Example Sheet 3 Solutions

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1 Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note that if domain was e.g. [, ] it would be singular. 2. This is a regular Sturm-Liouville problem the eigenvalues are real. Let us first consider λ >, then the lutions are with sin λx cos λx φx = A sin λx + B cos λx φ = B = φl = A sin λl = A = give the trivial lution therefore we have nπ λ = for n =, 2, 3,... L nπ 2 = λ = L Let us try the case of λ = : φ = B = φ = = π = Ax + B φl = AL = = A =. Which again results in the trivial lution, therefore there is no zero eigenvalue. Now let us try the case of λ < : We set λ = ζ for me ζ >, then the lutions are e x and e x then φ ζφ =, φx = Ae x + Be x } φ = A + B = φl = Ae L + Be L = A = B = Therefore there are no negative eigenvalues.

2 3. Let us assume theorem is satisfied. Then λ R. To illustrate this we lve the eigenvalue problem. First let λ >, then with which implies the following λ For the case of λ =, we have φx = A sin λx + B cos λx φ = B = φ L = λa cos λl = A = trivial lution 2n π or λ = > for n =, 2, 3,... 2 L = λ = n 2 π 2 2 L 2. φx = Ax + B } φ = B = φl = A = = φ = trivial lution, therefore there is no zero eigenvalue. For the case of λ <, we have φx = Ae x + Be x φ = A + B = = B = A φ L = Ae L Be L = = Ae L + e L = = A = trivial lution Therefore λ >. Theorem 2 holds since there are an infinite number of eigenvalues where the smallest is λ n = n 2 π 2 for n =, 2, 3,... 2 L λ = π2 4L 2. The first part of theorem 3 is satisfied since each eigenvalues has a corresponding unique eigenfunction asciated, i.e. φ n x = sin λ n x. In order to satisfy the second part we need to ensure each eigenfunction has n zeros in the open 2

3 interval,l. Let us see this for the first few eigenfunctions sketch the functions to see this easily: φ x = sin π x no zeros 2L φ 2 x = sin 3π x zero 2L φ 3 x = sin 5π x 2 zeros. 2L Theorem 4: The lution to the asciated BVP can be written ux = a n φ n x = n= n= [ a n sin n ] π 2 L x. This is a Fourier Sine series. From MATH24 we know with Fourier s convergence theorem that any piecewise smooth function can be represented as this. For theorem 5 we need to show orthogonality of the eigenfunctions. We have φ n, φ m = sin n 2 π L x sin m 2 π L x dx if m = n: φ n, φ m = = = L 2 sin 2 n π 2 L x dx cos2n π L x 2 dx if m n we have φ n, φ m = cosn m π 2 L x cosn + mπ L x dx = L [ sinn mπx 2π n m L sinn + mπx n + m L = since n m, n + m Z. Here we used the trig identity sin A sin B = cosa B cosa + B Using the definition of skew-adjoint we have f, Lf = L f, f = Lf, f = Lf, f ] L 3

4 f, Lf is purely imaginary. For an eigenvalue asciated with L we havelφ = λφ. Next we assume that the weighting µx = or take an inner product with that weighting, then λ is purely imaginary. λ φ, φ = φ, λφ = φ, Lφ = λ = purely imaginary {}}{ φ, Lφ φ, φ }{{} real 5. Since the problem is regular Sturm-Liouville we know that λ is real. First, let λ >, then φx = A sin λx + B cos λx φ x = λa cos λx B sin λx φ = λa = = A = φ L = B λ sin λl = = λ = nπ L n =, 2, 3,... λ n = n2 π 2 L 2 φ n x = cos λ n x Now let λ =, then φ = φx = Ax + B φ x = A φ = A = φ L = = φx = B there is a zero eigenvalue. If λ <, then φx = Ae x + Be x i φ x = Ae x Be x φ = A B = = A = B φ L = Ae L Be L = = Ae L e L = = A =, no negative eigenvalues. Hence, λ. So lution to BVP is ux = n= nπ a n cos L x. 4

5 6. Let λ m, λ n be distinct eigenvalues with distinct eigenfunctions φ m, φ n. Then = Lφ m + λ m e x φ m = Lφ n + λ n e x φ n = φ n Lφ m + λ m e x φ m φ m Lφ n + λ n e x φ n dx = φ n Lφ m φ m Lφ }{{ n +λ } m e x φ m φ n λ n e x φ m φ n dx = = by Q9, sheet 2 = λ m λ n e x φ m φ n dx = and since λ m λ n, we have that the eigenfunctions are orthogonal w.r.t. the weighting e x. 7. For now take λ R. Then if λ = we have φx = Ax + B φ = B = φπ = aφ = πa = aa there exists a lution only if a = π. Thus, φ = Ax for arbitrary A is an eigenfunction only if a = π, i.e. φ = x is an eigenfunction. If λ <, let λ = ζ for ζ >. Then we have φx = Ae ζx + Be ζx φ = A + B = φπ = Ae πζ + Be πζ φ = ζa B Thus, using the mixed boundary condition we have Ae πζ + Be πζ = aζa B. Then we have the following equations to lve for A and B: A + B = Aaζ e πζ Baζ + e πζ =, which results in aζ e πζ aζ + e πζ A B = Since the RHS is zero, a lution exists iff the determinant is zero. Hence lution exists iff aζ + e πζ aζ + e πζ = e πζ + e πζ = 5

6 no lution exists for any value of ζ. If λ >, φx = A cos λ + B sin λ φ = = A = φπ = B sin λπ φ = B λ applying the mixed boundary condition we have sin λπ = a λ. the eigenfunctions are then φx = sin λx. i If < a < π, a λ will intersect sin π λ at a finite number of points, thus there are a finite number of real eigenvalues and corresponding eigenfunctions. ii When a = π, the only intersection point is λ = which results in the trivial lution φ = λ = does not apply here. BUT if a = π we saw earlier that φ = x is an eigenfunction. iii If a > π, there are no points of intersection no real roots, but of course there could be complex roots λ. iv This is not a regular Sturm-Liouville problem due to the mixed boundary conditions therefore the theorem does not apply. 8. Assume λ > φ = A cos λx + B sin λx. The boundary condition φ L = φl gives We al have A cos λl B sin λl = A cos λl + B sin λl, λ = nπ L n =, 2, 3,.... φ x = λ A sin λx + B cos λx. Then applying the boundary condition φ L = φ L gives A sin λl + B cos λl = A sin λl + B cos λl, which al results in the same value for λ. Therefore, there are 2 eigenfunctions for each eigenvalue: nπ φ nc = cos L x, nπ φ ns = sin L x, i.e. the eigenfunctions are not unique. 6