本文探討Van der Pol方程式X"+μX'(X2— 1) + x = 0在相平面上一條特殊臨界曲線，記為 y∞ (χ ;μ )。它是Van der Pol方程式在相平面上特定區域對於極限環的漸進解。本研究證明在相平面的上半平面中，Van der Pol方程式的極限環與臨界曲線y∞ (χ ;μ )之差至多為Ο(μ −1/ 3 )，當−1 ≤ x ≤ 0，μ →+∞。更進一步，可以利用這個結果，證明當μ →+∞時，相平面上任一條Van der Pol方程式的解軌線從y軸出發且在極限環外部時，當第一次與x=1相交於第四象限之後，其與極限環的差至多為Ο(μ −1/ 3 )。

This article is concerned with the special trajectory y∞ (χ ;μ ) which is the leading term of the asymptotic solution of Van der Pol equation x"+μx' (x2 −1) + x = 0in the phase plane for some region. We show that in the phase plane, the difference of this asymptotic solution and the limit cycle of Van der Pol equation is not greater than Ο(μ −1/ 3 ) as μ →+∞ for all −1 ≤ x ≤ 0 . Using this result, we can show that every trajectory of Van der Pol equation starting from y-axis with initial value bigger than that of the limit cycle gets close to the limit cycle by Ο(μ −1/ 3 ) from its first time on intersecting x =1 in the four quadrant as μ →+∞ .