a. Compute all the eigenvalues and the corresponding eigenfunctions for the Sturm-Liouville problem $$u''(x)+\lambda u(x)= 0, \ u'(0)=u'(\frac{\pi}{2})=0.$$

b. Let $u_1$ and $u_2$ be the eigenfunctions corresponding to the two smallest eigenvalues. Find the projection (in $L^2(0,\frac{\pi}{2})$) of $f(x)=\sin2x$ onto the subspace spanned by $u_1$ and $u_2$.

a. From the eigenvalues I got that $m = \pm i\lambda$ thus the eigenfunction is $$u(x)=c_1\cos(mx)+c_2\sin(mx)$$ then $$u'(x)=-mc_1\sin(mx) +mc_2\cos(mx).$$

Substituting the inital gives $c_2=0$ and $$c_1 = 0 \implies \sin(m\frac{\pi}{2})\implies m=2n.$$

I understand this part. But I am confused on part b, I do no know where to start?

  • $\begingroup$ Shouldn't your $m$ be proportional to $\sqrt{\lambda}$? $\endgroup$ – Wright-Moran Sep 17 '14 at 5:20
  • $\begingroup$ @DanielCooney Can you elaborate please? $\endgroup$ – Robben Sep 17 '14 at 5:23
  • $\begingroup$ For example, if $u(x) = \cos(\lambda x)$, then $u'(x) = -\lambda \sin(\lambda x)$ and $u''(x) = - \lambda^2 \cos(\lambda x) = - \lambda^2 u(x)$. $\endgroup$ – Wright-Moran Sep 17 '14 at 5:25
  • $\begingroup$ @DanielCooney How did you obtain that? $\endgroup$ – Robben Sep 17 '14 at 5:47
  • $\begingroup$ By guessing $u(x)$ and differentiating twice. Clearly this doesn't work because of the $\lambda^2$, which is why the actual function you would want would have the form $u(x) = c_1 \cos(\sqrt{\lambda} x) + c_2 \sin(\sqrt{\lambda} x)$. $\endgroup$ – Wright-Moran Sep 17 '14 at 6:00

Your argument that $u(x)$ takes on the form $c_1 \cos(mx) + c_2 \sin(mx)$ relied on the fact that $\lambda$ was positive (as you know that $\sin$ and $\cos$ satisfy the fact that, for instance, taking two derivatives of $\sin(\sqrt{\lambda} x)$ gives you back $-\lambda \sin(\sqrt{\lambda} x)$). Going through the motions with $u(x) = c_1 \cos(\sqrt{\lambda} x) + c_2 \sin(\sqrt{\lambda} x)$ would give us that $\sqrt{\lambda} = 2n$, and therefore $\lambda = 4n^2$ for each $n \in \mathbb{N}$ with corresponding eigenfunction $u_{\lambda}(x) = \sin(\sqrt{4n^2} x) = \sin(2nx)$.

If $\lambda = 0$, then the differential equation reduces to $u''(x) = 0$, which has solution of the form $u(x) = c_1 x + c_2$. Applying the boundary condition $u'(0) = 0$ tells us that $c_1 = 0$, so $u(x) = c_2$, and therefore any constant function is an eigenfunction corresponding to $\lambda = 0$.

If $\lambda < 0$, let $s = - \lambda$, and so our differential equation becomes $$u''(x) - s u(x) = 0 \Longrightarrow u''(x) = s u(x)$$ Two functions that satisfy this differential equation are $e^{\sqrt{s} x}$ and $e^{- \sqrt{s} x}$, so we can say that an eigenfunction corresponding to eigenvalue $s$ has the form $$u(x) = c_1 e^{\sqrt{s} x} + c_2 e^{- \sqrt{s} x} \Longrightarrow u'(x) = c_1 \sqrt{s} e^{\sqrt{s} x} - c_2 \sqrt{s} e^{- \sqrt{s} x}$$ Applying the boundary condition for $u'(0)$ gives $$0 = u'(0) = c_1 + c_2 \Longrightarrow c_1 = - c_2$$ Applying the other boundary condition gives that $$0 = u'(\frac{\pi}{2}) = c_1 \sqrt{s} e^{\sqrt{s} \frac{\pi}{2}} - (-c_1) \sqrt{s} e^{-\sqrt{s} \frac{\pi}{2}} \Longrightarrow 0 = c_1 \sqrt{s} \left(e^{\sqrt{s} x} + e^{-\sqrt{s} \frac{\pi}{2}}\right)$$ But $s > 0$ by assumption, and the both terms involving $e$ are positive, so this boundary condition implies that $c_1 = 0$ (and thereby $c_2 = 0$ as well). Thus there are no nontrivial eigenfunctions for negative eigenvalues $\lambda$.

Now addressing part $b$, we have that the two smallest eigenvalues are $0$ and $4$, which have corresponding eigenfunctions $u_1(x) = c$ (for constant $c \in \mathbb{R}$ and $u_2(x) = \cos(2x)$. We wish to project $f(x) = \sin(2x)$ onto a basis consisting of these two eigenfunctions, so let's pick $c = 1$ to give us the basis $\{1, \cos(2x)\}$. In general, the projection of a function $f(x) \in L^2(0,\frac{\pi}{2})$ onto a basis vector $g(x)$ is given by $<f,g>_{L^2(0,\frac{\pi}{2})} =\displaystyle\int_0^{\frac{\pi}{2}} f(x) g(x) dx$. We see that

$$<\sin(2x),1> _{L^2(0,\frac{\pi}{2})} = \displaystyle\int_0^{\frac{\pi}{2}} \sin(2x) dx = - \frac{1}{2} \cos(2x) \bigg|_0^{\frac{\pi}{2}} = 1$$

and that

$$<\sin(2x),<\cos(2x)>_{L^2(0,\frac{\pi}{2})} = \displaystyle\int_0^{\frac{\pi}{2}} \sin{2x} \cos{2x} dx = \frac{1}{2} \displaystyle\int_0^{\frac{\pi}{2}} \sin(4x) dx = \frac{1}{8} \left(- \cos(4x) \bigg|_0^{\frac{\pi}{2}}\right) = 0$$

Furthermore, we can find that $1$ and $\cos(2x)$ are orthogonal by computing

$$<1,\cos(2x)> _{L^2(0,\frac{\pi}{2})} = \displaystyle\int_0^{\frac{\pi}{2}} \cos{2x} dx = \frac{1}{2} \left(- \sin{2x} \bigg|_0^{\frac{\pi}{2}} \right) = 0$$

Then, the projection of a function $f(x)$ onto an orthogonal basis $\{g_1(x),g_2(x)\}$ is given by $<f,g_1>_{L^2(0,\frac{\pi}{2})} + <f,g_2>_{L^2(0,\frac{\pi}{2})}$, and therefore the projection of $\sin(2x)$ onto $\{1, \cos(2x)\}$ is just equal to $1$.

  • $\begingroup$ Thank you very much!! How were you able to get that $0$ and $4$ as the smallest eigenvalues and their corresponding eigenfunctions? $\endgroup$ – Robben Sep 17 '14 at 7:26
  • $\begingroup$ We know, from the argument above, that there are no negative eigenvalues, that $\lambda = 0$ is an eigenvalue, and that the positive eigenvalues have the form $\lambda = 4n^2$ for $n \in \mathbb{Z}^{+}$. Therefore 0 and 4 are the smallest eigenvalues. $\endgroup$ – Wright-Moran Sep 17 '14 at 7:29
  • $\begingroup$ Thank you very much for clarifying! $\endgroup$ – Robben Sep 17 '14 at 19:23

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