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UDC 539.3
THEORY OF ELASTICITY
R. D. BANCHURI
SOLUTION OF THE FIRST FUNDAMENTAL PROBLEM OF THE THEORY OF ELASTICITY FOR A WEDGE HAVING A FINITE CUT
(Presented by Academician N. I. Muskhelishvili on 26 VII 1965)
Let an elastic body in the plane \(\zeta=\xi+i\eta\) occupy the angle
\(-\alpha<\arg \zeta<\alpha,\ 0<\alpha\leq \pi\), which is cut along the bisector from the vertex of the angle; we shall take the length of the cut to be equal to unity.
Let the boundary \(\arg \zeta=\pm\alpha\) be free of external stresses (this may be assumed without loss of generality), and let the stress components on the cut be prescribed:
\[ Y_y=p_1(\xi),\quad X_y=q_1(\xi) \quad \text{on the upper edge of the cut,} \]
\[ Y_y=p_2(\xi),\quad X_y=q_2(\xi) \quad \text{on the lower edge of the cut,} \tag{1} \]
where \(p_1(\xi), p_2(\xi), q_1(\xi), q_2(\xi)\) are absolutely continuous functions.
This problem, in the case when \(\alpha=\pi/2\) (a half-plane with a cut), was solved in work \((^1)\). We propose another way of solving the problem.
Map the region occupied by the body onto the infinite strip
\(-1\leq y\leq 1,\ -\infty<x<\infty\), cut along the negative half-axis
\(x<0,\ y=0\). The mapping is given by the function \(\zeta=e^{\alpha z}\).
Consider two strips
\(G_1(-\infty<x<\infty,\ 0<y<1)\) and
\(G_2(-\infty<x<\infty,\ -1<y<0)\). By the method of N. I. Muskhelishvili \((^2)\), the problem formulated above is reduced to the problem of determining functions \(\Phi_{01}(z), \Psi_{01}(z)\) and \(\Phi_{02}(z), \Psi_{02}(z)\), holomorphic in the strips \(G_1\) and \(G_2\), with the following boundary conditions:
\[ \Phi_{01}(z_0)+\overline{\Psi_{01}(z_0)} =\alpha\left[Y_y^{(1)}-iX_y^{(1)}\right]e^{\alpha z_0}, \quad \operatorname{Im} z_0=0, \]
\[ \Phi_{01}(z_0)+\overline{\Psi_{01}(z_0)} +\frac{1}{\alpha}\left(e^{2i\alpha}-1\right)\overline{\Phi_{01}(z_0)}=0, \quad \operatorname{Im} z_0=1; \tag{2} \]
\[ \Phi_{02}(z_0)+\overline{\Psi_{02}(z_0)} =\alpha\left[Y_y^{(2)}-iX_y^{(2)}\right]e^{\alpha z_0}, \quad \operatorname{Im} z_0=0; \]
\[ \Phi_{02}(z)+\overline{\Psi_{02}(z_0)} +\frac{1}{\alpha}\left(e^{-2i\alpha}-1\right)\overline{\Phi_{02}(z_0)}=0, \quad \operatorname{Im} z_0=-1; \tag{3} \]
\[ \Phi_{01}(z)-\Phi_{02}(z_0)=\Psi_{01}(z_0)-\Psi_{02}(z_0)=0 \quad \text{for } \operatorname{Im} z_0=0,\ \operatorname{Re} z_0>0; \tag{4} \]
\[ Y_y^{(1)}(z_0)-Y_y^{(2)}(z_0) =X_y^{(1)}(z_0)-X_y^{(2)}(z_0)=0 \quad \text{for } \operatorname{Im} z_0=0,\ \operatorname{Re} z_0>0, \tag{5} \]
where
\[ Y_y^{(k)}(z_0) =Y_y\left(e^{\alpha x}\cos\alpha y,\ e^{\alpha x}\sin\alpha y\right), \quad X_y^{(k)}(z_0) =X_y\left(e^{\alpha x}\cos\alpha y,\ e^{\alpha x}\sin\alpha y\right), \]
\[ (x,y)\in G_k, \]
\[ \Phi_{0k}(z)=\frac{d}{dz}\varphi(e^{\alpha z}), \quad \Psi_{0k}(z)=\frac{1}{\alpha}\Phi_{0k}(z)+\frac{d}{dz}\psi(e^{\alpha z}), \quad (x,y)\in G_k,\ k=1,2. \]
\(X_y^{(k)}, Y_y^{(k)}\) are the stress components in the strip \(G_k\), and \(\varphi(\zeta), \psi(\zeta)\) are the complex potentials for the wedge with a cut.
