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We have $$U_{tt}=U_{xx} \quad 0<x<{\pi \over 2}, \quad t>0$$ $$U(x,0)=U(0,t)=U_x({\pi \over 2},t)=0$$ $$U_t(x,0)=\cos(5x) \cdot \sin(x)$$

We are looking for a solution in the form $U(x,t)=X(x)T(t)$

Then $U_x=X'(x)T(t) \quad U_t=X(x)T'(t)$

and $U_{xx}=X''(x)T(t) \quad U_{tt}=X(x)T''(t)$

From what is given $X''(x)T(t)=X(x)T''(t)\Rightarrow \exists \lambda=\text{const}:{X''(x) \over X(x)}={T''(t) \over T(t)}=\lambda$

Now $U(0,t)=0=X(0)T(t)$ but $U \not =0 \Rightarrow T \not = 0 \Rightarrow X(0)=0$

And $U_x({\pi \over 2},t)=0=X'({\pi \over 2})T(t)$ but $U \not =0 \Rightarrow T \not = 0 \Rightarrow X'({\pi \over 2})=0$

We thus got the Sturm–Liouville problem ${X''(x) \over X(x)}=\lambda \land X(0)=0=X'({\pi\over2})$

We proceed to find $\lambda=-(2k+1)^2 \land X(x)=C_k\cdot \sin[(2k+1)x]$

We now examine ${T''(t) \over T(t)}=\lambda=-(2k+1)^2$

We get $T(t)=\hat A_k \cdot \cos (2k+1)t+\hat B_k \cdot \sin(2k+1)t$

$\Rightarrow U(x,t)=X(x)T(t)=\sin(2k+1)x[ A_k \cdot \cos (2k+1)t+B_k \cdot \sin(2k+1)t]$

Where $A_k=\hat A_kC_k \land B_k=\hat B_kC_k$

We are looking for a solution in the form $$U(x,t)=\sum_{k=0}^\infty \sin(2k+1)x[ A_k \cdot \cos (2k+1)t+B_k \cdot \sin(2k+1)t]$$

We have $U(x,0)=0=\sum_{k=0}^\infty \sin(2k+1)x\cdot A_k$

As we know the scalar product in $L_2$ we get $$A_k \int_0^{\pi \over 2} {\sin^2[(2k+1)x]\,\mathrm{d}x}=0$$

Let us examine $$\int_0^{\pi \over 2} {\sin^2[(2k+1)x]\,\mathrm{d}x} = \int_0^{\pi \over 2} {{1-\cos[(4k+2)x]\over 2}\,\mathrm{d}x}={\pi \over 4}-{{1 \over 2}\int_0^{\pi \over 2}} {\cos[(4k+2)x]\,\mathrm{d}x}={\pi \over 4}-{1 \over 2(4k-2)}\int_0^{\pi \over 2} {\cos[(4k+2)x]\,\mathrm{d}[(4k+2)x]}={\pi \over 4}-{1 \over 2(4k-2)}\int_0^{\pi \over 2} {\cos(x)\,\mathrm{d}x}={\pi \over 4}-{1 \over 8k-4}[\sin({\pi\over 2})-\sin(0)]={\pi \over 4}-{1 \over 8k-4}$$

This is my calculation. Note that $$\int {\cos[(4k+2)x]\,\mathrm{d}x}={1 \over 4k-2}\int {\cos[(4k+2)x]\,\mathrm{d}[(4k+2)x]}={1 \over 4k-2}\int {\cos(x)\,\mathrm{d}x}={1 \over 4k-2}\sin(x)$$ as $$\int {f(x)\,\mathrm{d}x}={1 \over a}\int {f(x)\,\mathrm{d}(ax)}$$

Here is where wolfram-alpha disagrees. They say that $$\int_0^{\pi \over 2} {\cos[(4k+2)x]\,\mathrm{d}x}=0$$ since $$\int {\cos(ax)\,\mathrm{d}x}={1 \over a} \sin(ax)$$

Either way all $A_k=0$. But to get $B_k$ we use $$U_t(x,0)=\cos(5x)\sin(x)=\sum_{k=0}^\infty \sin(2k+1)x\cdot B_k \cdot (2k+1)$$ $$\Rightarrow (2k+1)B_k\int_0^{\pi \over 2} {\sin^2[(2k+1)x]\,\mathrm{d}x} = \int_0^{\pi \over 2} \sin[(2k+1)x]\cos(5x){\sin(x)\,\mathrm{d}x}$$

1) Can you help me with those two integrals.

