Find the eigenvalues and eigenfunctions for y'' - 2y' + (1 + lambda)y = 0 with y(0) = y(1) = 0,

Question

Please could someone show me a step by step method that works for all these kind of problems.

Answer 1

Just as a preemptive notice, the solution for this equation is going to involve complex numbers and the Euler Identity. I'll use a bold i for the imaginary unit [i=√(-1)]. The Euler identity is

 

e^(±ikx) = cos(kx)±isin(kx).

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So, notice that this equation is 2nd order, linear, homogeneous, and constant coefficients. Even if λ is hanging around, it's still a number and doesn't depend on x. This is good, because it means we can just try out a simple solution right from the onset.

 

The simplest solution for most linear ODEs is the exponential function. We want to keep things general, so we'll say there's a constant factor k multiplying x. The assumed form of the solution is

 

y=e^(kx)

y'=k*e^(kx)

y''=k²*e^(kx).

 

Simple so far, even if it appears a bit arbitrary. Because we're dealing with a 2nd order equation, we need to have 2 independent solutions. If this "arbitrary" solution pans out into two possible choices that satisfy the ODE and boundary conditions, then we're all set--our arbitrary choice was good enough to be the right answer.

 

Plugging y, y', and y'' into the ODE we get

 

k²e^(kx) - 2ke^(kx) + (1+λ)e^(kx) = 0,

 

and we can divide out the common factor of e^(kx). This results in a quadratic equation in k that is exactly equivalent to the original ODE. If we find the values of k that satisfy the quadratic equation, we find the functions e^(kx) that satisfy the ODE. One more thing--because it's a quadratic equation, we anticipate 2 roots and thus the 2 solutions we were aiming for. We have

 

k² - 2k + (1+λ) = 0.

 

The quadratic equation with A=1, B=-2, and C=(1+λ) gives

 

k= [2±√(4-4(1+λ))]/[2].

 

After doing some algebra (this is where the complex numbers come in):

 

k₁= 1 + i√λ

k₂= 1 - i√λ.

 

The general solution is a linear combination of e^(kx) for the two k values

 

y= Ae^{([1 + i√λ]x)} + Be^{([1 - i√λ]x)}.

 

The next few steps take the general solution to its final form:

y= Ae^(x)e^(i√λx) + Be^(x)e^(-i√λx)

y= Ae^(x)[cos(√λx) + isin(√λx)] + Be^(x)[cos(√λx)- isin(√λx)]

 

We can go the usual way of distributing and collecting like terms. The "like terms" in this case are products of exp(...)cos(...) and exp(...)sin(...). It's perfectly okay to lump the i in with the other constants. If the solution is truly complex-valued, then we will recover the i when we apply the boundary conditions. For now, the general solution is 

 

y(x)= C₁ e^(x)cos(√λx) + C₂ e^(x)sin(√λx).

 

These are also the two eigenfunctions of the problem. 

 

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Moving on to the two boundary conditions, we'll arrive at a restriction on λ. This restriction will specify the allowable eigenvalues for this problem. The boundary condition at x=0 states that y=0:

 

0 = C1 e⁽⁰⁾cos(0) + C2 e⁽⁰⁾sin(0)

0 = C1(1)(1) + C2(1)(0)

0 = C1

and therefore

y(x)=C2 e^(x)sin(√λx)

 

The second boundary condition is y(x=1)=0

 

0 = C2 e⁽¹⁾sin(√λ)

We need to be careful here. If we go ahead and say C2=0, then our solution amounts to y(x)=0. This is the trivial solution. We want the non-trivial solutions, so we disallow C2=0. What's left? e⁽¹⁾ is just a number, and we know it isn't zero. That means sin(√λ) must equal zero. That's fine. We know that the sine function passes y=0 infinitely many times. We're looking for the roots of the sine function then. Simple: sin(...) = 0 whenever its argument is an integer multiple of pi. The argument in this case is √λ, so enforcing the second boundary condition is equivalent to stating

 

√λ = n*pi

or

λ = n² pi²

where n is just any integer [n=1,2,3,4,....]

 

The solution to the boundary value problem is

 

y(x) = C e^(x)sin(n*pi*x)

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A few things to take away from this:

• The eigenvalue, and by extension the eigenfunction, is restricted based on the integer n. For every integer there is a new eigenvalue, and each eigenvalue has a corresponding eigenfunction. Sometimes you see this written as λ_n and y_n(x) for emphasis.
• We can substitute the form of the eigenvalue directly into the differential equation: y'' - 2y'+ (1+n²pi²)y=0. Written this way, we can immediately say that the correct solution is y_n(x). For example we know that the solution of y'' - 2y'+ (1+81pi²)y=0 must be y₉(x)=Ce^(x)sin(9*pi*x).
• If we had an ODE that didn't exhibit a perfect square, then we could not claim to have a solution. Do not just take the square root of whatever number is multiplying the pi². This is because we placed a very particular restriction on λ when enforcing the boundary conditions. There is no simple way to accomodate this kind of situation without re-solving the entire problem from scratch. Sometimes it may not even have a solution that can be found by hand.
• λ=0 is another special case because it completely changes the structure of the ODE--we lose the λ term entirely. Again, the only way to find a solution in this instance would be to start from the beginning and solve (the much simpler) y'' - 2y' + y = 0. The solutions are no longer going to be e^(x)sin(n*pi*x). (And if they were, then we would have to let n=0....which just gives us y(x)=0, which we were avoiding.)
• Without any more information, we cannot identify the value of C. It's analogous to a constant of integration for an indefinite integral.