V. F. Dyachenko

ON CONDITIONS FOR UNIQUENESS OF A CONTINUOUS SOLUTION OF THE RIEMANN PROBLEM FOR A SYSTEM OF THREE EQUATIONS

(Presented by Academician M. V. Keldysh on 3 VII 1963)

Consider the hyperbolic system of three quasilinear equations

[
\frac{\partial u}{\partial t}+\frac{\partial f(u)}{\partial s}=0,
\tag{1}
]

where (u={u^{(1)},u^{(2)},u^{(3)}}), (f={f^{(1)},f^{(2)},f^{(3)}}). Hyperbolicity of the system means that the eigenvalues (\lambda_i) of the matrix (\partial f/\partial u) are all real. We shall assume that they are distinct and that the corresponding eigenvectors (l_i) are linearly independent,

[
\lambda_1<\lambda_2<\lambda_3;\qquad |\,l_1,l_2,l_3\,|\ne 0.
\tag{2}
]

The Riemann problem consists (see ((^1))) in finding a solution of the system (1), in general discontinuous, satisfying discontinuous initial data: for (t=0), (u=u^-) for (s<0), (u=u^+) for (s>0), where (u^-) and (u^+) do not depend on (s). We shall restrict ourselves to considering only continuous (for (t>0)) solutions depending on (s/t). When the number of equations in the system (1) is less than three, such a solution is unique. Below an example is constructed of a system of three equations having a nonunique solution, and conditions for uniqueness are given.

Putting (s/t=\lambda), we obtain from (1)

[
\lambda\frac{du}{d\lambda}=\frac{df(u)}{d\lambda}
\tag{3}
]

or

[
\frac{du}{d\lambda}=l_i,\qquad \lambda=\lambda_i.
\tag{3′}
]

The curves (u_i(\lambda)) in the space (u) satisfying (3′) are called characteristic curves of the (i)-th family; denote them by (L_i). A solution in the space (u) is represented by a sequence of three (or fewer) arcs (L_1,L_2,L_3), which together give the transition from (u^-) to (u^+). A continuous solution is unique if such a transition is unique.

As is known (see ((^1,{}^2))), a necessary, and sometimes sufficient, condition for uniqueness of a generalized solution is the convexity condition

[
(l_i\cdot\nabla\lambda_i)\ne 0,
\tag{4}
]

meaning that (\lambda_i) varies monotonically along (L_i).

We shall now construct a system for which, in some domain of the space (u), conditions (2), (4) are satisfied and nevertheless there is no uniqueness. Make a change of variables: from (u^{(1)},u^{(2)},u^{(3)}) pass to (x,y,z), defined by the equations

[
(l_2\cdot\nabla x)=0,\qquad (l_2\cdot\nabla y)=0,\qquad (l_2\cdot\nabla z)=1,
\tag{5}
]

i.e., we take for (x,y) a pair of independent Riemann invariants for (L_2), and (z=\lambda_2). Putting (\varphi=zu-f), by virtue of (3) and (5) we obtain (\varphi_z=u), i.e., system (1) can be written in the form

[
\frac{\partial \varphi_z}{\partial t}+\frac{\partial}{\partial s}(z\varphi_z-\varphi)=0,
\tag{6}
]

and the characteristic system (3) in the form

[
d\varphi+\delta\,d\varphi_z=\varphi_z\,dz,
\tag{7}
]

where (\delta=\lambda-z). In the new variables (l) and (L) we denote, respectively, by (m) and (M). (M_2) are straight lines parallel to the (z)-axis, to which (\delta=\delta_2=0) corresponds. (M_1) and (M_3) are determined from (7) for (\delta=\delta_1<0) and (\delta=\delta_3>0).

For simplicity let us require that (M_1) and (M_3) lie in the planes (z=\mathrm{const}); then (7) can be written in the form

[
d'\varphi+\delta\,d'\varphi_z=0,
\tag{8}
]

where (d') contains differentiation only with respect to (x) and (y). We need to construct three functions (\varphi(x,y,z)) satisfying (8) for (\delta=\delta_1) and (\delta=\delta_3). We take one of them, (\varphi^{(3)}=z^2/2), which obviously satisfies (8) for any (\delta). From the remaining, as yet undetermined, (\varphi^{(1)}(x,y,z)), (\varphi^{(2)}(x,y,z)), we form the matrix

[
K=
\begin{pmatrix}
\varphi_x^{(1)} & \varphi_y^{(1)}\
\varphi_x^{(2)} & \varphi_y^{(2)}
\end{pmatrix}
\tag{9}
]

and rewrite (8) in the form

[
(K+\delta K_z)m'=0,
\tag{10}
]

where (m') is the two-dimensional eigenvector of the first or the third family.

