MATHEMATICS

M. B. KAPILEVICH

ON THE EFFECTIVE SOLUTION OF SINGULAR TRICOMI PROBLEMS FOR THE CHAPLYGIN EQUATION

(Presented by Academician I. N. Vekua, 17 VII 1961)

In the theory of transonic gas flows an important role is played by the equation of S. A. Chaplygin
\(\eta z_{\theta\theta}+z_{\eta\eta}+b(\eta)z_\eta=0\), in which, near the line

\[ \eta=0 \qquad b(\eta)=\sum_{m=0}^{\infty} b_m \eta^m . \]

In the domain \(\eta<0\), in the variables \(x=\theta-\frac{2}{3}(-\eta)^{3/2}\), \(y=\theta+\frac{2}{3}(-\eta)^{3/2}\), it takes the form

\[ G[z]=z_{xy}+A(z_x-z_y)=0, \tag{1} \]

\[ A=\frac{1}{6s}+\sum_{m=0}^{\infty} a_m s^{(2m-1)/3},\qquad s=y-x;\qquad a_m=\frac14(-1)^{m+1}(3/4)^{(2m-1)/3}b_m . \]

We shall call \(z(x,y)\) and \(\bar z(x,y)\) solutions of the first and second singular Tricomi problems if these functions, in a certain domain \(D\) \((0\le x\le y\le x_0)\), satisfy equation (1) and the boundary conditions

\[ z(x,x)=\tau(x),\qquad \bar z_\eta(x,x)=\nu(x),\qquad z(0,y)=\bar z(0,y)=0, \tag{2} \]

where \(\tau(0)=0\), \(\tau(x)\) and \(\nu(x)\in C^2[0,x_0]\). Problems of this type occur, for example, in the calculation of the characteristics of a wing profile \((^{1-3})\). To solve problems (2) it is sufficient to find their resolvents of Duhamel type \(U(x,y)\) and \(\bar U(x,y)\)—discontinuous integrals of equation (1) with boundary data (see \((^4)\), formulas (5)):

\[ U(x,x)=\bar U_\eta(x,x)=1,\qquad U(0,y)=\bar U(0,y)=0. \tag{3} \]

For the purpose of constructing \(U(x,y)\), make in (1) the substitution \(s=y-x\), \(t=x/y\), and then, in the equation arising thereby,

\[ \begin{aligned} Q[z] &= s(1-t^2)z_{st}-t(1-t)^2z_{tt}-s^2z_{ss} \\ &\quad +[As(1-t^2)-(1-t)^2]z_t-2As^2z_s=0 \end{aligned} \tag{4} \]

put

\[ z=U(x,y)=\sum_{n=0}^{\infty}U_n(x,y)=\sum_{n=0}^{\infty}s^{n/3}f_n(t). \tag{5} \]

Then we obtain the recurrent system of ordinary differential equations

\[ \begin{aligned} L_{m+2}[f] &= t(1-t)^2 f_{m+2}''(t) -\frac16(1-t)[2m-1+(2m+11)t]f_{m+2}'(t) \\ &\quad +\frac19 m(m+2)f_{m+2}(t) =\sum_{n=0}^{m} a_{n/2}\left[(1-t^2)f_{m-n}'(t)-\frac23(m-n)f_{m-n}(t)\right]. \end{aligned} \tag{6} \]

Here \(m=-2,-1,0,1,2,\ldots;\ f_{-2}(t)=f_{-1}(t)=0;\ a_{1/2}=a_{3/2}=a_{5/2}=\cdots=0\). In addition, in order to ensure fulfillment of the boundary conditions (3), we shall assume that \(f_0(0)=0,\ f_0(1)=1\), and also \(f_n(0)=0\),

\[ \lim_{t\to 1}\left[(1-t)^{n/3} f_n(t)\right]=0,\qquad \text{if } n=1,2,\ldots \tag{7} \]

