I am trying to verify a claim from the book Numerical Partial Differential Equations: Conservation Laws and Elliptic Equations by JW Thomas. I am reading section 9.2.3, which considers discontinuous weak solutions to 1D problems of the form $v_t + \frac{\partial}{\partial x}F(v) = 0$. On page 96 the author gives the following example:
And finally, we include one last example where we have the continuous initial condition $v_0(x) = 1$ for $x \in \mathbb{R}$. We note that the function $v$ defined by
$$
v(x,t) =
\begin{cases}
\;\;\, 1 & \; x < -t/2 \\
-2 & \; -t/2 \leq x < 0 \\
\;\;\, 2 & \; 0 \leq x < 3t/2 \\
\;\;\,1 & \; 3t/2 \leq x
\end{cases}
$$
is a weak solution to the initial value problem given by Burgers' equation [$v_t + \left(\tfrac{1}{2}v^2 \right)_x = 0$] along with the initial condition $v_0(x) = 0$.
(I am already confused by whether $v_0 = 1$ or $v_0 = 0$, but that's not my main question…) Here's my work so far.
My attempt: I am trying to verify, by direct integration, that $v$ given above is a weak solution to Burgers' equation. To this end, I start by fixing finite sub-intervals $[a,b]$ and $[0,T]$ of $\mathbb{R}$ and $[0,\infty)$, respectively. Now let $\phi: \mathbb{R} \times [0,\infty) \to \mathbb{R}$ be a test function, i.e., $\phi$ is $C^1$ and $\{(x,t) \in \mathbb{R} \times [0,\infty): \phi(x,t) \neq 0 \} \subset [a,b] \times [0,T]$. By the discussion on page 81, to show that $v$ is a weak solution of
$$
\begin{cases}
v_t + (\frac{1}{2}v^2)_x = 0\\
v(x,0) = v_0(x)
\end{cases}
$$
it suffices to show that
\begin{align*}
\int_{0}^{T} \int_{a}^{b} (v \phi_t + \tfrac{1}{2}v^2 \phi_x)\,dx \,dt + \int_{a}^{b} v_0(x) \phi(x,0)\,dx = 0.
\end{align*}
To ease notation, let
\begin{align*}
(1) &:= \int_{0}^{T} \int_{a}^{b} v \phi_t \,dx \,dt \\[5pt]
(2) &:= \int_{0}^{T} \int_{a}^{b} \tfrac{1}{2}v^2 \phi_x \,dx\,dt \\[5pt]
(3) &:= \int_{a}^{b} v_0(x) \phi(x,0)\,dx
\end{align*}
So our goal is to show that $(1) + (2) + (3) = 0$. Now I made the following sketch of the domain of integration ($t$ is on the horizontal axis).
We can split up the domain of integration as $[a,b] \times [0,T] = D_1 \cup D_2 \cup D_3 \cup D_4$, where
\begin{align*}
D_1 &= \{(x,t)\,|\, 0 \leq t \leq T, \, a \leq x \leq -t/2 \} = \{(x,t)\,|\, a \leq x \leq 0, \, 0 \leq t \leq -2x \} \\[2pt]
D_2 &= \{(x,t)\,|\, 0 \leq t \leq T, \, -t/2 \leq x \leq 0 \} = \{(x,t)\,|\, a \leq x \leq 0, \, -2x \leq t \leq T \} \\[2pt]
D_3 &= \{(x,t)\,|\, 0 \leq t \leq T, \, 0 \leq x \leq 3t/2 \} \; = \{(x,t)\,|\, 0 \leq x \leq b, \, 2x/3 \leq t \leq T \} \\[2pt]
D_4 &= \{(x,t)\,|\, 0 \leq t \leq T, \, 3t/2 \leq x \leq b \} \; = \{(x,t)\,|\, 0 \leq x \leq b, \, 0 \leq t \leq 2x/3 \} \\[2pt]
\end{align*}
Now splitting up the double integral,
\begin{align*}
\int_{0}^{T} \int_{a}^{b} (v\phi_t + \tfrac{1}{2}v^2 \phi_x) \,dx \,dt = \underbrace{ \int_{0}^{T} \int_{a}^{b} v\phi_t \,dx \,dt}_{(1)} + \underbrace{\frac{1}{2} \int_{0}^{T} \int_{a}^{b}v^2 \phi_x \,dx \,dt}_{(2)}
\end{align*}
we have:
\begin{align*}
(1) &= \iint_{D_1} \phi_t \,dA – 2 \iint_{D_2} \phi_t \,dA + 2 \iint_{D_3} \phi_t \,dA + \iint_{D_4} \phi_t \,dA \\[5pt]
&= \int_{a}^{0} \int_{0}^{-2x} \phi_t \,dt \,dx – 2 \int_{a}^{0} \int_{-2x}^{T} \phi_t \,dt \,dx + 2 \int_{0}^{b} \int_{\frac{2x}{3}}^{T} \phi_t \,dt \,dx + \int_{0}^{b} \int_{0}^{\frac{2}{3}x} \phi_t \,dt \,dx \\[5pt]
&= \int_{a}^{0} \left[ \phi(x,-2x) – \phi(x,0) \right] dx – 2 \int_{a}^{0} \left[ \phi(x,T) – \phi(x,-2x) \right] dx + 2 \int_{0}^{b} \left[ \phi(x,T) – \phi(x,\tfrac{2}{3}x) \right] dx + \int_{0}^{b} \left[ \phi(x,\tfrac{2}{3}x) – \phi(x,0) \right] dx
\end{align*}
Now combining terms and using the fact that $\phi(x,T) = 0$, this simplifies to
\begin{align*}
(1) = 3 \int_{a}^{0} \phi(x,-2x)\,dx – \int_{a}^{b} \phi(x,0)\,dx – \int_{0}^{b} \phi(x,\tfrac{2}{3}x)\,dx.
