In this paper the uniqueness problem of the Chaplygin's equation K(y) u_(xx)+u_(yy)=0 (K(0)=0; dk/dy >0 for k≠0) is considered. The domain D is bounded by three curves showing in the figure, where Γ_1 is a characteristic define by the equation dy = —(—K)~(-1/2) dx, Γ_2 and Γ_3 are continuous and piecewise continuously differentiable curves, and moreover, Γ_2 satisfying the condition 0 ≤ dy ≤ (—K)~(-1/2) dx.We consider solution u whose first derivatives are continuous on the boundary except at O and A. When P ∈D, P→O and P→A, the conditions imposed on u_x, u_y are u_x= O (OP~α),u_y = O (OP~α) (—1<α≤0),u_x=O(AP~β),u_y=O(AP~β) (—1/2<β≤0).The result of this paper is as follows.Let two continuous and piecewise continuously differentiable functions B, C define in the upper half-plane and satisfying the following conditions: B = x, C = 0 (when 0≤x≤1, y=0),(1) C ≥ 0 (when y > 0),(2) B= O (OP), C = O (OP) (when P∈D,P→O),(3) B = O (1), C = O (AP) (when P ∈D, P→A),(4) (B_y,+K C_x)~2 ≤ (B_x—C_y) [(C K)_y—K B_x] .(5) be the polar coordinates of x,Y and It is easily seen from (2)and (5)that m≥0. otherwise let θ_o be the angle defined by the equation LetTheorem 1. The portion of Γ_3 lying in the region 0 < θ < θ_o, 0 < r < exp satisfying the condition B dy — C dx ≥ 0 and the solution u in D which vanishes on Γ_2 and Γ_3, then u = 0 in D.We consider two particular, cases of the theorem.Ⅰ. B=x, C=y satisfying (1)—(5).In this case m=0. Thus the uniqueness theorem holds if the portion of Γ_3 lying in the domain 0 < θ < θ_o, 0 < r < exp 1/n satisfying x dy — y dx ≥ 0 .This result is a little better than a theorem of Morawetz.Ⅱ. In the case of Tricomi's equation (K(y)=y), if l is a constant satisfying 1/2≤l≤1, then B=x, C=ly satisfying (1)—(5). In this case h(θ)=(cos~2 θ+9/4 l~2 sin~2 θ)~(-1/2),m=l/2-|3/2l-1|, θ_o is defined by the equation ln (1+m)~(1/m),n=min h(θ)≥min (1,2/3l).Thus the uniqueness theorem of Tricomi's equation holds if the portion of Γ_3 lying in the domain 0 < 0 <θ_o, 0 < r < satisfying x dy—ly dx≥0.Theorem 2. If the portion of Γ_3 lying in the region y>0, (x—1)~2+Y~2<1 satisfying the condition dy≥0 and the solution u in D which vanishes on Γ_2 and Γ_3, then u=0 in D.The method of proof is to consider the energy integral = 0 for suitable choice of a, b, c.