The Riemann problem for a generalised Burgers equation with spatially decaying sound speed. II General qualitative theory

Consider the set $\mathcal{S}$ of functions $v:D_{T^{*}}\to\mathbb{R}$ which satisfy (2.3) on $D_{T^{*}}$ and are such that

$$||v||_{\infty}\leq||u_{0}||_{\infty}+1.$$

(2.5)

Next, consider the mapping $M:\mathcal{S}\to\mathbb{R}(D_{T^{*}})$ given by $M[v]$ for $v\in\mathcal{S}$ where

	$\displaystyle M[v](x,t)$	$\displaystyle=\int_{-\infty}^{\infty}u_{0}(s)G(x,t;s,0)ds$
		$\displaystyle\quad+\int_{0}^{t}\int_{-\infty}^{\infty}\frac{v^{2}(s,\tau)}{2}\left(G(x,t;s,\tau)h_{\alpha}^{\prime}(s)+G_{s}(x,t;s,\tau)h_{\alpha}(s)\right)dsd\tau$		(2.6)

for all $(x,t)\in D_{T^{*}}$. Observe that the first term in the right hand side of (2.6) is the solution to the heat equation with measurable initial data $u_{0}\in L^{\infty}(\mathbb{R})$, and in particular, is contained in $C^{2,1}(D_{T^{*}})\cap L^{\infty}(D_{T^{*}})$, with bound $||u_{0}||_{\infty}$ on $D_{T^{*}}$. Also, using (A.1) and (A.2), it follows that the integrand of the second term on the right hand side of (2.6) is absolutely integrable, and hence the integral is well-defined, and bounded on $D_{T^{*}}$, for each $v\in\mathcal{S}$. Moreover, via (A.5), (A.6), (A.8) and (A.9), it follows that the second term in the right hand side of (2.6) is continuous on $D_{T^{*}}$ for each $v\in\mathcal{S}$. Furthermore, for each $v\in\mathcal{S}$, and all $(x,t)\in D_{T^{*}}$, observe, via (A.1) and (A.2), that

		$\displaystyle\left|\int_{0}^{t}\int_{-\infty}^{\infty}\frac{v^{2}(s,\tau)}{2}\left(G(x,t;s,\tau)h_{\alpha}^{\prime}(s)+G_{s}(x,t;s,\tau)h_{\alpha}(s)\right)dsd\tau\right|$
		$\displaystyle\quad\leq\frac{||v||_{\infty}^{2}}{2}\int_{0}^{t}\int_{-\infty}^{\infty}\left(||h_{\alpha}^{\prime}||_{\infty}G(x,t;s,\tau)+||h_{\alpha}||_{\infty}|G_{s}(x,t;s,\tau)|\right)dsd\tau$
		$\displaystyle\quad\leq||v||_{\infty}^{2}\left(\frac{||h_{\alpha}^{\prime}||_{\infty}}{2}\sqrt{t}+\frac{1}{\sqrt{\pi}}\right)\sqrt{t}$
		$\displaystyle\quad\leq 1.$		(2.7)

Consequently it follows from (2.5)-(2.7) that $M:\mathcal{S}\to\mathcal{S}$. Now for $v_{1},v_{2}\in\mathcal{S}$ we have

		$\displaystyle|M[v_{1}](x,t)-M[v_{2}](x,t)|$
		$\displaystyle\quad\leq\int_{0}^{t}\int_{-\infty}^{\infty}\frac{|v_{1}^{2}(s,\tau)-v_{2}^{2}(s,\tau)|}{2}\left|G(x,t;s,\tau)h_{\alpha}^{\prime}(s)+G_{s}(x,t;s,\tau)h_{\alpha}(s)\right|dsd\tau$
		$\displaystyle\quad\leq(||v_{1}||_{\infty}+||v_{2}||_{\infty})\left(\frac{||h_{\alpha}^{\prime}||_{\infty}}{2}\sqrt{t}+\frac{1}{\sqrt{\pi}}\right)\sqrt{t}||v_{1}-v_{2}||_{\infty}$		(2.8)

