Convexity of Self-Similar Transonic Shocks and Free Boundaries for the Euler Equations for Potential Flow

Abstract

We are concerned with geometric properties of transonic shocks as free boundaries in two-dimensional self-similar coordinates for compressible fluid flows, which are not only important for the understanding of geometric structure and stability of fluid motions in continuum mechanics, but are also fundamental in the mathematical theory of multidimensional conservation laws. A transonic shock for the Euler equations for self-similar potential flow separates elliptic (subsonic) and hyperbolic (supersonic) phases of the self-similar solution of the corresponding nonlinear partial differential equation in a domain under consideration, in which the location of the transonic shock is apriori unknown. We first develop a general framework under which self-similar transonic shocks, as free boundaries, are proved to be uniformly convex, and then apply this framework to prove the uniform convexity of transonic shocks in the two longstanding fundamental shock problems—the shock reflection–diffraction by wedges and the Prandtl–Meyer reflection for supersonic flows past solid ramps. To achieve this, our approach is to exploit underlying nonlocal properties of the solution and the free boundary for the potential flow equation.

Appendix A. Paths Connecting Endpoints of the Minimal and Maximal Chains

For $\varLambda \subset \mathbb {R}^n$, we denote

$$\begin{aligned} \varLambda _r:=\{\varvec{\xi }\in \varLambda \;: \;\mathrm{dist}(\varvec{\xi }, \partial \varLambda )>r\}. \end{aligned}$$

(A.1)

Lemma A.1

Let $\varLambda \subset \mathbb {R}^n$ be an open set such that $\varLambda _r$ is connected for each $r\in [0, r_0]$ with given $r_0>0$. Let $P, Q\in \overline{\varLambda }$ be such that $B_r(P)\cap \varLambda _\rho$ and $B_r(Q)\cap \varLambda _\rho$ are connected for each $0\le \rho < r\le r_0$. Then there exists a continuous curve $\mathcal {S}$ with endpoints P and Q such that $\mathcal {S}^0\subset \varLambda$. More precisely, $\mathcal {S}=g([0,1])$, where $g\in C([0, 1];\mathbb {R}^n)$, g is locally Lipschitz on (0, 1), $g(0)=P$, $g(1)=Q$, and $g(t)\in \varLambda$ for all $t\in (0,1)$.

Proof

We first note that, after points P and Q are fixed, we can assume that $\varLambda$ is a bounded set; otherwise, we replace $\varLambda$ by $\varLambda \cap B$, where B is an open ball and $P, Q\in B$.

We divide the proof into three steps.

1. We notice that, if $P, Q\in \varLambda _r$ for some $r\in [0, r_0)$, then there exists a piecewise-linear path $\mathcal {S}$ with a finite number of corner points connecting P to Q such that $\mathcal {S}\subset \varLambda _{r/2}$. This is obtained via covering $\overline{\varLambda _r}$ by balls $B_{r/2}(\xi _i)$, $i=1, \dots , N$, with each $\xi _i\in \overline{\varLambda }_r$ and via noting that, since $\varLambda _r$ is connected, then any $\xi _i$ and $\xi _j$ can be connected by a piecewise-linear path with at most N corners, each section of which is a straight segment connecting centers of two intersecting balls of the cover.

Thus, the path connecting $\xi _i$ to $\xi _j$ lies in $\cup _{k=1}^N B_{r/2}(\xi _k)\subset \varLambda _{r/2}$. Then we connect P to Q by first connecting P (resp. Q) to the nearest center of ball $\xi _i$ (resp. $\xi _j$) via a straight segment that lies in $B_{r_2}(\xi _i)$ (resp. $B_{r_2}(\xi _j)$), and next connect $\xi _i$ to $\xi _j$ as above. In this way, the whole path $\mathcal {S}$ between P and Q is Lipschitz up to the endpoints and lies in $\varLambda _{r/2}$. Clearly, there exists $g\in C^{0,1}([0,1]; \mathbb {R}^n)$ with $g(0)=P$, $g(1)=Q$, and $g(t)\in \varLambda _{r/2}$ for all $t\in [0,1]$ such that $\mathcal {S}=g([0, 1]$. Therefore, this lemma is proved for any $P, Q\in \varLambda$.

2. Now we consider the case when $P\in \partial \varLambda$ and $Q\in \varLambda$. Since $\varLambda$ is open, there exists a sequence $P_m\rightarrow P$ with $P_m\in \varLambda$ for $m=1,2,\dots$. Then $P_m\in \varLambda _{r_m}$ with $r_m>0$ and $r_m\rightarrow 0$. Thus, taking a subsequence, we can assume without loss of generality that $0<r_m< \frac{r_0}{m}$ for all m.

