Fourier Cosine Series — Question 3

Question 3 – Solution

Strategy. Check the forcing mean before dividing by eigenvalues; the constant mode has eigenvalue zero.

Step 1: Check compatibility before solving. Integration gives $\int_0^\pi f=-u'(\pi)+u'(0)=0$. The stated forcing satisfies this condition, whereas $\int_0^\pi x=\pi^2/2\ne 0$. The replacement is impossible regardless of the additive level of $u$.

Step 2: Solve the mode equations. The forcing has mean zero and coefficients $F_n=2((-1)^n-1)/(\pi n^2)$. Since $-(\cos nx)''=n^2\cos nx$, $$\boxed{a_n=\frac{2((-1)^n-1)}{\pi n^4}\quad(n\geq 1).}$$ The zero-mode equation is $0=0$ and leaves the mean free. The normalization sets $a_0=0$; dividing this equation by zero would be invalid.

Step 3: Verify an independent polynomial solution. Direct integration gives $$\boxed{u=-\frac{x^3}{6}+\frac{\pi x^2}{4}-\frac{\pi^3}{24} =-\frac 4\pi\sum_{j=0}^\infty\frac{\cos((2j+1)x)}{(2j+1)^4}.}$$ Its derivative is $x(\pi-x)/2$, zero at both ends, and $-u''=x-\pi/2$. The mean of the first two terms is $\pi^3/24$, canceled by the constant. Twice integrating its coefficient integral by parts using $u'=0$ at the endpoints gives $n^2a_n=F_n$, verifying the coefficients. The continuous piecewise smooth even extension identifies the series sum.

Step 4: Separate the free constant from compatibility. A difference satisfies $w''=0$ and zero endpoint derivatives, hence is constant. Its zero mean forces it to vanish. Without normalization, all solutions are the displayed polynomial plus any real constant. Adding a constant changes neither the equation nor its derivative conditions, so it cannot fix the integrated forcing balance.