## Linear Methods of Applied Mathematics Evans M. Harrell II and James V. Herod*

version of 24 March 2000

The following problem is best done with software because of some heavy algebra, particularly for part (b).
Find two Green functions for the ordinary differential equation

y'''' - y = g(x), 0 < x < 1, With the following boundary conditions (three different, independent problems!):
• (a) y(0) = y'(0) = y''(0) = y'''(0) = 0; or, respectively,
• (b) y(0) = y'(0) = y(1) = y'(1) = 0
• (c) y(0) = y''(0) = y() = y''() = 0.
• (solution)

Solution (parts a and b).

• (a) Since this is an initial-value problem, we know that G(x,t) = 0 for x < t. For x > t, G(x,t) will satisfy the homogeneous equation Gxxxx - G = 0 with the conditions that G(t+,t) = Gx(t+,t) = Gxx(t+,t) = 0,
Gxxx(t+,t) = 1.
We can easily find four independent solutions of Gxxxx - G = 0: sin(x), cos(x), sinh(x), and cosh(x) (Recall that sinh(x) := (exp(x)-exp(-x))/2 and cosh(x) := (exp(x)+exp(-x))/2.) There is actually a better choice for the solutions because the differential equation is unchanged when we change variables so that x -> x-t; the technical phrase for this is that the equation is autonomous. That is, we may write G(x,t) for x > t as c1 sin(x-t) + c2 cos(x-t) + c3 sinh(x-t) + c4 cosh(x-t). In this form we swiftly find from the initial conditions at x=t+ that c2 = c4 = c1-c3=0, and the final initial condition then gives us: G(x,t) = H(x-t) (sinh(x-t) - sin(x-t))/2. Here we have used the Heaviside function H(x-t) to write the formula compactly for x < t as well as x > t.
• (b) Let's make life easier for ourselves by using the solution to (a) to give us a head start on (b). We know that both Green functions have the same jump condition at x=t, so if F(x,t) := G(b> - G(a) then F solves Fxxxx-F = 0,
F(0,t) = Fx(0,t) = 0,
F(1,t) = -G(a)(1,t),
Fx(1,t) = -Gx(a)(1,t)
If we write F(x,t) = b1 sin(x) + b2 cos(x) + b3 sinh(x) + b4 cosh(x), we find from the boundary conditions at x=0 that b1 = -b3 and b2 = -b4. Then we apply the boundary conditions at x=1 to get the pair of equations: b3(sinh(1) - sin(1)) + b4(cosh(1)-cos(1)) = -(1/2) (sinh(1-t) - sin(1-t)),
b3(cosh(1)-cos(1)) + b4 (sinh(1) + sin(1)) = -(1/2) (cosh(1-t) - cos(1-t)).
The solution to this 2 by 2 system is:
```
Solve[{b3 (Sinh[1]-Sin[1]) + b4 (Cosh[1]-Cos[1]) == -(1/2) (Sinh[1-t] - Sin[1-t]), \
b3 (Cosh[1]-Cos[1]) + b4 (Sinh[1]+Sin[1]) == -(1/2) (Cosh[1-t] - Cos[1-t])}, {b3,b4}]

-(-Sin[1 - t] + Sinh[1 - t])
{{b3 -> ---------------------------- +
2 (-Sin[1] + Sinh[1])

((-Cos[1] + Cosh[1]) (2 (-Cos[1 - t] + Cosh[1 - t]) (-Sin[1] + Sinh[1]) -

2 (-Cos[1] + Cosh[1]) (-Sin[1 - t] + Sinh[1 - t]))) /

2
((-Sin[1] + Sinh[1]) (-4 (-Cos[1] + Cosh[1])  + 4 (-Sin[1] + Sinh[1]) (Sin[1] + Sinh[1])))\

, b4 -> -((2 (-Cos[1 - t] + Cosh[1 - t]) (-Sin[1] + Sinh[1]) -

2 (-Cos[1] + Cosh[1]) (-Sin[1 - t] + Sinh[1 - t])) /

2
(-4 (-Cos[1] + Cosh[1])  + 4 (-Sin[1] + Sinh[1]) (Sin[1] + Sinh[1])))}}

```
Hence the Green function is:
```

G[x_,t_] = H[x-t] (Sinh[x-t] - Sin[x-t])/2 + \
b3 (Sinh[x] - Sin[x]) + b4 (Cosh[x] - Cos[x])  /. %

{-(((-Cos[x] + Cosh[x]) (2 (-Cos[1 - t] + Cosh[1 - t]) (-Sin[1] + Sinh[1]) -

2 (-Cos[1] + Cosh[1]) (-Sin[1 - t] + Sinh[1 - t]))) /

2
(-4 (-Cos[1] + Cosh[1])  + 4 (-Sin[1] + Sinh[1]) (Sin[1] + Sinh[1]))) +

H[-t + x] (Sin[t - x] - Sinh[t - x])
------------------------------------ +
2

-(-Sin[1 - t] + Sinh[1 - t])
(---------------------------- + ((-Cos[1] + Cosh[1])
2 (-Sin[1] + Sinh[1])

(2 (-Cos[1 - t] + Cosh[1 - t]) (-Sin[1] + Sinh[1]) -

2 (-Cos[1] + Cosh[1]) (-Sin[1 - t] + Sinh[1 - t]))) /

2
((-Sin[1] + Sinh[1]) (-4 (-Cos[1] + Cosh[1])  + 4 (-Sin[1] + Sinh[1]) (Sin[1] + Sinh[1])))

) (-Sin[x] + Sinh[x])}

```