Strauss PDEs 2e: Section Exercise 6 Page 1 of 5

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1 Strauss PDEs 2e: Section Exercise 6 Page 1 of 5 Exercise 6 If a 0 = a l = a in the Robin problem, show that: (a) There are no negative eigenvalues if a 0, there is one if 2/l < a < 0, an there are two if a < 2/l. (b) Zero is an eigenvalue if an only if a = 0 or a = 2/l. Solution The eigenvalue problem to solve here is subject to the Robin bounary conitions, Part (a) X = λx X (0) ax(0) = 0 X (l) + ax(l) = 0. Suppose the eigenvalues are negative λ = γ 2. Then X = γ 2 X. The solution to this ifferential equation can be written in terms of hyperbolic sine an hyperbolic cosine. X(x) = C 1 cosh γx + C 2 sinh γx Apply the two bounary conitions here to etermine C 1 an C 2. X (0) ax(0) = γc 2 ac 1 = 0 X (l) + ax(l) = γ(c 1 sinh γl + C 2 cosh γl) + a(c 1 cosh γl + C 2 sinh γl) = 0 The first equation can be solve for C 2. C 2 = a γ C 1 Substitute this result into the secon equation. ( γ C 1 sinh γl + a ) ( γ C 1 cosh γl + a C 1 cosh γl + a ) γ C 1 sinh γl = 0 Divie both sies by γc 1 cosh γl. tanh γl + a γ + a γ + a2 tanh γl = 0 γ2 ) (1 + a2 γ 2 tanh γl = 2a γ (γ 2 + a 2 ) tanh γl = 2aγ

2 Strauss PDEs 2e: Section Exercise 6 Page 2 of 5 Divie both sies by γ 2 + a 2. tanh γl = 2aγ γ 2 + a 2 Let µ = γl in the equation to remove l from the argument of tanh. Then γ = µ/l. tanh µ = 2a µ l µ 2 l 2 + a 2 Multiply the numerator an enominator of the right sie by l 2 an group a an l together. Thus, the negative eigenvalues are λ = µ2 l 2, where µ satisfies tanh µ = µ 2 + (al) 2. The solutions to this transcenental equation occur where the graphs of these two functions intersect. If a = 0, then the equation simplifies to tanh µ = 0. To solve for µ, write tanh µ in terms of the exponential function. tanh µ = eµ e µ e µ + e µ = 0 eµ e µ = 0 e µ = e µ µ = 0 So there are no negative eigenvalues when a = 0. Suppose now that a > 0. Since tanh µ an /[µ 2 + (al) 2 ] are o functions of µ an λ is a function of µ 2, negative values of µ that satisfy the equation yiel reunant values of λ. The point is that only intersections for µ > 0 nee to be consiere. The hyperbolic tangent function is positive an the rational function is negative for µ > 0, so there are no intersections. Therefore, there are no eigenvalues for a 0. Figure 1: In blue is a plot of y = tanh µ, an in re is a plot of y = /[µ 2 + (al) 2 ]. There are no intersections for µ > 0, which means there are no negative eigenvalues.

3 Strauss PDEs 2e: Section Exercise 6 Page 3 of 5 Now we will show that there s one intersection (an hence one eigenvalue) if 2 < al < 0 an two intersections (an hence two eigenvalues) if al < 2. The erivative of tanh µ is (sech µ) 2, so tanh µ is an increasing function for all µ. Writing it in terms of exponentials, tanh µ = eµ e µ 1 e2µ e µ = + eµ 1 + e 2µ, we see that if µ = 0, then tanh µ = 0. In aition, as µ, tanh µ 1. Shifting our attention now to the rational function, the first erivative of it is (after simplifying) [ ] µ µ 2 + (al) 2 = 2(al)[µ2 (al) 2 ] [µ 2 + (al) 2 ] 2, which means the rational function is { increasing if µ < (al) ecreasing if µ > (al). The extrema occur when the first erivative is zero, that is, when µ = ±(al). µ 2 + (al) 2 = 1 (maximum) µ=(al) µ 2 + (al) 2 = 1 (minimum) µ=(al) The rational function goes to 1 at a finite value of µ unlike the hyperbolic tangent function. If the hyperbolic tangent function increases faster than the rational function initially, then there will be two intersections. If it oes not, then there will only be one intersection. [ µ (tanh µ) µ = 1 µ=0 ] µ 2 + (al) µ=0 2 = 2 al We see that if al > 2, or al < 2, then there are two intersections. If al < 2, or 2 < al < 0, then there is only one intersection. These ieas are illustrate in Figure 2 an Figure 3.

4 Strauss PDEs 2e: Section Exercise 6 Page 4 of 5 Figure 2: In blue is a plot of y = tanh µ an in re is a plot of y = /[µ 2 + (al) 2 ]. If 2 < al < 0, then the hyperbolic tangent function oes not increase faster than the rational function initially, an there is only one eigenvalue, λ 1. Figure 3: In blue is a plot of y = tanh µ an in re is a plot of y = /[µ 2 + (al) 2 ]. If al < 2, then the hyperbolic tangent function increases faster than the rational function initially, an there are two eigenvalues, λ 1 an λ 2.

5 Strauss PDEs 2e: Section Exercise 6 Page 5 of 5 Part (b) Suppose that λ = 0. Then the ifferential equation for X simplifies to X = 0, which has the general solution X(x) = C 3 x + C 4. Apply the bounary conitions here to etermine C 3 an C 4. X (0) ax(0) = C 3 ac 4 = 0 X (l) + ax(l) = C 3 + a(c 3 l + C 4 ) = 0 Solve the first equation for C 3 C 3 = ac 4 an plug it into the secon equation. ac 4 + a(ac 4 l + C 4 ) = 0 Divie both sies by C 4 an expan the left sie. a 2 l + 2a = 0 Factor a. a(al + 2) = 0 Therefore, zero is an eigenvalue if an only if a = 0 or al + 2 = 0, that is, a = 2/l.