Strauss PDEs 2e: Section Exercise 1 Page 1 of 7

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1 Strauss PDEs 2e: Section Exercise 1 Page 1 of 7 Exercise 1 Find the eigenvaues graphicay for the boundary conditions X(0) = 0, X () + ax() = 0. Assume that a 0. Soution The aim here is to determine the vaues of λ (eigenvaues) in the ODE, X = λx, that satisfy the provided boundary conditions. This probem comes about as a resut of separating variabes in a PDE. Determination of Positive Eigenvaues: λ = µ 2 Assuming λ is positive, the differentia equation for X becomes X = µ 2 X. The genera soution can be written in terms of sine and cosine. X(x) = C 1 cos µx + C 2 sin µx Now use the boundary conditions to determine C 1 and C 2. x(0) = 0 C 1 = 0 X () + ax() = 0 µ( C 1 sin µ + C 2 cos µ) + a(c 1 cos µ + C 2 sin µ) = 0 From the second condition we see that C 2 cances out, eaving a transcendenta equation for µ. µ cos µ + a sin µ = 0 Divide both sides by a cos µ. µ + tan µ = 0 a µ a = tan µ In order to group the constants together, make the substitution m = µ. Then µ = m/. m a = tan m The vaues of m that sove the equation are the m-coordinates of the intersections in the graphs of these two functions. Since the eigenvaues are given by ( m ) 2 λ = µ 2 = and both functions, m/(a) and tan m, are odd, negative vaues of m give redundant vaues for λ. The vaue of a infuences the ocations of the intersections as iustrated in the foowing figures.

2 Strauss PDEs 2e: Section Exercise 1 Page 2 of 7 Figure 1: This is a pot of the ine (in red) and the tangent function (in bue) for a < 1. Figure 2: This is a pot of the ine (in red) and the tangent function (in bue) for a = 1.

3 Strauss PDEs 2e: Section Exercise 1 Page 3 of 7 Figure 3: This is a pot of the ine (in red) and the tangent function (in bue) for 1 < a < 0. Note that in this case there is an intersection on the curve of tangent going through the origin, which gives an extra eigenvaue λ 0. Figure 4: This is a pot of the ine (in red) and the tangent function (in bue) for a > 0. The extra eigenvaue λ 0 is gone in this case.

4 Strauss PDEs 2e: Section Exercise 1 Page 4 of 7 Determination of the Zero Eigenvaue: λ = 0 Assuming λ is zero, the differentia equation for X becomes X = 0. The genera soution is a inear function. X(x) = C 3 x + C 4 Appy the boundary conditions here to determine C 3 and C 4. x(0) = 0 C 4 = 0 X () + ax() = 0 C 3 + a(c 3 + C 4 ) = 0 The second equation simpifies to C 3 (1 + a) = 0, so C 3 = 0. The trivia soution is obtained for X(x), so zero is not an eigenvaue. Determination of Negative Eigenvaues: λ = η 2 Assuming η is positive, the differentia equation for X becomes X = η 2 X. The genera soution can be written in terms of hyperboic sine and hyperboic cosine. X(x) = C 5 cosh ηx + C 6 sinh ηx Now use the boundary conditions to determine C 5 and C 6. x(0) = 0 C 5 = 0 X () + ax() = 0 η(c 5 sinh η + C 6 cosh η) + a(c 5 cosh η + C 6 sinh η) = 0 From the second condition we see that C 6 cances out, eaving a transcendenta equation for η. η cosh η + a sinh η = 0 Divide both sides by a cosh η. η + tanh η = 0 a η = tanh η a In order to group the constants together, make the substitution n = η. Then η = n/. n a = tanh n The vaues of n that sove the equation are the n-coordinates of the intersections in the graphs of these two functions. Since the eigenvaues are given by ( n ) 2 λ = η 2 = and both functions, n/(a) and tanh n, are odd, negative vaues of n give redundant vaues for λ. The vaue of a infuences the ocations of the intersections as iustrated in the foowing figures.

5 Strauss PDEs 2e: Section Exercise 1 Page 5 of 7 Figure 5: This is a pot of the ine (in red) and the hyperboic tangent function (in bue) for a < 1. There is one eigenvaue η 1. Figure 6: This is a pot of the ine (in red) and the hyperboic tangent function (in bue) for a = 1. The eigenvaue η 1 is gone in this case.

6 Strauss PDEs 2e: Section Exercise 1 Page 6 of 7 Figure 7: This is a pot of the ine (in red) and the hyperboic tangent function (in bue) for 1 < a < 0. Figure 8: This is a pot of the ine (in red) and the hyperboic tangent function (in bue) for a > 0.

7 Strauss PDEs 2e: Section Exercise 1 Page 7 of 7 In concusion, the positive eigenvaues of the ODE, X = λx, subject to the boundary conditions, X(0) = 0 and X () + ax() = 0, are where m is a soution to the equation m a λ = m2 2, = tan m. The eigenfunctions associated with these positive eigenvaues are X(x) = C 2 sin µx X k (x) = sin m kx, where k is an index going over a the positive eigenvaues. The negative eigenvaues of the ODE subject to the boundary conditions are λ = n2 2, where n is a soution to the equation n = tanh n. a The eigenfunctions associated with these negative eigenvaues are X(x) = C 6 sinh ηx X 1 (x) = sinh n 1x. Since there s ony one eigenvaue depending on a as indicated by the graphs, there s ony one possibe eigenfunction.