For \(m=-1,1,3,5,\ldots\) the right-hand sides of formulas (7) contain only the functions \(f_k(t)\) with odd indices \(k=1,3,5,\ldots,m\), i.e., these equations have the form \(L_{m+2}[f_{m+2}]=\varphi_{m+2}(f_1,f_3,f_5,\ldots,f_m)\), where \(\varphi_{m+2}(0,0,0,\ldots,0)=0\). In the first of these equations \(L_1[f_1]=0\) \((m=-1)\) the right-hand side \(\varphi_1\equiv 0\), and, as follows from the known properties of Gauss functions, for its solution
\(f_1(t)=B_1\sqrt{t}\,F(^{4}/_{3},\,^{5}/_{6},\,^{3}/_{2};t)\), satisfying the requirement \(f_1(0)=0\), the second of conditions (7) can be fulfilled only for \(B_1=0\), i.e., when \(f_1(t)\equiv0\). This, in turn, means that \(\varphi_3\equiv0\), and \(f_3(t)\) is likewise determined from the homogeneous equation \(L_3[f]=0\). On the other hand, it is not difficult to see that for any values \(m=1,2,\ldots\) the homogeneous equations \(L_{m+2}[f]=0\) cannot yield nontrivial solutions \(f_{m+2}^{(0)}(t)\) satisfying both requirements (7) simultaneously. Indeed, in a neighborhood of the point \(t=0\)

\[ \begin{aligned} f_{m+2}^{(0)}(t) &=(1-t)^{-m/3}\Bigl[A_{m+2}F(^{5}/_{6},-m/3,\,^{1}/_{6}-m/3;t) \\ &\qquad\qquad +B_{m+2}t^{5/6+m/3}F(^{5}/_{3}+m/3,\,^{5}/_{6},\,^{11}/_{6}+m/3;t)\Bigr]. \end{aligned} \tag{8} \]

Putting here \(A_{m+2}=0\), we obtain the zero value \(f_{m+2}^{(0)}(0)=0\); however, continuing the remaining hypergeometric function into a neighborhood of the point \(t=1\), we find that the limit

\[ \lim_{t\to 1}\left[(1-t)^{(m+2)/3}f_{m+2}^{(0)}(t)\right] = B_{m+2}\Gamma(^{2}/_{3})\Gamma(^{11}/_{6}+m/3)\, /\, \Gamma(^{5}/_{6})\Gamma(^{5}/_{3}+m/3) \]

turns to zero only when \(B_{m+2}=0\), i.e., if \(f_{m+2}^{(0)}(t)\equiv0\). Thus, step by step we arrive at the conclusion \(f_1(t)\equiv f_3(t)\equiv f_5(t)\equiv\cdots\equiv0\), and, consequently, the summation in formula (5) must be carried out only over even indices \(n\). In other words, the function \(U\) admits an expansion in a series of the form (5) in positive integral powers of the variable \(\eta=-(^{3}/_{4}s)^{2/3}\). The integral of the first equation \(L_0[f]=0\) \((m=-2)\) of system (6) with the boundary data \(f_0(0)=0,\ f_0(1)=1\) is the Joukovsky resolvent \(U_0(x,y)\) of the Euler–Poisson equation (4):

\[ U_0(x,y)=f_0(t)=B_0t^{1/6}F(^{1}/_{6},\,^{1}/_{3},\,^{7}/_{6};t)=I_t(^{1}/_{6},\,^{2}/_{3}), \tag{9} \]

where \(B_0\Gamma(^{7}/_{6})\Gamma(^{2}/_{3})=\Gamma(^{5}/_{6})\). Further, in finding a suitable integral of the second \((m=0)\) of equations (6),

\[ L_2[f]=t(1-t)f_2''(t)+{}^{1}/_{6}(1-11t)f_2'(t)=a_0(1+t)f_0'(t) \tag{10} \]

one may disregard the solution \(f_2^{(0)}(t)\) corresponding to the homogeneous equation \(L_2[f]=0\), since the function needed by us with boundary data (7) is contained among the particular solutions \(f_2^{(1)}(t)\) of equation (10) and has the form \(f_2(t)=-{}^{3}/_{2}a_0f_0(t)\). In a similar way, from conditions (6), (7) the functions \(f_4(t), f_6(t),\ldots\) are determined uniquely. Formula (9) shows that as \(t\) increases from \(t=0\) to \(t=1\), \(f_0(t)\) varies within the limits \(0\leq f_0(t)\leq1\). Thus, \(|f_2(t)|\leq{}^{3}/_{2}|a_0|\), when \(0\leq t\leq1\), and, consequently, in a neighborhood of the line \(\eta=0\) the second term \(U_2=s^{2/3}f_2(t)\) of expansion (5) will be a quantity of at least order \(|\eta|\). In an analogous way, estimating \(f_n(t)\) for \(0\leq t\leq1\), we arrive at the conclusion that near the sonic line \(\eta=0\)

\[ U=U_0+O(|\eta|). \]

