Use the Laplace Transform to solve the wave equation

Question

We are given

$u_{tt}-u_{xx}=0$, $\forall t>0,x \in \mathbb R$

subject to

$u(x,0)=sin\pi x, u_t(x,0)=0$, $\forall x\in \mathbb R$

Use the Laplace Transform to solve the wave equation.

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This problem is slightly different to my examples I have been given previously, in that the $u(x,0)$ term is not equal to zero.

So far here is what I have tried:

$u_{tt}-u_{xx}=0$

$\implies s^2\tilde{u}(x,s)-su(x,0)-u_t(x,0)-\tilde{u_{xx}}(x,s)=0$

$\implies s^2\tilde{u}(x,s)-ssin\pi x-\tilde{u_{xx}}(x,s)=0$

From here it can rearranged to the form of

$s^2\tilde{u}(x,s)-\tilde{u_{xx}}(x,s)=ssin\pi x$

But it is here that I am struggling with how to go about the Laplace Transform part of the solution as all my other examples have the LHS equal to zero at this point!

Thanks for any help!

Answer 1

I think I have worked out the answer!

As @Aryadeva correctly suggested we are looking for solutions of the form $\tilde{u}=Asin(\pi x)$.

We then take this to find the first and second order partial differentials of $\tilde{u}$ with respect to $x$.

$\tilde{u}_x=A\pi cos(\pi x) \implies \tilde{u}_{xx}=-A\pi^2sin(\pi x)$

Then, substituting into the PDE we had before we get

$s^2Asin(\pi x)+A\pi^2sin(\pi x)=ssin(\pi x)$

$\implies s^2A+A\pi^2=s$

$\implies A(s^2+\pi^2)=s \implies A=\frac{s}{s^2+\pi^2}$

Then we get our solution for $\tilde{u}$ as being

$\tilde{u}(x,s)=\frac{s}{s^2+\pi^2}sin(\pi x)$

Which we can then transform into the correct form through recognising that from our well known Laplace Transforms

$\frac{s}{s^2+\pi^2}$ transforms to $cos(\pi t)$

And hence our solution is

$u(x,t)=cos(\pi t)sin(\pi x)$

Answer 2

$$s^2\tilde{u}(x,s)-\tilde u_{xx}(x,s)=s \sin (\pi x)$$ Note that this is simply a differential equation of second order with constants coefficients. Since we differentiate wrt $x$ not $s$: $$s^2 \tilde u- \tilde u''= s \sin(\pi x)$$ Solve the homogeenous equation first: $$\tilde u''-s^2 \tilde u=0$$ The characteristic equation is: $$r^2-s^2=0 \implies r= \pm s$$ The solution is: $$\tilde u_h(x,s)=c_1(s)e^{xs}+c_2(s)e^{-xs}$$

Use method of undetermined coefficients for the inhomogeneous DE. Your guess should be : $$\tilde u_p =A \sin ( \pi x)$$