for all $(x,t)\in D_{T^{*}}$, again using (A.1) and (A.2). It follows from (2.8) and (2.4) that

$$||M[v_{1}]-M[v_{2}]||_{\infty}\leq\frac{1}{2}||v_{1}-v_{2}||_{\infty}$$

(2.9)

for all $v_{1},v_{2}\in\mathcal{S}$, and hence, $M$ is a contraction mapping. Since the metric space $(\mathcal{S},||\cdot||_{\infty})$ is complete, it follows that there exists $u^{*}\in\mathcal{S}$ such that $u^{*}=M(u^{*})$, i.e. $u^{*}:D_{T^{*}}\to\mathbb{R}$ is a solution to (2.2) and (2.3), as required. ∎

Let $u:D_{T^{*}}\to\mathbb{R}$ be the solution to (2.2) and (2.3) given by Proposition 2.1. Then via (A.5), for each $\beta\in(0,1)$, it follows that

$$\left|\int_{-\infty}^{\infty}u_{0}(s)(G(x_{1},t;s,0)-G(x_{2},t;s,0))ds\right|\leq||u_{0}||_{\infty}c(\beta)\left(\frac{|x_{1}-x_{2}|}{\sqrt{t}}\right)^{\beta}$$

(2.12)

for all $(x_{1},t),(x_{2},t)\in D_{T^{*}}$. Moreover, it follows from (A.5), (A.6) and (2.4) that

		$\displaystyle\bigg{|}\int_{0}^{t}\int_{-\infty}^{\infty}\frac{u^{2}(s,\tau)}{2}[(G(x_{1},t;s,\tau)-G(x_{2},t;s,\tau))h_{\alpha}^{\prime}(s)$
		$\displaystyle\quad\quad\quad\quad+(G_{s}(x_{1},t;s,\tau)-G_{s}(x_{2},t;s,\tau))h_{\alpha}(s)]dsd\tau\bigg{|}$
		$\displaystyle\quad\leq\frac{(||u_{0}||_{\infty}+1)^{2}}{2}\int_{0}^{t}\left(||h_{\alpha}^{\prime}||_{\infty}c(\beta)\left(\frac{|x_{1}-x_{2}|}{\sqrt{t-\tau}}\right)^{\beta}+c(\beta)\left(\frac{|x_{1}-x_{2}|}{\sqrt{t-\tau}}\right)^{\beta}\frac{1}{\sqrt{t-\tau}}\right)d\tau$
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)|x_{1}-x_{2}|^{\beta}{T^{*}}^{(1-\beta)/2}$		(2.13)
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)\left(\frac{|x_{1}-x_{2}|}{\sqrt{t}}\right)^{\beta}$		(2.14)

for all $(x_{1},t),(x_{2},t)\in D_{T^{*}}$. Inequality (2.11) follows from (2.12) and (2.14), as required.

∎

For $u_{n}:D_{T^{*}}\to\mathbb{R}$ given by (2.10), since $u_{0}$ is measurable and $u_{0}\in L^{\infty}(\mathbb{R})$, $u_{nx}:D_{T^{*}}\to\mathbb{R}$ exists, is continuous, and is given by

	$\displaystyle u_{nx}(x,t)$	$\displaystyle=\int_{-\infty}^{\infty}u_{0}(s)G_{x}(x,t;s,0)ds$
		$\displaystyle\quad+\int_{0}^{t_{n}}\int_{-\infty}^{\infty}\frac{u^{2}(s,\tau)}{2}\left(G_{x}(x,t;s,\tau)h_{\alpha}^{\prime}(s)+G_{sx}(x,t;s,\tau)h_{\alpha}(s)\right)dsd\tau,$		(2.17)

for all $(x,t)\in D_{T^{*}}$. After a change of variables $s=x+2\sqrt{t-\tau}\lambda$, the second integral in (2.17) can be expressed as

