Question

Consider the differential equation, dx2/d2y​(x)−k2y(x)=0, subject to the boundary conditions

y(0)=a,y(L)=b Express your answer in the form, y(x)=aY1​(x)+bY2​(x). The functions Y1​(x) and Y2​(x) do not depend on a or b. In other words Y1​(x) is describing the influence of the boundary value a, and Y2​(x) is describing the influence of boundary value b. You may also find it interesting to compare your results with those in part (c) of exercise 2. Provide an interpretation of the result by describing the behavior of the functions Y1​(x) and Y2​(x). Answer: The two functions have special properties: Y1​(0)=1,Y1​(L)=0; Y2​(0)=0,Y2​(L)=1

Answer 1

Given y(0) = a and y(L) = b, and:

Y1(x) = 1 - x/L (which satisfies Y1(0) = 1, Y1(L) = 0)
Y2(x) = x/L (which satisfies Y2(0) = 0, Y2(L) = 1)

To find a solution y(x) that satisfies the boundary conditions, we can construct it as a linear combination of these two functions:

y(x) = a*Y1 + b*Y2

Plugging in the definitions of Y1 and Y2, we get:

y(x) = a*(1 - x/L) + b*(x/L)
= a - ax/L + bx/L

Then, we can distribute the 'a' and 'b' to get:

y(x) = a - ax/L + bx/L
= a - (ax - bx)/L
= a - (a - b)x/L

So, the solution to the differential equation is:

y(x) = a - (a - b)x/L

Interpretation:
Both Y1(x) and Y2(x) represent the effects of the boundary conditions a and b on the solution.

Y1(x) starts at 1 and linearly decreases to 0 as x increases from 0 to L. This means that at the starting point, the solution is strongly influenced by the boundary condition at x = 0 (value a), but this influence gradually decreases as we move away from this point.

On the other hand, Y2(x) starts at 0 and linearly increases to 1 as x increases from 0 to L. This means that initially, the solution is not influenced by the boundary condition at L (value b), but this influence increases as x gets closer to L.

Therefore, the solution at any point x is a balance between the diminishing influence of the initial boundary condition and the increasing influence of the boundary condition at the other end.