As proved in Step 1, $P_1$ can be connected to Q by a Lipschitz curve that is parameterized by $g\in C^{0,1}([\frac{1}{2},1]; \mathbb {R}^n)$ with

$$\begin{aligned} g(\frac{1}{2})=P_1,\,\,\,\, g(1)=Q, \quad \,\,\, g(t)\in \varLambda _{\tilde{r}} \,\, \text{ for } \text{ all } t\in [0,1], \end{aligned}$$

where $\tilde{r}>0$. Since $(B_r(P)\cap \varLambda )_\varepsilon = B_{r-\varepsilon }(P)\cap \varLambda _\varepsilon$ for all $\varepsilon \in [0, \frac{r}{2})$, then the assumptions of this lemma allow to apply the result of Step 1 to sets $B_{r_0/m}(P)\cap \varLambda$ for $m=1, 2, \dots$. Thus, for each $m=1, 2, \dots$, we obtain a Lipschitz path between $P_m$ and $P_{m+1}$ which lies in $B_{r_0/m}(P)\cap \varLambda$ and is parameterized by $\displaystyle g\in C^{0,1}([\frac{1}{m+2},\frac{1}{m+1}]; \mathbb {R}^n)$ with

$$\begin{aligned}&g(\frac{1}{m+1})=P_m,\;\;\;\; g(\frac{1}{m+2})=P_{m+1},\\&g(t)\in B_{\frac{r_0}{m}}(P)\cap \varLambda _{\tilde{r}_m} \,\,\,\, \text{ for } \text{ all } t\in \big [\frac{1}{m+2},\frac{1}{m+1}\big ]. \end{aligned}$$

Combining the above together, we obtain a function $g: (0, 1]\rightarrow \mathbb {R}^n$ such that $\displaystyle g\in C([0,1]; \mathbb {R}^n)\cap C^{0,1}_\mathrm{loc}((0,1]; \mathbb {R}^n)$ with

$$\begin{aligned} \lim _{t\rightarrow 0+} g(t)=P,\quad g(1)=Q,\;\quad \; g(t)\in \varLambda \; \text{ for } \text{ all } t\in (0,1]. \end{aligned}$$

This completes the proof for the case when $P\in \partial \varLambda$ and $Q\in \varLambda$.

3. The remaining case for both $P, Q\in \partial \varLambda$ now readily follows, by connecting each of P and Q to some $C\in \varLambda$ and taking the union of the paths. $\quad \square$

Lemma A.2

Let $\varOmega \subset \mathbb {R}^2$ satisfy the conditions stated at the beginning of §3.3, and let $r^*$ be the constant from Lemma 3.10. Let $\varOmega _{\rho }$ be defined as in (A.1). Then there exists $r_0\in (0, \frac{r^*}{10}]$ such that sets $\varOmega _\rho$ are connected for all $\rho \in [0, r_0]$, and sets $B_r(E)\cap \varOmega _\rho$ are connected for any $E\in \overline{\varOmega }$ and $0\le \rho <r\le r_0$. Moreover, if $0\le \rho < r\le 2r_0$, $P\in \overline{\varOmega }$, and $\mathrm{dist}(P, \partial \varOmega )<r$, then

$$\begin{aligned} \mathrm{dist}(E, \partial \varOmega \cap B_r(P))\le C\rho \qquad \; \text{ for } \text{ each } E\in \partial \varOmega _\rho \cap B_r(P), \end{aligned}$$

(A.2)

where C depends only on the constants in the assumptions on $\varOmega$.

Proof

Throughout this proof, C denotes a universal constant, depending only on $\varOmega$. We divide the proof into two steps.

1. We first describe the structure of $\partial \varOmega _\rho$ for sufficiently small $\rho >0$ and show that $\varOmega _\rho$ is connected for such $\rho$ and (A.2) holds.