We shall obtain nonuniqueness of the solution if (M_1), taken at (z=z_1), and (M_3), taken at (z=z_3>z_1), do not form a regular coordinate net when superposed in the plane (x,y). But in this case there must exist a point (x_0,y_0) (say (0,0)) at which the curves of the resulting net are tangent.

The matrix

[
K^0=
\begin{pmatrix}
1+\cos z & 1+\sin z\
-1+\sin z & 1-\cos z
\end{pmatrix}
\tag{11}
]

realizes this possibility for (z_1=0,\ z_3=\frac{3}{2}\pi). Indeed, by direct substitution we verify that for

[
\delta_{1,3}=\mp 1;\qquad
m'{1,3}={-\sin \tfrac{1}{2}(z-z)}}),\ \cos \tfrac{1}{2}(z-z_{1,3
\tag{12}
]

the relation (10) is identically fulfilled, while the vectors (m'_1(z_1)) and (m'_3(z_3)) are equal to ({0,1}), i.e., they coincide. We have defined (K) on the straight line (x=0,\ y=0). It remains to define it in a neighborhood of this line so that the convexity condition (4) be satisfied; it is now written in the form

[
(m'\cdot\nabla\delta)\ne 0.
\tag{13}
]

Differentiate (10) with respect to (x); we obtain

[
(K_x+\delta K_{zx})m' + \delta_x K_zm' + (K+\delta K_z)m'_x=0.
\tag{14}
]

We define (m^*) by the condition

[
(K^+\delta K_z^)m^*=0.
\tag{15}
]

Multiplying (14) scalarly by (m^*), we obtain

[
\bigl((K_x+\delta K_{zx})m'\cdot m^\bigr)+\delta_x(K_zm'\cdot m^)=0.
\tag{16}
]

Differentiating (10) with respect to (y), we obtain an analogous equality, a linear combination of which with (16) gives

[
(m'\cdot \nabla\delta)
=
-\frac{
m^{(1)}\bigl((K_x+\delta K_{zx})m'\cdot m^\bigr)
+
m^{(2)}\bigl((K_y+\delta K_{zy})m'\cdot m^\bigr)
}{
(K_zm'\cdot m^*)
}.
\tag{17}
]

In order that ((m'\cdot \nabla\delta)) not change sign for (x=0,\ y=0,\ z_1\leq z\leq z_3), it is necessary that the numerator and denominator of the expression on the right-hand side of (17) not vanish. The denominator has already been determined, and since

[
m^*{1,3}={-\sin \tfrac12(z+z)},}),\ \cos \tfrac12(z+z_{3,1
\tag{18}
]

we easily obtain that

[
(K_z^0m'\cdot m^*)_{1,3}=\mp \frac{1}{\sqrt2}.
]

From the definition of the matrix (K) (9) it follows that the second column of (K_x) must coincide with the first column of (K_y); otherwise they are arbitrary. Setting

[
K_x^0=
\begin{pmatrix}
-1 & 0\
1 & 0
\end{pmatrix},
\qquad
K_y^0=q
\begin{pmatrix}
0 & 1\
0 & 1
\end{pmatrix},
]

where (q=\tfrac12(e^{-z}-e^{z-z_3})), and using (12), (18), (19), we obtain

[
\tfrac12(m'\cdot\nabla\delta){1,3}
=
\pm \sin^3\left(\frac{z-z\right)}}{2
+
e^{\mp z-z_{1,3}}\cos^3\left(\frac{z-z_{1,3}}{2}\right).
]

It is not difficult to verify that the last expression is different from zero for
(0=z_1\leq z\leq z_3=\tfrac32\pi).

