Eigenvalues and Eigenfunctions — Question 2

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Question 2

Study the derivative boundary conditions −y″=λy,0≤x≤L,y′(0)=y′(L)=0,L>0.-y''=\lambda y,\qquad 0\leq x\leq L,\qquad y'(0)=y'(L)=0, \quad L>0. A calculation that divides by λ\sqrt\lambda can lose an important mode.

Tasks

  1. Find every real eigenvalue and eigenspace, including a separate check at zero and an exclusion of negative eigenvalues.

  2. Normalize each mode by ∫0Ly2dx=1\int_0^L y^2dx=1 and y(0)>0y(0)>0. Explain why the zero mode has a different normalization factor.

  3. Integrate the differential equation to prove that every nonzero-eigenvalue mode has zero mean. Decide what happens to the zero eigenspace if zero mean is imposed as an extra constraint.

  4. For L=2L=2, test the claims that xx is a zero-mode eigenfunction and that 1+cos⁡(πx/2)1+\cos(\pi x/2) is an eigenfunction. Distinguish satisfying both endpoint conditions from belonging to a single eigenspace.

Original worksheet page 1: question and worked solution for 8-2-002
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Question 2 – Solution

Strategy. Preserve the constant solution by checking the zero parameter before dividing by a frequency.

Step 1: Find the complete spectrum. Integration by parts for any eigenfunction gives λ∫0Ly2dx=∫0L(y′)2dx≥0\lambda\int_0^L y^2dx=\int_0^L(y')^2dx\geq 0, so negative eigenvalues are impossible. At zero, y=A+Bxy=A+Bx and the derivative conditions force B=0B=0; nonzero constants are eigenfunctions. For λ=k2>0\lambda=k^2>0, y=Acos⁡(kx)+Bsin⁡(kx)y=A\cos(kx)+B\sin(kx); the left derivative condition forces B=0B=0 and the right requires sin⁡(kL)=0\sin(kL)=0. Thus λn=(nπ/L)2,En=span⁡{cos⁡(nπx/L)},n=0,1,2,….\boxed{\lambda_n=(n\pi/L)^2,\quad E_n=\operatorname{span}\{\cos(n\pi x/L)\}, \quad n=0,1,2,\ldots.}

Step 2: Normalize the modes. The constant has squared integral LL, while every positive-frequency cosine has squared integral L/2L/2. The positive value at zero fixes the signs: ϕ0=1/L,ϕn=2/Lcos⁡(nπx/L)(n≥1).\boxed{\phi_0=1/\sqrt L,\qquad \phi_n=\sqrt{2/L}\cos(n\pi x/L)\quad(n\geq 1).}

Step 3: Interpret the mean constraint. Integrating gives λ∫0Lydx=−y′(L)+y′(0)=0\lambda\int_0^L y\,dx=-y'(L)+y'(0)=0. Thus all modes with λ≠0\lambda\ne 0 have zero mean. The only constant with zero mean is the zero function, so the additional constraint removes the zero eigenspace from the admissible nonzero functions; it preserves every higher mode.

Step 4: Test the two proposed functions. The function xx has derivative one at both ends, so it fails the boundary conditions. The function g=1+cos⁡(πx/2)g=1+\cos(\pi x/2) satisfies both derivative conditions, but −g″=(π2/4)cos⁡(πx/2)-g''=(\pi^2/4)\cos(\pi x/2) is not any constant multiple of gg. Indeed, matching the constant term would require λ=0\lambda=0, which fails for the cosine term. A sum from different eigenspaces need not be an eigenfunction.

Original worksheet page 2: question and worked solution for 8-2-002

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