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Let $u(x,t)$ be solution of initial boundary value problem $$u_{tt}=u_{xx},\quad 0<x<\infty,\quad t>0 $$ $$u(x,0)=\cos\left(\frac{\pi x}{2}\right),\quad 0<x<\infty$$ $$u_{t}(x,0)=0,\quad 0<x<\infty$$ $$u_{x}(0,t)=0,\quad t\geq0 $$ then

  1. $u(2,2)=?$
  2. $u(1/2,1/2)=?$

I am stuck on this problem . Can anyone help me please........

I can't solve it with separation of variable........ I don't know where to begin?

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how can we apply d'Alambert's formula, not getting sir.......... – pankaj Dec 18 '12 at 12:36

The problem is posed in the positive real line. The boundary condition suggests to consider the initial value problem in the while real line extending the initial data to be even. Once you have that, use d'Alambert's formula.

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how can we apply d'Alambert's formula, not getting sir.......... – pankaj Dec 18 '12 at 12:16

One approach is to assume that you can split $u(x,t)=X(x)\cdot T(t)$. I don't know the english expression, but it should be something like "separation approach". Using this for the equation $u_{tt}=u_{xx}$, we get \begin{align} \frac{d^2T}{dt^2} \cdot X = T \cdot \frac{d^2X}{dx^2} \\ \Leftrightarrow \frac{d^2T}{dt^2} \cdot T^{-1} = X^{-1} \frac{d^2X}{dx^2}X \equiv -\lambda^2 \end{align} Where $\lambda \in C$ is an arbitrary constant. The reason, why the expression are constant, is that the left side only depends on $t$ while the right side only depends on $x$. If now the i.e. the left side would not be constant, we couldn't reproduce that change on the right side as it is only depend on $x$. The reason why I choose $\lambda^2$ and not $-\lambda$ or $+\lambda^2$ becomes clear in the next step. From the above it follows \begin{align} &\frac{d^2T}{dt^2} = -\lambda^2 T \Rightarrow T(t) = a \sin(\lambda t) + b \cos(\lambda t) \\& \frac{d^2X}{dt^2} = -\lambda^2 X \Rightarrow X(x)=c \sin(\lambda x) + d \cos(\lambda x) \end{align} If we had chosen $\lambda^2$ instead of $-\lambda^2$ we would have got another solution to the two ODEs, i.e. the $\exp()$. But as the initial condition suggest trigonometric functions, $\sin()$ and $\cos()$ fit better. Okay. So we have $u(x,t) =X(x)T(t)$ Now we check for initial condition. \begin{align} u(x,0) = X(x)\cdot T(0) = \cos(\frac{\pi x}{2}) \end{align} This suggests $c=0,\lambda=\frac{\pi}{2}$ and with $T(0) = b$ it follows $d\cdot b=1$ and $X(x) =d\cdot \cos(\frac{\pi x}{2})$.

Next initial condition is $u_t(x,0)=0$. \begin{align} u_t(x,0) &= T'(t)X(0) =(\frac{\pi}{2} a \cos(\frac{\pi}{2} 0) -\frac{\pi}{2} b \sin(\frac{\pi}{2} 0)) \cdot d\cos(\frac{\pi x}{2}) = \\ &=(\frac{\pi}{2} a)\cdot d\cos(\frac{\pi x}{2})=0 \end{align} As $d\neq0$, it follows $a=0$ and therefore $T(t) = b\cos(\frac{\pi t}{2})$. Combining this, we have \begin{align} u(x,t) = b\cos(\frac{\pi t}{2})\cdot d \cos(\frac{\pi x}{2})=e\cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2}) \end{align} With $e=b\cdot d$. But as $b\cdot d = 1$ as stated above, the solution becomes \begin{align} u(x,t) = \cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2}) \end{align} Now we still have to check the last initial condition $u_x(0,t)=0$. \begin{align} u_x(0,t) = X'(x)T(0) = \frac{\pi}{2} \sin(0)T(0) = 0 \end{align} So this one also holds. Therefore we have the solution to the PDE as $u(x,t) = \cos(\frac{\pi t}{2})\cdot \cos(\frac{\pi x}{2})$.

Note: In my opinion we could have gone a lot faster. Why? Except for the first initial condition the problem looks quite symmetric. So thought that maybe also $T(t) = \cos(\frac{\pi t}{2})$. And if you check this with the first initial condition it holds.

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if there is any shortcut please explain sir......... – pankaj Dec 18 '12 at 12:49
Sorry, what do you mean exactly? – k1next Dec 18 '12 at 13:18

Note that when without the condition $u_x(0,t)=0$ this is in fact a just-determining problem and the solution can be expressed by using D’Alembert’s formula $u(x,t)=\dfrac{1}{2}\left(\cos\dfrac{\pi(x+t)}{2}+\cos\dfrac{\pi(x-t)}{2}\right)=\cos\dfrac{\pi x}{2}\cos\dfrac{\pi t}{2}$

Check for $u_x(0,t)$ :

$u_x(x,t)=-\dfrac{\pi}{2}\sin\dfrac{\pi x}{2}\cos\dfrac{\pi t}{2}$

$u_x(0,t)=-\dfrac{\pi}{2}\sin\dfrac{\pi\times0}{2}\cos\dfrac{\pi t}{2}=0$

$\therefore u(x,t)=\cos\dfrac{\pi x}{2}\cos\dfrac{\pi t}{2}$ is already an exact solution of this problem, we don't need to use the result in

Hence $u(2,2)=\cos\dfrac{\pi\times2}{2}\cos\dfrac{\pi\times2}{2}=\cos\pi\cos\pi=1$

Hence $u\left(\dfrac{1}{2},\dfrac{1}{2}\right)=\cos\dfrac{\pi\times\dfrac{1}{2}}{2}\cos\dfrac{\pi\times\dfrac{1}{2}}{2}=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{4}=\dfrac{1}{2}$

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This is a wave equation on the half-line with Neumann boundary condition. You can use even extension(fortunately, everything you need to extend is automatically even) and d'Alambert's Formula to solve it.

I think the answer should be $ u(x,t) =\frac { 1 }{ 2 } \left[ cos\frac { \pi (x+t) }{ 2 } +cos\frac { \pi (x-t) }{ 2 } \right] $ with x and t positive real numbers. Plugin $(2,2)$ and $( \frac{1}{2} , \frac{1}{2} )$ you'll get 1 and 1/2

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