What is the solution to the boundary-value problem?

Question

y"-2y'+ (λ+1)y=0, y(0)=0, y'(L)=0

Answer 1

This is a homogeneous, linear, 2nd-order equation with constant coefficients. That means we can attempt a trial solution of the form y=e^Ax, where A is a constant.

 

Substituting expressions for y, y', and y'' into the differential equation results in the characteristic equation:

A²-2A+(λ+1)=0. 

 

Use the quadratic formula to solve for the possible values of A:

A={2±√[4-4(λ+1)]}/2

A=1±√[-λ]

A=1±i√λ (this is a complex number, which is perfectly okay)

 

The two possible values of A are complex conjugates:

A₁=1+i√λ

A₂=1-i√λ

 

And our general solution is 

y(x)=Be^(1+i√λ)x + Ce^(1-i√λ)x, where B and C are our constants of integration.

 

Notice that we can simplify each exponential term:

e^(1+i√λ)x = e^x+ix√λ = (e^x)(e^ix√λ),

 

Next use Euler's identity [e^ikx=cos(kx)+i sin(kx) for any constant k]:

e^(1+i√λ)x --> e^x[cos(√[λ]x)+i sin(√[λ]x)]

e^(1-i√λ)x --> e^x[cos(√[λ]x)-i sin(√[λ]x)]

 

Then,

y(x)-->Be^x[cos(√[λ]x)+i sin(√[λ]x)] + Ce^x[cos(√[λ]x)-i sin(√[λ]x)]

 

Regroup terms:

y(x)=(B+C)e^xcos(√[λ]x) + i(B-C)e^xsin(√[λ]x)

 

Rename constant coefficients (notice we can lump the i into the second constant--that is always allowed):

y(x)=a*e^xcos(√[λ]x)+b*e^xsin(√[λ]x).

 

All that remains is to apply the two boundary conditions:

• y(0)=0

0=a*e⁰cos(0)+b*e⁰sin(0)=a

 

With a=0 the solution y(x) is reduced to

y(x)=b*e^xsin(√[λ]x)

 

• y'(L)=0

y'(x)=b*{√[λ]e^xcos(√[λ]x) + e^xsin(√[λ]x)}

0=b*{√[λ]e^Lcos(√[λ]L)+e^Lsin(√[λ]L)}

0=be^L{√[λ]cos(√[λ]L)+sin(√[λ]L)}

We want to avoid setting b=0 because it reduces our solution to y(x)=0. Notice that e^L is just a constant, so we can divide it out of the equation. That means that the trig term inside {...} must equal 0:

 

√[λ]cos(√[λ]L)+sin(√[λ]L)=0

√[λ]cos(√[λ]L)=-sin(√[λ]L)

√[λ]=-tan(√[λ]L)

At this point we must plot the left side and right side of the above bolded equality as functions of λ and graphically find all intersections (this will depend on L and cannot be solved generally--but you can try plotting for L=1,2,3,etc to observe patterns). Here's an example when L=2: https://www.wolframalpha.com/input/?i=plot%5Bsqrt(x),+-tan(2sqrt(x))%5D

 

 

 

 

So we can say that the solution to the BVP is y(x)=b e^xsin(√[λ]x), where b is still arbitrary (and possibly complex) and where λ satisfies the equation √[λ]=-tan(√[λ]L) for a given value of L.