		$\displaystyle\int_{0}^{t_{n}}\int_{-\infty}^{\infty}\frac{u^{2}(s,\tau)}{2}G_{x}(x,t;s,\tau)h_{\alpha}^{\prime}(s)dsd\tau$
		$\displaystyle\quad-\int_{0}^{t_{n}}\int_{-\infty}^{\infty}u^{2}(x+2\sqrt{t-\tau}\lambda,\tau)\frac{\left(\lambda^{2}-1/2\right)}{\sqrt{\pi}(t-\tau)}e^{-\lambda^{2}}h_{\alpha}(x+2\sqrt{t-\tau}\lambda)d\lambda d\tau,$		(2.18)

for all $(x,t)\in D_{T^{*}}$. Now, the second integral in (2.18) can be expressed as

		$\displaystyle\int_{0}^{t_{n}}\int_{-\infty}^{\infty}\left(u^{2}(x+2\sqrt{t-\tau}\lambda,\tau)h_{\alpha}(x+2\sqrt{t-\tau}\lambda)-u^{2}(x,\tau)h_{\alpha}(x)\right)$
		$\displaystyle\quad\quad\quad\quad\quad\times\frac{\left(\lambda^{2}-1/2\right)}{\sqrt{\pi}(t-\tau)}e^{-\lambda^{2}}d\lambda d\tau$		(2.19)

for all $(x,t)\in D_{T^{*}}$. Using Proposition 2.2, $h_{\alpha}^{\prime}\in C(\mathbb{R})\cap L^{\infty}(\mathbb{R})$, and the mean value theorem, it follows that for each $\beta\in(0,1)$, there exists a constant $c$ such that

		$\displaystyle|u^{2}(x+2\sqrt{t-\tau}\lambda,\tau)h_{\alpha}(x+2\sqrt{t-\tau}\lambda)-u^{2}(x,\tau)h_{\alpha}(x)|$
		$\displaystyle\quad\leq 2||u(\cdot,\tau)||_{\infty}||h_{\alpha}||_{\infty}|u(x+2\sqrt{t-\tau}\lambda,\tau)-u(x,\tau)|$
		$\displaystyle\quad\quad+||u(\cdot,\tau)||_{\infty}^{2}|h(x+2\sqrt{t-\tau}\lambda)-h(x)|$
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)\left(\frac{\sqrt{t-\tau}|\lambda|}{\sqrt{\tau}}\right)^{\beta}$		(2.20)

for all $(x,t;\lambda,\tau)\in\mathcal{X}_{T^{*}}$. Therefore, the absolute value of the integral in (2.19) is bounded above by

		$\displaystyle\int_{0}^{t_{n}}\int_{-\infty}^{\infty}c(||u_{0}||_{\infty},\alpha,\beta)\left(\frac{\sqrt{t-\tau}|\lambda|}{\sqrt{\tau}}\right)^{\beta}\left|\lambda^{2}+1/2\right|\frac{1}{(t-\tau)}e^{-\lambda^{2}}dsd\tau$
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)\int_{0}^{t_{n}}\frac{1}{\tau^{\beta/2}(t-\tau)^{1-\beta/2}}d\tau$
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)$		(2.21)

for all $(x,t)\in D_{T^{*}}$. Therefore, via (2.17), (2.18), (2.19) and (2.21), it follows from (A.2) that

$$||u_{nx}(\cdot,t)||_{\infty}\leq\frac{c(||u_{0}||_{\infty},\alpha,\beta)}{\sqrt{t}}$$

(2.22)

for all $t\in(0,T^{*}]$. We now demonstrate that $u_{nx}$ converges uniformly on compact subsets of $D_{T^{*}}$, to a the continuous limit $u_{x}$, as $n\to\infty$. It follows from (A.2) and (2.20) that

