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The displacement of an infinite string obeys the wave equation: $$ \frac{\partial ^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$$

Find the solution in the form: $$u(x,t) = f(x-ct) + g(x+ct) $$

where $g=-f$ and the initial conditions are given by: $$u(x,0)=0$$ $$\frac{\partial}{\partial t} u(x,0) = \frac{x}{(1+x^2)^2}$$

The second condition leads to: $$\frac{\partial u}{\partial t}=-cf'(x,0)+cg'(x,0)=-2cf'(x,0)=\frac{x}{(1+x^2)^2}$$ $$f(x,0) = - \frac{1}{2c} \left( \frac{x}{(1+x^2)^2} \right) t + h(x)$$ Now, this is the solution for $t=0$. Is it mathematically correct to simply substitute $x$ back with $x \pm ct$? i.e.:

$$u(x,t) = - \frac{1}{2c} \left( \frac{(x-ct)}{(1+(x-ct)^2)^2} \right) t +\frac{1}{2c} \left( \frac{(x+ct)}{(1+(x+ct)^2)^2} \right)t$$

In principle the initial conditions would be satisfied for any function of $t$ that yields $h(0)=0$, but since I am supposed to find a solution of that particular form ($f(x\pm ct)$), I suppose it is a correct solution?

alkamid
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2 Answers2

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Let's be careful:

The second condition leads to: $$\left.\frac{\partial u}{\partial t}\right|_{t=0}=-cf'(x)+cg'(x)=-2cf'(x)=\frac{x}{(1+x^2)^2}$$ Thus $$f(x) = \frac{1}{4c} \cdot\frac{1}{1+x^2}+c_1,$$ where $c_1$ is a constant, as $$ f'(x)=-\frac{1}{2c}\cdot\frac{x}{(1+x^2)^2}. $$

Yiorgos S. Smyrlis
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  • But we differentiate with respect to t, not x, don't we? I should have written $\dot{f}$ instead of $f'$. – alkamid Mar 10 '14 at 10:08
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    $f$ is a function of one variable, and in the solution expression is estimated at $x+t$, while at $t=0$ it is estimated at $x$. So $$\left.\frac{\partial}{\partial t}f\right|_{t=0}=f'(x).$$ – Yiorgos S. Smyrlis Mar 10 '14 at 10:10
  • To convince yourself, for verification, try $$u(x,t)=\frac{1}{4c}\cdot\frac{1}{1+{(x+t)}^2}-\frac{1}{4c}\cdot\frac{1}{1+{(x-t)}^2}. $$ – Yiorgos S. Smyrlis Mar 10 '14 at 10:12
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    I convinced myself that your result is correct, but I am still confused by your first comment. Is it generally true that $\left. \frac{\partial}{\partial x} f(x,y) \right|_{x=x_1} = \frac{df(x)}{dy}$? It looks reasonable (at constant x there is only change in y), but I'm not able to derive it. Could you point me to a description of this particular property of partial differentials? – alkamid Mar 10 '14 at 10:47
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Note:
There is a difference between $f(x,y)$ and $f(x+y)$. The first is a function of two variables and the second is a function of one variable $z=x+y$. For example $$f(x+0)=f(x)=x^2\implies f(x+y)=(x+y)^2\\f(x,0)=x^2\implies \, \text{nothing} $$Also there is a meaning of $\frac{\partial f(x,y)}{\partial x}$ but it doesn't mean any thing for $f(x+y)$ despite $$w=f(x+y)\implies \frac{\partial w}{\partial x}=f'(x+y)\times \frac{\partial (x+y)}{\partial x}+=f'(x+y)\times 1=f'(x+y)$$

Semsem
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