Comparing the Riemann resolvents \(U\) and \(U_0\) makes it possible to give the found connection formulas, which transform the solutions \(z_0\) and \(\bar z_0\) of the Euler–Poisson equation into the integrals \(z,\bar z\) of the Chaplygin equation. Such connections, as in (5), are established for the case of transformed initial values \(\tau(x)=F(x)\tau_0(x)\), \(\nu(x)=P(x)\nu_0(x)\). Let us also note that the well-known maximum principle for solutions \(z(x,y)\), as well as the lemma proved in (6), characterizing the behavior of the function \(z(x,y)\) at its extremal points, are direct consequences of the corresponding inequalities \(0 \leq U(x,y)\leq 1\), \(U_n(x,x)>0\), \((x,y)\in \bar D\), for the resolvent \(U(x,y)\).

Let us turn, finally, to the construction of the Riemann kernel \(\bar U(x,y)\) of the second singular Tricomi problem. For this purpose we shall seek the solution of equation (4) in the form of the series

\[ \bar z=\bar U(x,y)=\sum_{n=2}^{\infty}s^{n/3}f_n(t) =\sum_{n=2}^{\infty}(4/3)^{n/3}(-\eta)n/2\,f_n(t), \tag{11} \]

where, in order to satisfy the conditions (2), we require that \((n=3,4,5,\ldots)\)

\[ f_2(0)=0,\qquad \lim_{t\to 1}\left[{}^{3}/_{2}(1-t)f'_2(t)-f_2(t)\right]=(^{3}/_{4})^{2/3}, \tag{12} \]

\[ f_n(0)=0,\qquad \lim_{t\to 1}(1-t)^{(n-2)/3}\left[nf_n(t)-3(1-t)f'_n(t)\right]=0. \tag{13} \]

Then here too we arrive at the system (6), in which this time \(f_0(t)\equiv 0\), so that \(f_2(t)\) satisfies the homogeneous equation \(L_2[f]=0\). Its solution with the boundary data (12) is the function \(f_2(t)=B_2t^{5/6}F(5/3,5/6,11/6;t)\), where \(2B_2\Gamma(1/3)\Gamma(11/6)=(4/3)^{1/3}\Gamma(1/6)\). Further it is not difficult to see that in the present case as well \(f_3=f_5=f_7=\cdots=0\). On the other hand, in computing the function \(f_4(t)\) we can no longer confine ourselves to the particular solution \(f_4^{(1)}(t)=-{}^{3}/_{2}a_0f_2(t)\) of the nonhomogeneous equation
\[ L_4[f]=t(1-t)^2f''_4(t)-{}^{1}/_{2}(1-t)(1+5t)f'_4(t)+{}^{8}/_{9}f_4(t) =a_0\left[(1-t^2)f'_2(t)-{}^{4}/_{3}f_2(t)\right], \]
so that now \(f_4(t)\) must be constructed in the form of the sum \(f_4(t)=f_4^{(0)}(t)+f_4^{(1)}(t)\), where \(f_4^{(0)}(t)\) is that integral of the equation \(L_4[f]=0\) for which \(f_4^{(0)}(0)=0\), namely
\[ f_4^{(0)}(t)=B_4t^{7/2}(1-t)^{-2/3}F(7/3,5/6,5/2;t), \]
and the second of the conditions (13) gives \(3^{4/3}\Gamma(5/6)B_4=-8\sqrt{\pi}\,a_0\). Continuing this process further, in an analogous way we uniquely determine \(f_6(t), f_8(t),\ldots\), and thereby prove the regularity of the expansion (11). In the case of infinitely differentiable initial functions \(\tau(x)\) and \(\nu(x)\) for the Euler–Poisson equation \((A=1/6s)\), there hold the symbolic expansions (5):

\[ z_0=B_0t^{1/6}(1-t)^{2/3}\Xi_1(5/6,1/6,1,7/6;t,-\delta_x)\tau(x), \]

\[ \bar z_0=\bar B_0\eta t^{5/6}(1-t)^{-2/3}\Xi_1(1/6,5/6,1,11/6;t,-\delta_x)\nu(x), \]

where \(\Xi_1\) is Humbert’s confluent hypergeometric function, \(\delta_x=xD_x\), and \(\bar B_0=-(4/3)^{2/3}B_2\). In constructing analogous resolving operators \(z=V(x,y,D_x)\tau(x)\) and \(z=V(x,y,D_x)\nu(x)\) in the case (1), one may use the following device (4).