Denote by $\varGamma _i$, $i=1,\dots , m$, the smooth regions of $\partial \varOmega$ up to the corner points. Then, for $P\in \varOmega$, we have

$$\begin{aligned} {\mathrm{dist}}(P, \partial \varOmega )=\min _{i=1,\dots , m}{\mathrm{dist}}(P, \varGamma _i). \end{aligned}$$

Denote

$$\begin{aligned} \varOmega _i=\{P\in \varOmega \;:\; \mathrm{dist}(P, \partial \varOmega )=\mathrm{dist}(P, \varGamma _i)\}. \end{aligned}$$

Using that each $\varGamma _i$ is $C^{1,\alpha }$ up to the corner points, and the angles at the corner points are between $(0, \pi )$, we now show that there exists $r_0>0$ such that, for any $\rho \in (0, r_0)$ and $i=1,\dots , m$, the set:

$$\begin{aligned} \varGamma _i^\rho :=\{P\in \varOmega _i\;:\; \mathrm{dist}(P, \partial \varOmega )=\rho \} \end{aligned}$$

is a Lipschitz curve. In addition, $\varGamma _i^\rho$ is close to $\varGamma _i$ in the Lipschitz norm in the sense described bellow.

Consider first a curve $\varGamma =\{(s,t)\in \mathbb {R}^2\;:\; s=g(t)\}$ for some $g\in C^{1,\alpha }(\mathbb {R})$. Let $\rho >0$ and $\varGamma ^\rho =\{(s,t)\in \mathbb {R}^2\;:\; s>g(t), \;\mathrm{dist}((s,t), \varGamma )=\rho \}$. Fix $t_0\in \mathbb {R}$ and $r>10\rho$, and denote $L:=\Vert g\Vert _{C^{0,1}([t_0-2r, t_0+2r])}$. Then we find that, for any $t_1\in [t_0-r, t_0+r]$, there exists $s_1\in [g(t_1)+\rho , \;g(t_1)+\rho \sqrt{L^2+1}]$ such that $(s_1, t_1)\in \varGamma ^\rho$ and

$$\begin{aligned} \varGamma ^\rho \cap \{(s,t)\in \mathbb {R}^2\;:\;|t-t_0|\le r,\; s>s_1+L|t-t_1|\}=\emptyset \end{aligned}$$

by noting that $B_\rho (s_1, t_1)\cap \varGamma =\emptyset$. From this,

$$\begin{aligned} \varGamma ^\rho =\{(s,t)\in \mathbb {R}^2\;:\; s=g^\rho (t)\} \end{aligned}$$

with $g^\rho \in C^{0,1}_\mathrm{loc}(\mathbb {R})$ and $\Vert g-g^\rho \Vert _{L^\infty ([-r,r])}\le \rho \sqrt{L^2+1}$. Moreover, fix $P\in \varGamma ^\rho$. Then there exists $Q\in \varGamma$ such that $\mathrm{dist}(P, Q)=\rho$. It follows that

$$\begin{aligned} B_\rho (P)\cap \varGamma =\emptyset ,\quad \;B_\rho (Q)\cap \varGamma ^\rho =\emptyset . \end{aligned}$$

From this, for any $r\in (0,1)$, we find that there exists $r_0\in (0,\frac{r}{10}]$ depending only on r, $\alpha$, and $\hat{L}:=\Vert g\Vert _{C^{1,\alpha }([3r, 3r])}$ such that, if $\rho \in (0, r_0]$, then, for any $P=(g^\rho (t_P), t_P)\in \varGamma ^\rho \cap \{t\in [-r, r]\}$, we have

$$\begin{aligned} g^\rho (t)&\ge g^\rho (t_P)+ g'(t_Q)(t-t_P)-\hat{L} r^\alpha |t-t_P|\\&\ge g^\rho (t_P)+ g'(t_P)(t-t_P)-\hat{L} (r^\alpha +\rho ^\alpha )|t-t_P| \end{aligned}$$

for any $t\in [-2r, 2r]$, where $Q:=(g(t_Q), t_Q)$ a point such that $\mathrm{dist}(P, Q)=\rho$.

Then, noting that

$$\begin{aligned} |g(t)- g(t_P)- g'(t_P)(t-t_P)|\le L r^\alpha |t-t_P|\qquad \,\text{ for } \text{ any } t\in [-2r, 2r], \end{aligned}$$

and $\Vert g-g^\rho \Vert _{L^\infty ([-r,r])}\le \rho \sqrt{\hat{L}^2+1}$, we have

$$\begin{aligned} \Vert g-g^\rho \Vert _{C^{0,1}([-r,r])}\le \hat{L} \rho ^\alpha +\rho \sqrt{\hat{L}^2+1}. \end{aligned}$$

Thus, for any $\varepsilon \in (0,1)$, reducing $r_0$, we obtain

$$\begin{aligned} \Vert g-g^\rho \Vert _{C^{0,1}([-r,r])}\le \varepsilon \qquad \; \text{ if } \rho \le r_0. \end{aligned}$$

