Determining a Trigonometric Solution with Collocation

[roldy]

Homework Statement

Determine a three-parameter solution, with trigonometric functions using collocation at x=1/4,1/2, and 3/4.

Given: $$u=\phi_0 + c_1\phi_1 + c_2\phi_2 + c_3\phi_3$$

$$\phi_0=0, \phi_1=sin\pi x, \phi_2=sin2\pi x, \phi_3=sin3\pi x$$

Homework Equations

$$R=-\frac{d^2u}{dx^2}-cos(\pi x)=\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)$$

The Attempt at a Solution

So what I basically do here is substitute the three different values for x into
$$R=\Sigma_{i=1}^n c_i(i\pi)^2sin(i\pi x)-cos(\pi x)$$
for i=1,2,3

$$0=R(x=1/4)=\Sigma_{i=1}^n c_i(i\pi)^2sin(\frac{i\pi}{4})-cos(\frac{\pi}{4})$$
$$0=c_1(\pi)^2sin\frac{\pi}{4}+c_2(2\pi)^2sin\frac{ \pi}{2}+c_3(3\pi)^2sin\frac{3\pi}{4}-cos\frac{\pi}{4}$$
$$0=c_1\pi^2 \frac{\sqrt{2}}{2} + c_2(2\pi)^2 + c_3(3\pi)^2\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}$$

$$0=R(x=1/2)=\Sigma_{i=1}^n c_i(i\pi)^2sin(\frac{i\pi}{2})-cos(\frac{\pi}{2})$$
$$0=c_1(\pi)^2sin\frac{\pi}{2}+c_2(2\pi)^2sin \pi+c_3(3\pi)^2sin\frac{3\pi}{2}-cos\frac{\pi}{2}$$
$$0=c_1\pi^2 + c_2(0) + c_3(3\pi)^2-(0)$$

$$0=R(x=3/4)=\Sigma_{i=1}^n c_i(i\pi)^2sin(\frac{3i\pi}{4})-cos(\frac{3\pi}{4})$$
$$0=c_1(\pi)^2sin\frac{3\pi}{4}+c_2(2\pi)^2sin\frac{3\pi}{2}+c_3(3\pi)^2sin\frac{9\pi}{4}-cos(\frac{3\pi}{4})$$
$$0=c_1\pi^2 \frac{\sqrt{2}}{2} - c_2(2\pi)^2 + c_3(3\pi)^2\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}$$


Putting the coefficients into a 3X3 matrix and solving for the constants,

$$c_1=0, c_2=\frac{\sqrt{2}\pi^2}{8}, c_3=0$$

so finally

$$u=\frac{\sqrt{2}\pi^2}{8}sin(2\pi x)$$

I checked with the answers in the book and these are correct. What I need to do now is compare this against the solution of
$$u_0=\pi^{-2}(cos\pi x +2x -1)$$

I let x=1/4 and solved each of the equations

u(collocation)=1.74472
u(exact)=0.020984

Is this correct? Does this just mean that the collocation method does not produce a good enough approximation?

 

[mighty2000]

Your solution looks correct. The discrepancy between the collocation method and the exact solution could be due to the fact that the collocation method only uses a finite number of points (in this case, three) to approximate the solution, while the exact solution takes into account all points on the interval. This means that the collocation method may not be able to capture all the details of the exact solution. However, in general, the collocation method is still a valid and useful approach for solving differential equations.