We shall seek the analytic functions \(\Phi_{0k}(z), \Psi_{0k}(z),\ z\in G\ (k=1,2)\), in the following form:
\[ \Phi_{0k}(z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{A_k(t)}{t}e^{-itz}\,dt -\frac{1}{\sqrt{2\pi}}\frac{d}{dz}\int_{-\infty}^{\infty}\frac{A_k(t)}{t}e^{-itz}\,dt -c_k,\qquad z\in G_k, \tag{6} \]
\[ \Psi_{0k}(z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{B_k(t)}{t}e^{-itz}\,dt -\frac{1}{\sqrt{2\pi}}\frac{d}{dz}\int_{-\infty}^{\infty}\frac{B_k(t)}{t}e^{-itz}\,dt +\bar c_k,\qquad z\in G_k, \tag{7} \]
where
\[ c_k=\lim_{x\to-\infty}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{A_k(t)}{t}e^{-itz}\,dt; \]
at the point \(t=0\) the integrals are understood in the sense of the Cauchy principal value.
Below we shall show that the desired functions \(A_k(t)\), \(B_k(t)\) \((k=1,2)\) are Fourier transforms of summable functions, continuous on the entire axis, except, possibly, at the point \(t=0\). In what follows the class of such functions will be denoted by \(R_0'\).
In the representations (6) and (7) passage to the limit is possible both under the sign of differentiation and under the sign of integration (as \(z\) tends to the boundary \(G_k\)), and differentiation may be performed any number of times under the integral sign when \(\operatorname{Im} z\ne 0\).
Let us introduce some further definitions and notation. Denote by \(R_0\) the totality of all functions
\[ \Omega(t)=\int_{-\infty}^{\infty}\omega(x)e^{itx}dx, \quad\text{where}\quad \omega(x)\in L(-\infty,\infty); \]
\(R_0\) is a certain ring of continuous functions on the closed line \((^3)\). Further, denote by \(R_0^+\) (\(R_0^-\)) the subring of \(R_0\) composed of functions
\[ \Omega^+(t)=\int_0^\infty \omega(x)e^{itx}dx \quad \left(\Omega^-(t)=\int_{-\infty}^0 \omega(x)e^{itx}dx\right). \]
The ring obtained by extending \(R_0\) (\(R_0^+\), \(R_0^-\)) by adjoining the identity to it will be denoted by \(R\) (\(R^+\), \(R^-\)).
Denote by \(S_k(t)\), \(T_k(t)\), \(P_k(t)\), \(Q_k(t)\) the Fourier transforms of the functions \(ae^{\alpha x}Y_y^{(k)}(x,0)\), \(ae^{\alpha x}X_y^{(k)}(x,0)\), \(ae^{\alpha x}p_k(e^{\alpha x})\), \(ae^{\alpha x}q_k(e^{\alpha x})\) \((k=1,2)\).