\end{align*}
And for (2) we have
\begin{align*}
(2) &= \frac{1}{2} \left[ \iint_{D_1} \phi_x \,dA + \iint_{D_2} 4 \phi_x \,dA + \iint_{D_3} 4 \phi_x \,dA + \iint_{D_4} \phi_x \,dA \right] \\[5pt]
&= \frac{1}{2} \int_{0}^{T} \int_{a}^{-t/2} \phi_x \,dx\,dt + 2 \int_{0}^{T} \int_{-t/2}^{0} \phi_x \,dx\,dt + 2 \int_{0}^{T} \int_{0}^{\frac{3t}{2}} \phi_x \,dx \,dt + \frac{1}{2} \int_{0}^{T} \int_{\frac{3t}{2}}^{b} \phi_x \,dx\,dt \\[5pt]
&= \frac{1}{2} \int_{0}^{T} \left[ \phi(-\tfrac{t}{2},t) – \phi(a,t) \right] dt + 2 \int_{0}^{T} \left[ \phi(0,t) – \phi(-\tfrac{t}{2},t) \right] dt + 2 \int_{0}^{T} \left[ \phi(\tfrac{3t}{2},t) – \phi(0,t) \right] dt + \tfrac{1}{2} \int_{0}^{T} \left[ \phi(b,t) – \phi(\tfrac{3t}{2},t) \right] dt.
\end{align*}
Now using the fact that $\phi(a,t) = \phi(b,t) = 0$ and combining terms, this simplifies to
\begin{align*}
(2) = -\frac{3}{2} \int_{0}^{T} \phi(-\tfrac{t}{2},t)\,dt + \frac{3}{2} \int_{0}^{T} \phi(\tfrac{3t}{2},t) \,dt
\end{align*}
Thus,
\begin{align*}
(1) + (2) + (3) &= 3 \int_{a}^{0} \phi(x,-2x)\,dx – \int_{a}^{b} \phi(x,0)\,dx – \int_{0}^{b} \phi(x,\tfrac{2}{3}x)\,dx \\[5pt]
&\quad -\frac{3}{2} \int_{0}^{T} \phi(-\tfrac{t}{2},t)\,dt + \frac{3}{2} \int_{0}^{T} \phi(\tfrac{3t}{2},t) \,dt + \int_{a}^{b} \phi(x,0)\,dx \\[5pt]
&= \underbrace{3 \int_{a}^{0} \phi(x,-2x)\,dx}_{\text{(A)}} – \underbrace{\int_{0}^{b} \phi(x,\tfrac{2}{3}x)\,dx}_{\text{(B)}} – \underbrace{\frac{3}{2} \int_{0}^{T} \phi(-\tfrac{t}{2},t)\,dt}_{\text{(C)}} + \underbrace{\frac{3}{2} \int_{0}^{T} \phi(\tfrac{3t}{2},t) \,dt}_{\text{(D)}}
\end{align*}
Then for (C) I made the substitution $x = -\frac{t}{2}, \, dx = -\frac{1}{2}\,dt$ to obtain:
\begin{align*}
\text{(C)} = 3 \int_{0}^{-T/2} \phi(x,-2x)\,dx.
\end{align*}
And for (D) I made the substitution $x = \frac{3t}{2}, \, dx = \frac{3}{2}\,dt$ to obtain
\begin{align*}
\text{(D)} &= \int_{0}^{3T/2} \phi(x,\tfrac{2x}{3})\,dx = \int_{0}^{T} \phi(x,\tfrac{2x}{3})\,dx,
\end{align*}
where the last equality is by the fact that $\phi$ vanishes when $t \geq T$. At this point, (A) and (C) now have matching integrands, as do (B) and (D). But I don't see how the integrals cancel because the bounds of the corresponding integrals are different (one pair is in terms of $a$ and $b$, the other is in terms of $T$). How should I proceed? Or did I make a mistake somewhere in my work?
Any help would be greatly appreciated.
Best Answer
I figured it out:
A weak solution $u$ satisfies $$ \int_{0}^{\infty} \int_{-\infty}^{\infty} \left[ u \phi_t + \tfrac{1}{2}u^2 \phi_x \right] \,dx \,dt + \int_{-\infty}^{\infty} u_0(x) \phi(x,0) \,dx = 0 $$ for all test functions $\phi$, so taking $a = -\infty$ and $b = T = \infty$ in my argument gives \begin{align*} (1) + (2) + (3) &= 3 \int_{-\infty}^{0} \phi(x,-2x)\,dx - \int_{0}^{\infty} \phi(x,\tfrac{2x}{3})\,dx \\[5pt] &\quad + 3 \int_{0}^{-\infty} \phi(x,-2x)\,dx + \int_{0}^{\infty} \phi(x,\tfrac{2x}{3})\,dx \\[5pt] &= -3 \int_{0}^{-\infty} \phi(x,-2x)\,dx + 3 \int_{0}^{-\infty} \phi(x,-2x)\,dx \\[5pt] &= 0, \end{align*}
as desired. (By the assumptions on $\phi$ we have $\phi(-\infty,t) = \phi(\infty,t) = 0$ for all $t \geq 0$, and $\phi(x,\infty) = 0$ for all $x \in \mathbb{R}$.)