(A.3)

From this, under the conditions of Case (a) in the proof of Lemma 3.10, when (3.17) holds, we follow the argument in the proof of Lemma 3.10 and choosing sufficiently small $r_0$ and $\varepsilon$ in (A.3) to obtain that, for any positive $\rho <\min \{r, r_0\}$,

$$\begin{aligned}&\varOmega _\rho \cap Q_{\frac{3r}{2}}=\{(s,t)\in Q_{\frac{3r}{2}}\;:\; s>g^\rho (t)\},\nonumber \\&\partial \varOmega ^\rho \cap Q_{\frac{3r}{2}}=\{(s,t)\in Q_{\frac{3r}{2}}\;:\; s=g^\rho (t)\}. \end{aligned}$$

(A.4)

Furthermore, under the conditions of Case (b) in the proof of Lemma 3.10, when (3.18)–(3.19) hold, we repeat the argument there by choosing small $r_0$ and $\varepsilon$, and conclude that, for any positive $\rho <\min \{r, r_0\}$,

$$\begin{aligned}&\varOmega ^\rho \cap Q_{3N r}=\{(s,t)\in Q_{3N r}\;:\; s>\max (g_1^\rho (t), g_2^\rho (t))\},\nonumber \\&\partial \varOmega ^\rho \cap Q_{3N r}=\{(s,t)\in Q_{3Nr}\;:\; s=\max (g_1^\rho (t), g_2^\rho (t))\}, \end{aligned}$$

(A.5)

where $g_1^\rho$ and $g_2^\rho$ satisfy (A.3) with $g_1$ and $g_2$, respectively, and that there exists $t_\rho \in (-C\rho , C\rho )$ such that

$$\begin{aligned} g_1^\rho (t)>g_2^\rho (t)\; \text{ for } t<t_\rho ,\qquad g_1^\rho (t)<g_2^\rho (t)\; \text{ for } t>t_\rho . \end{aligned}$$

(A.6)

We adjust $r_0$ so that $r_0\le \frac{r^*}{10}$. Then, from (A.4)–(A.6) with $r=r^*$, we obtain that, for each $\rho \in (0, r_0]$, $\partial \varOmega _\rho$ is a Lipschitz curve without self-intersection. It follows that $\varOmega _\rho$ is simply-connected.

Also, combining (A.4) with (3.17) and (A.5)–(A.6) with (3.18)–(3.19) for $r=r_0$, choosing $\varepsilon$ small in (A.3) for $g, g_1$, and $g_2$, and adjusting $r_0$, we have

$$\begin{aligned} \mathrm{dist}(\partial \varOmega _\rho , \partial \varOmega )\le C\rho \qquad \text{ for } \text{ each } \rho \in (0, r^0). \end{aligned}$$

Then we conclude (A.2).

2. Now we show that $B_r(E)\cap \varOmega _\rho$ is connected for any $E\in \varOmega$ and $0\le \rho <r\le r_0$.

Assume that ${\mathrm{dist}}(E, \partial \varOmega )<2r$ (otherwise, the result already holds). Since $r_0\le \frac{r^*}{10}$, we obtain (3.17)–(3.19) for 2r instead of r, so that (A.4)–(A.6) hold for 2r instead of r. Then, arguing as in the proof of Lemma 3.10 and possibly reducing $r_0$, we obtain the following:

• If $B_r(E)\cap \varOmega$ has expression (3.21), then

  $$\begin{aligned} \varOmega _\rho \cap B_{r}(E)=\{(s,t)\;:\;t\in (t^-_\rho , t^+_\rho ),\; \max (f^-(t), g^\rho (t))<s<f^+(t)\}, \end{aligned}$$

  where $t^+_\rho \in (\frac{9r}{10}, r]$ and $t^-_\rho \in [-r, -\frac{9r}{10})$ with $|t^\pm _\rho -t^\pm |\le C\rho$, $f^+>g^\rho$ on $(t^-_\rho , t^+_\rho )$, and $f^+<g^\rho$ on $[-r, r]\setminus [t^-_\rho , t^+_\rho ]$;

• If $B_r(E)\cap \varOmega$ has expression (3.28), then

  $$\begin{aligned} \varOmega _\rho \cap B_{r}(E){=}\{(s,t)\;:\;t\in (t^-_\rho , t^+_\rho ),\; \max (f^-(t), g_1^\rho (t), g_2^\rho (t))<s<f^+(t)\}, \end{aligned}$$

  where $t^-_\rho \in [t^*- r, t^*)$ and $t^+_\rho \in (t^*, t^*+r]$ with $|t^\pm _\rho -t^\pm |\le C\rho$, and $f^+(t)>\max (g_1^\rho (t), g_2^\rho (t))$ on $(t^-_\rho , t^+_\rho )$.