	$\displaystyle\left|\int_{t_{n}}^{t}\int_{-\infty}^{\infty}\frac{u^{2}(s,\tau)}{2}\left(G_{x}(x,t;s,\tau)h_{\alpha}^{\prime}(s)+G_{sx}(x,t;s,\tau)h_{\alpha}(s)\right)dsd\tau\right|$
	$\displaystyle\quad\leq\frac{(||u_{0}||_{\infty}+1)^{2}}{2}||h_{\alpha}^{\prime}||_{\infty}\int_{t_{n}}^{t}\frac{1}{\sqrt{\pi(t-\tau)}}d\tau$
	$\displaystyle\quad\quad+\bigg{|}\int_{t_{n}}^{t}\int_{-\infty}^{\infty}\left(u^{2}(x+2\sqrt{t-\tau}\lambda,\tau)h_{\alpha}(x+2\sqrt{t-\tau}\lambda)-u^{2}(x,\tau)h_{\alpha}(x)\right)$
	$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\frac{\left(\lambda^{2}-1/2\right)}{\sqrt{\pi}(t-\tau)}e^{-\lambda^{2}}dsd\tau\bigg{|}$
	$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha)(2n)^{-1/2}$
	$\displaystyle\quad\quad+\int_{t_{n}}^{t}\int_{-\infty}^{\infty}c(||u_{0}||_{\infty},\alpha,\beta)\left(\frac{\sqrt{t-\tau}|\lambda|}{\sqrt{\tau}}\right)^{\beta}|\lambda^{2}-1/2|\frac{1}{(t-\tau)}e^{-\lambda^{2}}d\lambda d\tau$
	$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)\left((2n)^{-1/2}+\int_{1-1/(2n)}^{1}\frac{1}{q^{\beta/2}(1-q)^{1-\beta/2}}dq\right)$

for all $(x,t)\in D_{T^{*}}$. Therefore, via (2.17), it follows that $u_{nx}$ is uniformly convergent on (compact subsets of) $D_{T^{*}}$. It thus follows that there exists a continuous limit of $u_{nx}$ on $D_{T^{*}}$, which coincides with the derivative $u_{x}$. The bound in (2.15) follows immediately from (2.22). As a consequence $u_{x}:D_{T^{*}}\to\mathbb{R}$ can be represented, alternatively, as

	$\displaystyle u_{x}(x,t)$	$\displaystyle=\int_{-\infty}^{\infty}u_{0}(s)G_{x}(x,t;s,0)ds$
		$\displaystyle\quad+\int_{0}^{t}\int_{-\infty}^{\infty}\frac{u^{2}(s,\tau)}{2}\left(G_{x}(x,t;s,\tau)h_{\alpha}^{\prime}(s)+G_{sx}(x,t;s,\tau)h_{\alpha}(s)\right)dsd\tau$
		$\displaystyle=\int_{-\infty}^{\infty}u_{0}(s)G_{x}(x,t;s,0)ds-\int_{0}^{t}\int_{-\infty}^{\infty}(uu_{s})(s,\tau)G_{x}(x,t;s,\tau)h_{\alpha}(s)dsd\tau$		(2.23)

for all $(x,t)\in D_{T^{*}}$. Finally, from (2.23), (2.15) and (A.6) it follows that

		$\displaystyle|u_{x}(x_{1},t)-u_{x}(x_{2},t)|$
		$\displaystyle\quad\leq\frac{c(||u_{0}||_{\infty},\beta)}{\sqrt{t}}\left(\frac{|x_{1}-x_{2}|}{\sqrt{t}}\right)^{\beta}+\int_{0}^{t}\frac{c(||u_{0}||_{\infty},\alpha,\beta)}{\sqrt{\tau(t-\tau)}}\left(\frac{|x_{1}-x_{2}|}{\sqrt{t-\tau}}\right)^{\beta}d\tau$
		$\displaystyle\quad\leq\frac{c(||u_{0}||_{\infty},\beta)}{\sqrt{t}}\left(\frac{|x_{1}-x_{2}|}{\sqrt{t}}\right)^{\beta}+\frac{c(||u_{0}||_{\infty},\alpha,\beta)}{t^{\beta/2}}\int_{0}^{1}\frac{1}{\sqrt{q}(1-q)^{(1+\beta)/2}}dq|x_{1}-x_{2}|^{\beta}$
		$\displaystyle\quad\leq\frac{c(||u_{0}||_{\infty},\alpha,\beta)}{t^{(1+\beta)/2}}|x_{1}-x_{2}|^{\beta}$		(2.24)