Substituting the representations (6) and (7) into the boundary conditions (2) and (3), performing the Fourier transform and taking into account that \(S_k(t)=\overline{S_k(-t)}\), \(T_k(t)=\overline{T_k(-t)}\) \((k=1,2)\), we obtain
\[ A_1(t)= -\frac{ t\left[2\operatorname{sh}t\,e^{-t}-it\,\dfrac{e^{2i\alpha}-1}{\alpha}\right] }{ 4\left[\operatorname{sh}^2 t-\left(\dfrac{\sin\alpha}{\alpha}\right)^2t^2\right](1+it) }S_1(t) + \frac{ t\left[2\operatorname{sh}t\,e^{-t}+it\,\dfrac{e^{2i\alpha}-1}{\alpha}\right] }{ 4\left[\operatorname{sh}^2 t-\left(\dfrac{\sin\alpha}{\alpha}\right)^2t^2\right](1+it) }iT_1(t), \tag{8} \]
\[ A_2(t)= \frac{ t\left[2\operatorname{sh}t\,e^{t}+it\,\dfrac{e^{-2i\alpha}-1}{\alpha}\right] }{ 4\left[\operatorname{sh}^2 t-\left(\dfrac{\sin\alpha}{\alpha}\right)^2t^2\right](1+it) }S_2(t) - \frac{ t\left[2\operatorname{sh}t\,e^{t}-it\,\dfrac{e^{-2i\alpha}-1}{\alpha}\right] }{ 4\left[\operatorname{sh}^2 t-\left(\dfrac{\sin\alpha}{\alpha}\right)^2t^2\right](1+it) }iT_2(t); \tag{9} \]
\[ B_k(t)=\frac{t}{1+it}\,[S_k(t)+iT_k(t)]+\bar A_k(-t)\qquad (k=1,2). \tag{10} \]
By virtue of condition (4) we have
\[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{A_1(t)-A_2(t)}{t}e^{-itx}\,dt -\frac{1}{\sqrt{2\pi}}\frac{d}{dx}\int_{-\infty}^{\infty}\frac{A_1(t)-A_2(t)}{t}e^{-itx}\,dt +c_2-c_1=0 \]
\[ \text{for } x>0, \]
\[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{B_1(t)-B_2(t)}{t}e^{-itx}\,dt -\frac{1}{\sqrt{2\pi}}\frac{d}{dx}\int_{-\infty}^{\infty}\frac{B_1(t)-B_2(t)}{t}e^{-itx}\,dt +\bar c_1-\bar c_2 \]
\[ \text{for } x>0; \]
whence we easily conclude that if \(A_k(t), B_k(t)\in R_0\) \((k=1,2)\), then
\(A_1(t)-A_2(t)\in R_0^{-}\), \(B_1(t)-B_2(t)\in R_0^{-}\).
Applying condition (5), we obtain
\[ T_1(t)=T^{+}(t)+Q_1^{-}(t),\qquad T_2(t)=T^{+}(t)+Q_2^{-}(t); \tag{11} \]
\[ S_1(t)=S^{+}(t)+P_1^{-}(t),\qquad S_2(t)=S^{+}(t)+P_2^{-}(t), \tag{12} \]
where \(S^{+}(t)\), \(T^{+}(t)\) are the sought functions from the class \(R_0^{+}\).
From conditions (8)—(10) we shall have
\[ \begin{aligned} A_1(t)-A_2(t) &= -\frac{t\left[\operatorname{sh}t\,\operatorname{ch}t+ \frac{\sin 2\alpha}{2\alpha}\,t\right]} {\left[\operatorname{sh}^{2}t-\left(\frac{\sin\alpha}{\alpha}\right)^2 t\right](1+it)} \,S^{+}(t) \\ &\quad+ \frac{t\left[\operatorname{sh}t\,\operatorname{ch}t- \frac{\sin 2\alpha}{2\alpha}\,t\right]} {\left[\operatorname{sh}^{2}t-\left(\frac{\sin\alpha}{\alpha}\right)^2 t^2\right](1+it)} \,iT^{+}(t)+f_1(t); \end{aligned} \tag{13} \]
\[ \begin{aligned} B_1(t)-B_2(t) &= -\frac{t\left[\operatorname{sh}t\,\operatorname{ch}t+ \frac{\sin 2\alpha}{2\alpha}\,t\right]} {\left[\operatorname{sh}^{2}t-\left(\frac{\sin\alpha}{\alpha}\right)^2 t^2\right](1+it)} \,S^{+}(t) \\ &\quad- \frac{t\left[\operatorname{sh}t\,\operatorname{ch}t- \frac{\sin 2\alpha}{2\alpha}\,t\right]} {\left[\operatorname{sh}^{2}t-\left(\frac{\sin\alpha}{\alpha}\right)^2 t^2\right](1+it)} \,iT^{+}(t)+f_2(t), \end{aligned} \tag{14} \]
where \(f_1(t)\), \(f_2(t)\) are prescribed functions.