Thus, we have constructed the matrix

[
K=K^0+xK_x^0+yK_y^0
]

and, consequently, a pair of functions (\varphi^{(1)},\varphi^{(2)}), which, together with (\varphi^{(3)}=z^2/2), generate a system of the form (6) having, in a neighborhood of the segment (x=0,\ y=0,\ 0\leq z\leq \tfrac32\pi), a nonunique solution. In this case conditions (2), (4) are satisfied, and the Jacobian of the transformation from (u) to (x,y,z), equal to (|K_z|), is different from zero.

The most essential point in the constructed example is the realization of the case (m'_1(z_1)=m'_3(z_3)). We shall now show that the absence of the effect (m'_1=m'_3) ensures uniqueness of the solution of the posed problem.

Suppose that one can pass from (u^-) to (u^+) in two ways. Then on the portion not coinciding in these two transitions one of the following three cases occurs: a) each transition consists entirely of one arc (L); b) one of them consists of two arcs (L) of different families; c) both consist of two arcs. Consider case a): the points (u^-) and (u^+) are joined, for example, both by an arc (L_1) and by an arc (L_3). We pass to the variables (x,y,z) defined above. Projecting (M_1) and (M_3) onto the plane (z=\mathrm{const}), we obtain (M'_1) and (M'_3). The closed curve (M'_1) and (M'_3) bounds a domain (\Sigma'), which is the projection of the surface (\Sigma) spanned by the arcs (M_1) and (M_3). On (\Sigma+M_1+M_3) a continuous vector field (m_1) is defined, which generates a vector field (m'_1) on (\Sigma'+M'_1+M'_3). The latter is continuous everywhere except at the self-intersection points (M_1+M_3), and has no singular points, since (m_1) and (m_2) are not collinear. Let us first suppose that (M'_1+M'_3) has no self-intersections. Under the displacement

along (M_1' + M_3') the field vector (m_1') cannot change its orientation (from internal to external or vice versa), since otherwise at the point where it changes (m_1, m_2, m_3) would prove to be coplanar. Consequently, in going around (M_1' + M_3') the vector (m_1') turns through a nonzero angle, and this is impossible, since in (\Sigma') there are no singular points of the field (m_1'), i.e. (M_1' + M_3') must have a point of self-intersection (let (x = 0), (y = 0)). This point is the projection of a segment of the (z)-axis, which is a characteristic curve of the second family (M_2), and we obtain case b). Consider one of the parts (\Sigma') into which it is divided by the self-intersection of (M_1' + M_3'), with vertex at the point (0,0). We shall denote this part also by (\Sigma'). Remove from it a small neighborhood (\varepsilon') of the point (0,0). Then either for (\Sigma' - \varepsilon') the case already considered applies, or on the portion of the boundary (\varepsilon') lying in (\Sigma') there occurs a change in the orientation of the field (m_1') relative to (M_3'). But displacement along this portion of the boundary (\varepsilon') is equivalent to displacement along the segment of the (z)-axis projected to (0,0). Therefore on this segment there will be found a point where (m_1') is tangent to (M_3'). Consequently, on the (z)-axis there are two points giving the effect (m_1' = m_3'). It is not difficult to show that case c) leads to the same conclusion. The presence or absence of the effect (m_1' = m_3') can be related to the possibility or impossibility of constructing, for each point (x,y), a pair of vectors separating (m_1') and (m_3') for all (z). If such vectors exist, then the vector fields generated by them may be taken as coordinate fields, i.e. the coordinates (x,y) can be chosen so that the vectors (m_1') and (m_3') will always lie in different octants.

Thus, for uniqueness of the continuous solution of the problem posed, it is necessary and sufficient that among the solutions of the equation

[
(l_2 \cdot \nabla \psi) = 0
]

there be found two linearly independent solutions (\psi_1, \psi_2) such that

[
(l_1 \cdot \nabla \psi_1)(l_1 \cdot \nabla \psi_2)(l_3 \cdot \nabla \psi_1)(l_3 \cdot \nabla \psi_2) < 0.
]

Received
20 VI 1963

References

(^{1}) I. M. Gelfand, UMN, 14, no. 2(86), 87 (1959).
(^{2}) P. D. Lax, Comm. Pure and Appl. Math., 10, No. 4, 537 (1957).