D}{cos{x_0}} - 1,$ at some $t.$ Therefore, we need $D leq 2$ for this to happen at all.

I am a little surprised about this, that the RHS forcing term changes this answer from the normal result, i.e. blow up when $t = frac{-1}{cos{(x_0)}}.$

This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The method of characteristics gives

• 
  $frac{text d t}{text d s} = 1$. Letting $t(0) = 0$, we have $t=s$.


• 
  $frac{text d u}{text d s} = -Du$. Letting $u(0) = sin x_0$, we have $u = sin (x_0), e^{-Ds}$.


• 
  $frac{text d x}{text d s} = u$. Letting $x(0) = x_0$, we have $x = -frac{1}{D} sin(x_0)(e^{-Ds}-1) + x_0$.

This is sometimes written in implicit form as $$u = sinleft(x - frac{e^{Dt}-1}{D} uright) e^{-Dt}, .$$
Now, let us compute the breaking time. Differentiating $x$ w.r.t. $x_0$ , we have $frac{text d x}{text d x_0} = -frac{1}{D} cos(x_0)(e^{-Dt}-1) + 1$. This quantity vanishes at $t = frac{1}{D}lnleft(frac{cos x_0}{D + cos x_0}right)$, where characteristics intersect. The smallest such positive time corresponds to the breaking time $T_s$, i.e.
$$
T_s = frac{1}{D}lnleft(frac{1}{1-D}right) ,quad text{if}quad D < 1 , .
$$
Otherwise, if $Dgeqslant 1$, no shock occurs.

Alternatively, one can follow the method leading to Eq. (6.10) p. 36 of (1). Differentiating the PDE with respect to $x$ gives
$$
u_{tx} + (u_x)^2 + u u_{xx} = -Du_x , .
$$
Now, we introduce $q = u_x$ and the directional derivative $q' = q_t + u q_x$, so that the previous equation rewrites as
$q' + q^2 = -D q$.
This Bernoulli differential equation has the solution
$$
q(t) = frac{D q_0}{(D + q_0) e^{Dt}- q_0} , ,
$$
where $q_0 = q(0) = cos x_0$. Therefore, $q$ blows up at $t = frac{1}{D}lnleft(frac{cos x_0}{D + cos x_0}right)$, and the same expression of $T_s$ is obtained.

(1) P.D. Lax, Hyperbolic systems of conservation laws and the mathematical theory of shock waves. SIAM, 1973. doi:10.1137/1.9781611970562.ch1