Adding equalities (12) and (13) and introducing the notation
\[ \Phi^{-}(t)=\left[A_1(t)-A_2(t)+B_1(t)-B_2(t)\right]\sqrt{1+it}; \tag{15} \]
\[ \Phi^{+}(t)=-2S^{+}(t)\sqrt{1-it}, \tag{16} \]
we obtain
\[ \Phi^{+}(t)= \frac{\left[\operatorname{sh}^{2}t-\left(\frac{\sin\alpha}{\alpha}\right)^2 t^2\right]\sqrt{1+t^2}} {\left(\operatorname{sh}t\,\operatorname{ch}t+\frac{\sin 2\alpha}{2\alpha}\,t\right)t} \,\Phi^{-}(t)+g(t). \tag{17} \]
It is easy to see that the free term \(g(t)\in R_0'\), while the coefficient of the Hilbert boundary-value problem (17) belongs to the ring of functions \(R\). Since the coefficient is positive on the entire axis \(-\infty<t<\infty\), the index of the boundary-value problem (17) is \(\chi=0\).
On the basis of work [4], the Hilbert boundary-value problem (17) in the class of functions \(\Phi^{\pm}(t)\in R_0^{\pm}\) has the unique solution
\[ \Phi^{\pm}(t_0)= \frac{X^{\pm}(t_0)}{2} \left[ \pm\frac{g(t_0)}{X^{+}(t_0)} +\frac{1}{\pi i}\int_{-\infty}^{\infty} \frac{g(t)\,dt}{X^{+}(t)(t-t_0)} \right], \tag{18} \]
where
\[ \begin{aligned} X^{\pm}(t_0) &=\exp \frac{1}{2} \Bigg[ \pm\ln \frac{\left(\operatorname{sh}^{2}t_0-\left(\frac{\sin\alpha}{\alpha}\right)^2 t_0^2\right)\sqrt{1+t_0^2}} {t_0\left(\operatorname{sh}t\,\operatorname{ch}t+\frac{\sin 2\alpha}{2\alpha}\,t_0\right)} \\ &\qquad\qquad +\frac{1}{\pi i}\int_{-\infty}^{\infty} \frac{ \ln \frac{\left(\operatorname{sh}^{2}t-\left(\frac{\sin\alpha}{\alpha}\right)^2 t^2\right)\sqrt{1+t^2}} {t\left(\operatorname{sh}t\,\operatorname{ch}t+\frac{\sin 2\alpha}{2\alpha}\,t\right)} }{t-t_0}\,dt \Bigg]. \end{aligned} \tag{19} \]
From equality (16) we have
\[ S^{+}(t)=-\frac{1}{2\sqrt{1-it}}\,\Phi^{+}(t). \]
Subtracting equality (14) from equality (13), we obtain a boundary-value problem analogous to problem (17), from which \(T^{+}(t)\) is determined. From equalities (11)—(12) it follows that \(S_k(t)\sqrt{1-it}\in R_0\), \(T_k(t)\sqrt{1-it}\in R_0\) \((k=1,2)\). Substituting the found \(S_k(t)\), \(T_k(t)\) \((k=1,2)\) into (8)—(10), we determine \(A_k(t)\), \(B_k(t)\) \((k=1,2)\). Thus, the problem is solved.