The facts above imply that sets $B_r(E)\cap \varOmega _\rho$ are connected. $\quad \square$

In the next lemma, we use the minimal and maximal chains in the sense of Definition 3.7.

Lemma A.3

Let $\varOmega \subset \mathbb {R}^2$ satisfy the conditions stated at the beginning of §3.3, and let $r_0$ be the constant from Lemma A.2. Let $E_1, E_2\in \overline{\varOmega }$, and let there exist a minimal or maximal chain $\{E^i\}_{i=1}^N$ of radius $r_1\in (0, r_0]$ connecting $E_1$ to $E_2$ in $\varOmega$, i.e., $E^0=E_1$ and $E^N=E_2$. Denote

$$\begin{aligned} \varLambda =\bigcup _{i=0}^N B_{r_1}(E^i)\cap \varOmega \end{aligned}$$

so that $E_1, E_2\in \partial \varLambda$. Then there exists $\hat{r}_0>0$ such that set $\varLambda$ and points $\{E_1, E_2\}$ satisfy the conditions of Lemma A.1 with radius $\hat{r}_0$.

Proof

We divide the proof into two steps.

1. We first show the existence of $\hat{r}_0\in (0, r_1)$ such that $\varLambda _\rho$ is connected for each $\rho \in (0, \hat{r}_0]$. We recall that $r_1\le r_0\le r^*$ so that the conclusions of Lemma 3.10 hold for $B_{r_1}(E^i)$.

Since, for each $P\in \varLambda$,

$$\begin{aligned} {\mathrm{dist}}(P, \partial \varLambda )=\min \big \{ {\mathrm{dist}}\big (P, \; \partial \big (\bigcup _{i=0}^N B_{r_1}(E^i)\big )\big ),\; {\mathrm{dist}}(P, \;\partial \varOmega ) \big \}, \end{aligned}$$

then

$$\begin{aligned} \varLambda _\rho = \bigcup _{i=0}^N B_{r_1-\rho }(E^i) \cap \varOmega _\rho . \end{aligned}$$

(A.7)

By Lemma 3.10(ii) and property (b) of Definition 3.7, we see that, if $r_1\le r^*$, then $B_{r_1}(E^i) \cap B_{r_1}(E^{i+1}) \cap \varOmega \ne \emptyset$ for $i=0, \dots , N-1$. Note that all the sets in the last intersection are open. Then, recalling that $r_1\le r_0$ and using (A.2) in Lemma A.2, we obtain that there exists $\hat{r}^0\in (0, r_1)$ such that, for any $\rho \in (0, \hat{r}_0)$,

$$\begin{aligned} B_{r_1-\rho }(E^i)\cap B_{r_1-\rho }(E^{i+1}) \cap \varOmega _\rho \ne \emptyset \qquad \, \text{ for } i=0, \dots , N-1. \end{aligned}$$

Also, from Lemma A.2, sets $B_{r_1-\rho }(E^i)\cap \varOmega _\rho$ are connected, since $r_1\le r_0$. Then we obtain that $\displaystyle \bigcup _{i=0}^N B_{r_1-\rho }(E^i) \cap \varOmega _\rho$ is connected by using the argument in the proof of Lemma 3.12(i). Thus, by (A.7), we conclude that $\varLambda _\rho$ is connected for all $\rho \in (0, \hat{r}_0)$.

2. Since $B_{r_1}(E^0)\cap \varOmega \subset \varLambda$, then we use (A.7) to obtain

$$\begin{aligned} B_{r}(E^0)\cap \varLambda _\rho =B_{r}(E^0)\cap \varOmega _\rho \qquad \text{ for } \text{ all } r\in \left( \right. 0, \frac{r_1}{10}\left. \right] \text{ and } \rho \in (0, r). \end{aligned}$$

Sets $B_{r}(E^0)\cap \varOmega _\rho$ with r and $\rho$ as above are connected by Lemma A.2. Thus, the assumptions of Lemma A.1 with radius $\frac{r_1}{10}$ hold for point $E_1=E^0$. For point $E_2=E^N$, the argument is similar. $\quad \square$