2) Do I need to verify that the series converges, and how? Which criterion should I use? Does it not follow from the general theory of these equations that the series converges? When does it?

Librecoin
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superAnnoyingUser
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1 Answers1

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$$X'' + \lambda^2 X = 0$$

subject to $X(0) = X'(\pi/2) = 0$. When you write the solution as

$$X(x) = A \cos{\lambda x} + B \sin{\lambda x}$$

you see that the condition at $x=0 \implies A=0$. The condition at $x=\pi/2$ implies that, for $k \in \mathbb{Z}$:

$$\frac{\pi}{2} \lambda = (2 k+1) \pi/2 \implies \lambda = 2 k + 1 $$

so the spatial solution is

$$X(x) = B \sin{(2 k+1) x}$$

We also see that $T'' + (2 k+1)^2 T = 0$ subject to $T(0)=0$ and $T'(0) = \cos{5 x} \sin{x} = (\sin{6 x} - \sin{4 x})/2$

So you are looking for a solution of the form

$$U(x,t) = \sum_k B_k \sin{(2 k+1) x} \sin{(2 k+1) t}$$

The $B_k$ are found by applying the second initial condition (the derivative at $t=0$):

$$\sum_k ((2 k+1) B_k \sin{(2 k+1) x} = \frac{\sin{6 x} - \sin{4 x}}{2} $$

Because the Fourier series forms an orthogonal basis, we may find that the $B_K$ satisfy

$$\begin{align}(2 k+1) B_k \int_0^{\pi/2} dx \: \sin^2{[(2 k+1) x]} &= \frac{1}{2} \int_0^{\pi/2} dx \: \sin{[(2 k+1) x]} (\sin{6 x} - \sin{4 x})\\ (2 k+1) B_k \frac{\pi}{4} &= (-1)^{k+1}\left ( \frac{3}{ 4 k^2 +4 k -35} + \frac{2 }{ 4 k^2 +4 k -15} \right ) \end{align}$$

The integrals on the right are accomplished by repeated application of the sine product formula I used above. The integral on the left may be found by using $\sin^2{[(2 k+1) x]} = (1-\cos{[2 (2 k+1) x]})/2$.

Because the $B_k \sim 1/k^3$ for large $k$, even without the alternating series, the series would converge.

Ron Gordon
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  • Yes, thank you sir. If you look carefully, you will notice that $X'(\frac \pi 2) = 0$, hence my solution to the Sturm–Liouville problem is correct. I guess the questions stand. I would appreciate any help on those integrals. – superAnnoyingUser Feb 13 '13 at 14:53
  • I mean if ${e_k}{k=0}^\infty$ is an ortho-normal basis, then $(e_i,e_j)=\delta{ij}$ where $(\cdot ,\cdot )$ is the scalar product and $\delta$ is Kroneker's symbol. However, the second integral I suggested is not of this kind. The first one $\int_0^\frac\pi2 sin^2[(2k+1)x]dx$ should be the identity. A question then is which the identity of $L_2(0,\frac\pi2)$ is. The correct answer is either wolfram's or mine. I don't know what I'm missing. – superAnnoyingUser Feb 13 '13 at 14:53
  • @George: fixed, integrals evaluated and questions answered. – Ron Gordon Feb 13 '13 at 16:09
  • Thank you, sir. You have a funny way of writing the integrand out of the integral. It got me confused one time. I thought you meant ∫d[xf(x)] instead of ∫f(x)dx. – superAnnoyingUser Feb 14 '13 at 09:34
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    Yeah, I hear about that a lot here. It's a habit I got from my training in optics; it just makes multiple integrals much clearer, and treats the integral as an operator. – Ron Gordon Feb 14 '13 at 10:08