for all $(x,t)\in D_{T^{*}}$, from which we arrive at (2.16), as required. ∎

For $u_{n}:D_{T^{*}}\to\mathbb{R}$ given by (2.10), since $u_{0}$ is measurable and $u_{0}\in L^{\infty}(\mathbb{R})$, $u_{nxx}:D_{T^{*}}\to\mathbb{R}$ exists, is continuous, and is given by

$$u_{nxx}(x,t)=\int_{-\infty}^{\infty}u_{0}(s)G_{xx}(x,t;s,0)ds-\int_{0}^{t_{n}}\int_{-\infty}^{\infty}(uu_{s})(s,\tau)G_{xx}(x,t;s,\tau)h_{\alpha}(s)dsd\tau,$$

(2.26)

for all $(x,t)\in D_{T^{*}}$. After a change of variables $s=x+2\sqrt{t-\tau}\lambda$, the second integral in (2.26) can be expressed as

$$\int_{0}^{t_{n}}\int_{-\infty}^{\infty}(uu_{s})(x+2\sqrt{t-\tau}\lambda,\tau)\frac{\left(\lambda^{2}-1/2\right)}{\sqrt{\pi}(t-\tau)}e^{-\lambda^{2}}h_{\alpha}(x+2\sqrt{t-\tau}\lambda)d\lambda d\tau,$$

(2.27)

for all $(x,t)\in D_{T^{*}}$. From (A.4) it follows that the second integral in (2.18) can be expressed as

		$\displaystyle\int_{0}^{t_{n}}\int_{-\infty}^{\infty}\left((uu_{s})(x+2\sqrt{t-\tau}\lambda,\tau)h_{\alpha}(x+2\sqrt{t-\tau}\lambda)-(uu_{s})(x,\tau)h_{\alpha}(x)\right)$
		$\displaystyle\quad\quad\quad\quad\quad\times\frac{\left(\lambda^{2}-1/2\right)}{\sqrt{\pi}(t-\tau)}e^{-\lambda^{2}}d\lambda d\tau$		(2.28)

for all $(x,t)\in D_{T^{*}}$. Using Propositions 2.2 and 2.3, $h_{\alpha}^{\prime}\in C(\mathbb{R})\cap L^{\infty}(\mathbb{R})$, and the mean value theorem, it follows that for each $\beta\in(0,1)$, there exists a constant $c$ such that

		$\displaystyle|(uu_{s})(x+2\sqrt{t-\tau}\lambda,\tau)h_{\alpha}(x+2\sqrt{t-\tau}\lambda)-(uu_{s})(x,\tau)h_{\alpha}(x)|$
		$\displaystyle\quad\leq||u_{s}(\cdot,\tau)||_{\infty}||h_{\alpha}||_{\infty}|u(x+2\sqrt{t-\tau}\lambda,\tau,\tau)-u(x,\tau)|$
		$\displaystyle\quad\quad+||u(\cdot,\tau)||_{\infty}||h_{\alpha}||_{\infty}|u_{s}(x+2\sqrt{t-\tau}\lambda)-u_{s}(x)|$
		$\displaystyle\quad\quad+||u(\cdot,\tau)||_{\infty}||u_{s}(\cdot,\tau)||_{\infty}|h(x+2\sqrt{t-\tau}\lambda)-h(x)|$
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)\left(\frac{1}{\sqrt{\tau}}\left(\frac{\sqrt{t-\tau}|\lambda|}{\sqrt{\tau}}\right)^{\beta}\right)$		(2.29)

for all $(x,t;\lambda,\tau)\in\mathcal{X}_{T^{*}}$. Therefore, the absolute value of the integral in (2.28) is bounded above by