Now let us show that \(A_k(t)\), \(B_k(t)\in R_0'\). To this end we add to both sides of equalities (8) and (9), respectively, expressions of the form
\[ \frac{t e^{-t}}{(1+it)\,2\,\operatorname{ch} t}\,[S_1(t)-iT_1(t)],\qquad -\frac{t e^{t}}{(1+it)\,2\,\operatorname{ch} t}\,[S_2(t)-iT_2(t)]. \tag{20} \]
It is easy to see that the right-hand sides of the resulting equalities belong to the class \(R_0'\). In order to prove that \(A_k(t)\), \(B_k(t)\) belong to the class \(R_0'\), it is enough to show that the expressions (20) belong to the class \(R_0'\). This follows from the following lemma.
Lemma. Let \(f(t+i\tau)\) be an analytic function in the strip \(-h<\tau<h\), satisfying the conditions \(|f(t+i\tau)|<A/|t|^\beta\), \(|f'(t+i\tau)|<B/|t|^\gamma\) as \(|t|\to\infty\), \(\beta>0\), \(\gamma>1\); then \(f(t)\in R_0'\).
Proof. The function \(f(s)e^{-isx}\), \(s=t+i\tau\), for \(x\le 0\) is analytic in the strip \(0\le \tau<h_1\) \((h_1<h)\) and vanishes at infinity. The same is true for \(x\ge 0\) in the strip \(-h_1<\tau\le 0\). Therefore, using Cauchy’s theorem and the Sokhotski–Plemelj formulas, we obtain
\[ \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{f(t)}{t}e^{-itx}\,dt = \begin{cases} e^{h_1x}\Phi_1(x)+f(0)/2,\\ e^{-h_1x}\Phi_2(x)-f(0)/2, \end{cases} \tag{21} \]
where
\[ \Phi_1(x)=\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{f(t+ih_1)}{t+ih_1}e^{-itx}\,dt,\qquad \Phi_2(x)=\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{f(t-ih_1)}{t-ih_1}e^{-itx}\,dt. \tag{22} \]
According to the conditions of the lemma, \(\Phi_1(x)\), \(\Phi_2(x)\in H\) (\(H\) is the class of Hölder functions) and \(\Phi_1(0)-\Phi_2(0)=f(0)\). Hence it follows that both parts of equality (21) satisfy the Hölder conditions. Let us show that \(\Phi_1(x)\) and \(\Phi_2(x)\) have everywhere a continuous derivative, except possibly at the point \(x=0\).
We rewrite the first expression in equality (22) as follows:
\[ x\Phi_1(x)=-\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\frac{f(t+ih_1)}{t+ih_1}\right)'e^{-itx}\,dt, \]
whence, by differentiation, we obtain
\[ x\Phi_1'(x)=-\frac{1}{2\pi i}\int_{-\infty}^{\infty} t\left(\frac{f(t+ih_1)}{t+ih_1}\right)'e^{-itx}\,dt-\Phi_1(x). \]
According to the conditions of the lemma, the right-hand side of the last equality belongs to the class \(H\) and is equal to zero for \(x=0\). Therefore \(\Phi_1'(x)\) is a continuous function for \(x\ne 0\) and \(|\Phi_1'(x)|<c/|x|^\delta\), as \(x\to 0\), \(\delta<1\). It is clear that \(\Phi_2'(x)\) will have the same properties. Then
\[ \mu(x)\equiv\frac{d}{dx}\left[\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{f(t)}{t}e^{-itx}\,dt\right]\in L(-\infty,\infty), \]
whence, using the inversion formula, we obtain
\[ f(t)=\int_{-\infty}^{\infty}\mu(x)e^{itx}\,dx. \]
A. M. Razmadze Mathematical Institute
Academy of Sciences of the Georgian SSR
Received
8 VII 1965
CITED LITERATURE
- L. Wigglesworth, Mathematika, 4, p. 1, No. 7 (1957).
- N. I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity, Moscow—Leningrad, 1954.
- E. Titchmarsh, Introduction to the Theory of Fourier Integrals, Moscow—Leningrad, 1948.
- R. D. Bantsuri, G. A. Dzhanashia, DAN, 155, No. 2, 251 (1964).