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Consider the wave equation

$$u_{tt}=u_{xx}$$ with boundary conditions $$u\left(x,t\right)=X\left(x\right)T\left(t\right)$$ $$u\left(\pm a,t\right)=0$$ $$u\left(x,0\right)=\max\left(0,1-\left(x/b\right)^2\right)$$ $$u_t\left(x,0\right)=0$$ $$-a\le x\le a$$ Let $$u\left(x,t\right)=X\left(x\right)T\left(t\right)$$ $$\frac{X^{\prime\prime}}{X}={\frac{T}{T}}^{\prime\prime}=-\lambda$$ $$X(x)=A\cos\sqrt\lambda x+B\sin\sqrt\lambda x$$ $$X\left(-a\right)=A\cos\sqrt\lambda (-a)+B\sin\sqrt\lambda(-a)=0$$ $$\Longrightarrow\sqrt\lambda\left(-a\right)=n\pi$$ with $$A=0$$

Or $$\sqrt\lambda\left(-a\right)=\frac{n\pi}{2}$$ with $$B=0$$ Which seems slightly ambiguous. Just wondering whether there is a clue I am not seeing which would allow us to conclude that $$X(0) = 0$$ which would confirm that $$\sqrt\lambda\left(-a\right)=n\pi$$ with $$A\ =0.$$

Thank you in advance for your time.

Rócherz
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Gabsmacked
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1 Answers1

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There is no ambiguity. The nontrivial solutions to $X(\pm a)=0$, where $$ X(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x), \tag{1} $$ are \begin{align} X_n^{(1)}(x)&=A_n\cos\left(\sqrt{\lambda_n^{(1)}}x\right)\qquad \left(\lambda_n^{(1)}=\left(\frac{(2n+1)\pi}{2a}\right)^2,\, n\in\mathbb{N}\right), \tag{2a} \\ X_n^{(2)}(x)&=B_n\sin\left(\sqrt{\lambda_n^{(2)}}x\right)\qquad \left(\lambda_n^{(2)}=\left(\frac{n\pi}{a}\right)^2,\, n\in\mathbb{N}^*\right). \tag{2b} \end{align} It is, however, possible to rewrite $(2)$ in a more compact way. To do this, first rewrite $X(x)$ as $$ X(x)=C\cos(\sqrt{\lambda}(x+a))+D\sin(\sqrt{\lambda}(x+a)), \tag{3} $$ which is equivalent to $(1)$. In this representation, the nontrivial solutions to $X(\pm a)=0$ are given by $$ X_k(x)=D_k\sin\left(\sqrt{\lambda_k}(x+a)\right)\qquad\left(\lambda_k=\left(\frac{k\pi}{2a}\right)^2,\,k\in\mathbb{N}^*\right). \tag{4} $$ These functions are equivalent to $(2\text{a})$ if $k$ is odd, and to $(2\text{b})$ if $k$ is even.

Gonçalo
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