		$\displaystyle\int_{0}^{t_{n}}\int_{-\infty}^{\infty}c(||u_{0}||_{\infty},\alpha,\beta)\frac{1}{\sqrt{\tau}}\left(\frac{\sqrt{t-\tau}|\lambda|}{\sqrt{\tau}}\right)^{\beta}\left|\lambda^{2}+1/2\right|\frac{1}{(t-\tau)}e^{-\lambda^{2}}dsd\tau$
		$\displaystyle\quad\leq c(||u_{0}||_{\infty},\alpha,\beta)\int_{0}^{t_{n}}\frac{1}{\tau^{(\beta+1)/2}(t-\tau)^{1-\beta/2}}d\tau$
		$\displaystyle\quad\leq\frac{c(||u_{0}||_{\infty},\alpha,\beta)}{\sqrt{t}}$		(2.30)

for all $(x,t)\in D_{T^{*}}$. Therefore, via (2.26), (2.27), (2.28) and (2.30), it follows from (A.3) that

$$||u_{nxx}(\cdot,t)||_{\infty}\leq\frac{c(||u_{0}||_{\infty},\alpha,\beta)}{t}$$

(2.31)

for all $t\in(0,T^{*}]$. It follows, as in the proof of Proposition 2.3 that $u_{nxx}$ converges uniformly on compact subsets of $D_{T^{*}}$, to the continuous limit $u_{xx}$, as $n\to\infty$ with the bound on $u_{xx}$ in (2.25) following immediately from (2.31). Consequently $u_{xx}:D_{T^{*}}\to\mathbb{R}$ can be represented, as

$$u_{xx}(x,t)=\int_{-\infty}^{\infty}u_{0}(s)G_{xx}(x,t;s,0)ds-\int_{0}^{t}\int_{-\infty}^{\infty}(uu_{s})(s,\tau)G_{xx}(x,t;s,\tau)h_{\alpha}(s)dsd\tau$$

(2.32)

for all $(x,t)\in D_{T^{*}}$. Furthermore, $u_{n}:D_{T^{*}}\to\mathbb{R}$ given by (2.10), since $u_{0}$ is measurable and $u_{0}\in L^{\infty}(\mathbb{R})$, $u_{nt}:D_{T^{*}}\to\mathbb{R}$ exists, is continuous, and is given by

$$u_{nt}(x,t)=u_{nxx}(x,t)-\int_{-\infty}^{\infty}(uu_{s})(s,t_{n})G(x,t;s,t_{n})h_{\alpha}(s)ds,$$

(2.33)

for all $(x,t)\in D_{T^{*}}$. From (A.1), it follows that $G(x,t;s,t_{n})$ forms a $\delta$-sequence as $n\to\infty$, and since $u$ and $u_{x}$ are continuous on $D_{T^{*}}$ it follows that

$$\int_{-\infty}^{\infty}(uu_{s})(s,t_{n})G(x,t;s,t_{n})h_{\alpha}(s)ds\to u(x,t)u_{x}(x,t)h_{\alpha}(x)$$

(2.34)

for all $(x,t)\in D_{T^{*}}$. Moreover, on any compact subset of $D_{T^{*}}$ the convergence in (2.34) is uniform. Finally, it follows, as in the proof of Proposition 2.3 that $u_{nt}$ converges uniformly on compact subsets of $D_{T^{*}}$, to the continuous limit $u_{t}$, as $n\to\infty$. As a consequence $u_{t}:D_{T^{*}}\to\mathbb{R}$ is continuous and can be represented, as

$$u_{t}(x,t)=u_{xx}(x,t)-u(x,t)u_{x}(x,t)h_{\alpha}(x)$$

(2.35)

for all $(x,t)\in D_{T^{*}}$. The bound on $u_{t}$ in (2.25) now follows from Propositions 2.1 and 2.3, and (2.35), as required. ∎