Evans M. Harrell II and James V. Herod*
1. (Essentially same as Herod Section 1.5, #4)
In this problem we wish to solve y = K y + x for y(x), 0 <= x <= 1.
a) Is this equation separable? NO. If the limits in x and t were not coupled, it would be separable, but the restrictions 0 < t < x < 1 destroy this. The integral equation is of Volterra type.
b) If there is a solution, it will be a continuous function on 0 < x < 1. Is this, however, guaranteed by one of the conditions for K to be "small"? NO (check one)
We would need 1 > maxx of (integral of 1.2 dt from 0 to x). But this is 1.2.
c) If there is a solution, it will be square-integrable on 0 < x < 1. Is this, however, guaranteed by one of the conditions for K to be "small"? YES
The condition for this is that the double integral of the kernel squared be less than 1. We check integral 0 to 1 of (integral 0 to x of 1.44 dt) dx, which is 0.72 < 1.
d) Solve the problem or show that it cannot be solved:
This problem can be solved by iteration, but the easier way to get an exact answer is to differentiate it: y'(x) = 1.2 y(x) + 1. Since this is the same as z' = 1.2 z, with z = (y(x) + 1/1.2), it is easy to find that y(x) = (1/1.2) (exp(1.2 x) - 1) (Checking the integral equation reveals the boundary condition y(0) = 0)
2. Let
 := (1/5)Proj_f(h) + (x/2) Proj_1(h) for 0 < x < 1, where f(x) = x<sup>2</sup>.](exam12.GIF)
Background. Recall that for any function h, the orthogonal projection onto another function f is defined by
In the first contribution to K, f is the function x2 and in the second contribution it is the constant function 1.
In this problem we wish to solve y = K y + x2 for y(x), 0 <= x <= 1.
a) Write the explicit expression for the kernel of K:
K(x,t) = x2t2 + x/2, 0 < x < 1, 0 < t < 1.
b) Find the adjoint K*:
K*(x,t) = x2t2 + t/2, 0 < x < 1, 0 < t < 1.
c) What is the condition to check for the Fredholm alternative theorem?
Whether the homogeneous equation z = K z (or z = K* z) has a nontrivial solution.
d) What does it tell us here?
In this case, K is a contraction, since we can see that it shrinks the L2 norm of any function h to at most (1/5)+(1/2) ||h||, and (1/5)+(1/2) < 1. Therefore the only fixed point is z=0.
e) Find all possible solutions, or show that there are no solutions.
This problem is separable, so the solution must be of the form
&160;&160;&160; a x2 + b x.
Substituting and a bit of algebra shows that &160;&160;&160; y(x) = 90 x2/67 + 20 x/67.
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