Series Solutions — Question 8

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Question 8

Investigate the forced singular equation xy‴−y″=x,x>0.x y'''-y''=x,\qquad x>0. One wishes to continue a solution to x=0x=0. Analytic continuation, twice continuous differentiability and three times continuous differentiability are different requirements here.

Tasks

  1. Attempt an ordinary power series at zero. Identify the precise coefficient equation that obstructs a formal Taylor solution.

  2. Solve the equation on x>0x>0 by first setting z=y″z=y''. Give the complete three-constant family.

  3. Determine the highest integer kk for which the solutions extend as CkC^k functions to [0,∞)[0,\infty). Explain why no choice of the homogeneous constants removes the obstruction.

  4. Select the constants giving y=x3log⁡x/6y=x^3\log x/6. Find its endpoint limit, its minimum on (0,1](0,1] and its value at one. Sketch it, showing zero as an excluded endpoint of the original domain.

Original worksheet page 1: question and worked solution for 7-7-008
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Question 8 – Solution

Strategy. Let the failed coefficient equation identify a logarithmic term forced by resonance.

Step 1: Locate the formal obstruction. For y=∑anxny=\sum a_nx^n, matching xnx^n gives (n+2)(n+1)(n−1)an+2={1,n=1,0,n≠1.(n+2)(n+1)(n-1)a_{n+2}=\begin{cases}1,&n=1,\\0,&n\ne 1.\end{cases} At n=0n=0 this requires a2=0a_2=0, but at n=1n=1 it requires 0=10=1. No formal ordinary power series, and hence no analytic solution, exists at zero.

Step 2: Solve the reduced equation. For z=y″z=y'', (z/x)′=1/x(z/x)'=1/x, giving z=xlog⁡x+Cxz=x\log x+Cx. Integrating twice yields y=A+Bx+x3log⁡x6+(C6−536)x3.\boxed{y=A+Bx+\frac{x^3\log x}{6} +\left(\frac C6-\frac 5{36}\right)x^3.} The three homogeneous modes are 1,x,x31,x,x^3; the forced logarithm is additional.

Step 3: Classify endpoint smoothness. As x↓0x\downarrow 0, y→Ay\to A, y′→By'\to B and y″=xlog⁡x+Cx→0y''=x\log x+Cx\to 0. These limits give a C2C^2 extension. But y‴=log⁡x+1+C→−∞y'''=\log x+1+C\to-\infty, so no C3C^3 extension exists. Constants can change finite terms, not cancel the coefficient of log⁡x\log x. This is a formal inconsistency, unlike a solvable but divergent coefficient recurrence.

Step 4: Verify the illustrated member. Set A=B=0A=B=0, C=5/6C=5/6. Then y′=x2(3log⁡x+1)/6y'=x^2(3\log x+1)/6, so xmin=e−1/3,ymin=−118e,y(1)=0,limx↓0y(x)=0.\boxed{x_{\min}=e^{-1/3},\qquad y_{\min}=-\frac 1{18e},\qquad y(1)=0,\quad\lim_{x\downarrow 0}y(x)=0.} The derivative changes from negative to positive at this unique minimum. The open marker records the original domain x>0x>0, even though a C2C^2 extension to zero is possible.

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Original worksheet page 2: question and worked solution for 7-7-008

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