By the substitution \(z(x,y)=e^{kx}V(x,y)\) we transform (1) to the form

\[ G[V]+k(V_y+AV)=0. \tag{14} \]

Then, in order to find the function \(V(x,y,D_x)\), it is enough to determine the solution \(V(x,y,k)\) \((k=D_x=\mathrm{const})\) of equation (14) under the boundary conditions \(V(x,x)=1,\ V(0,y)=0\), i.e. to find the Riemann resolvent of the first singular Tricomi problem for equation (14). The operator \(V(x,y,k)\) will be...

where it should be sought in the form of the series \(V(x,y,k)=\sum_{n=0}^{\infty} k^n W_n(x,y)\), substitution of which into (14) gives the system

\[ G[W_0]=0;\qquad G[W_{m+1}]+W_{my}+A W_m=0\quad (m=0,1,2,3,\ldots). \tag{15} \]

To this system we adjoin the equalities \(W_0(0,y)=W_m(0,y)=W_m(x,x)=0\), \(W_0(x,x)=1\) for \(m=1,2,3,\ldots\). Then it is clear that \(W_0\) coincides with the resolvent \(U(x,y)\) computed above. The subsequent functions \(W_m(x,y)\), naturally, like \(W_0\), should be sought in the form of expansions \(W_m=s^m\sum_{n=0}^{\infty}s^{n/3}\Phi_n^{(m)}(t)\) from the equations \((s=y-x,\ t=x/y)\):

\[ Q[W_{m+1}]=st(1-t)W_{mt}-s^2 W_{ms}-A s^2 W_m\quad (m=0,1,2,\ldots). \tag{16} \]

This leads to ordinary differential equations for the functions \(\Phi_n^{(m)}(t)\). Thus, for example, in the case when \(A=1/6s\), one may put \(W_m=s^m\Phi_m(t)\), and then (16) gives
\[ \Delta_{m+1}[\Phi]=t(1-t)^2\Phi_{m+1}''-(1-t)\cdot [m+1/6+(m+13/6)t]\Phi_{m+1}'-(m+1/3)(m+1)\Phi_{m+1} =(m+1/6)\Phi_m-t(1-t)\Phi_m', \]
where \(m=-1,0,1,2,\ldots\), with \(\Phi_{-1}(t)\equiv0\). At the same time the boundary values for \(W_m\) are ensured by the equalities \(\Phi_0(0)=\Phi_m(0)=0\), \(\Phi_m(1)=1\), \(\lim_{t\to1}[(1-t)^m\Phi_m(t)]=0\), if \(m=1,2,3,\ldots\). Hence, first of all, it follows that \(\Phi_0(t)=f_0(t)\). Next we establish that for \(m=0\) and \(m=1\) the inhomogeneous equations and the required boundary conditions can be satisfied by setting
\[ \Phi_1(t)=\frac12\Phi_0(t)-3t\Phi_0'(t) \]
and
\[ \Phi_2(t)=\frac{7}{16}\Phi_1(t)-\frac38 t\Phi_1'(t). \]
Finally, by induction we find the recurrence formula

\[ m(m-2/3)\Phi_m(t)=(m-5/6)\Phi_{m-1}(t)-t\Phi_{m-1}'(t)\quad (m=1,2,3,\ldots), \]

from which, taking into account the known Gauss relations for contiguous hypergeometric functions, we obtain

\[ \Phi_m(t)=\mu_m t^{m+1/6}F(m+1/6,\ m+1/3,\ m+7/6;\ t) =\overline{\mu}_m B_t(m+1/6,\ 2/3-m), \]

where \(\mu_m(m+1/6)=(-1)^m\overline{\mu}_m\); \(m!\Gamma(1/6)\Gamma(2/3)\overline{\mu}_m=\Gamma(5/6)\), and \(B_t(p,q)\) is the incomplete Euler beta-function. Analogous constructions are also carried out in constructing the resolving operator \(V(x,y,D_x)\).

Received
11 VII 1961

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