Laplace transform of the wave equation

Question

I started of with the wave equation $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}$$ with boundary conditions $u=0$ at $x=0$ and $x=1$ and initial condition $u=sin(\pi x)$ and $\frac{\partial u}{\partial t}=0$ when $t=0$

I have gotten to the stage $$\frac{d^2 v}{dx^2}-s^2v=-s* sin(\pi x)$$ Where do I go from here?

My notes say find the complimentary function using the axillary function $m^2-s^2$ to get $$v_c=Ae^{sx}+Be^{-sx}$$ Then it says find the particular integral for the right hand side and use the trial function $$v_p=a*sin(\pi x)+b*cos(\pi x)$$ But I do not quite understand this. Why do we need the complimentary function? What is the trial function and why do we need it?

Answer 1

First of all, the point of using the Laplace transform was to turn the PDE into a family of ODEs for each $s$. In this case the ODEs are linear. The general recipe for these linear ODEs is that a solution is a sum of a solution to the homogeneous equation, where the right side is zero, and a particular solution to the given equation.

In your case the homogeneous equation is $v'' - s^2 v = 0.$ The general recipe for a homogeneous linear ODE with constant coefficients is an exponential solution. So you assume $v=e^{\lambda t}$ and after dividing out the $e^{\lambda t}$ you get the quadratic equation $\lambda^2-s^2=0$. So $\lambda=\pm s$, and the general solution to the homogeneous equation is $c_1 e^{st} + c_2 e^{-st}$ for constants $c_1,c_2$.

Now you need a particular solution. There a number of general recipes for finding one of these. One convenient one in a lot of simple cases is called undetermined coefficients, where you assume the function has a certain form with unknown coefficients and then find the coefficients. This method is more art than science, and depends on knowing how the elementary functions behave when they are differentiated. In your case, you want the right side to be a sinusoid. Since derivatives of sinusoids are sinusoids, you have some hope of having this work if $v$ is a sinusoid. Checking it by plugging into the equation, you find that it works out.

Answer 2

$$\frac{d^2 v}{dx^2}-s^2v=-s* sin(\pi x)$$

Once you have this, you have a differential equation of variable $x$. So you will solve it assuming $s$ is a constant. What you have is a linear second order constant coefficient inhomogeneous ordinary differential equation. It will have a complementary (homogeneous) and a particular component.

The auxiliary (characteristic) equation is actually $1-s^2=0$, so the solution will be a linear combination of the exponentials of the solutions of the characteristic equation. Solutions are $\pm s$, exponentials of the solutions are $e^s$ and $e^{-s}$. $A$ and $B$ are constant coefficients.

For the particular solution, the method you describe is using the method of undetermined coefficients. It is pointless to explain it here, there are quite a few things you need to be careful about and you need to study it properly. You can't solve partial differential equations without knowing such fundamentals about ordinary differential equations. But to find $a$ and $b$, you will insert $v_p$ into the equation (keep in mind $s$ is constant) and you will equate coefficients of $\sin$'s and $\cos$'s

Answer 3

Your equation $$\dfrac{d^2 v}{dx^2} - s^2 v = - s \sin(\pi x)$$ is a second-order linear differential equation. Like any linear differential equation, its general solution is of the form (particular solution) + (general solution of the homogeneous equation). Your "complementary function" is the general solution of the homogeneous equation $$ \dfrac{d^2 v}{dx^2} - s^2 v = 0$$ One way to find a particular solution (the method of "undetermined coefficients") is to make an educated guess at the form of a solution and then find what coefficients are needed to make this work. In this case the right side involves $\sin(\pi x)$, so we try a linear combination of $\sin(\pi x)$ and $\cos(\pi x)$ (the latter turns out not to be needed).

Once you find your general solution, the next step will be to look at it at $x=0$ and $x=1$ and see what $A$ and $B$ you need to make the boundary conditions true.

Answer 4

The language seems to be geared toward the annihilator method. $$ (D^{2}-s^{2})v = -s \sin(\pi x) \\ (D^{2}+\pi^{2})(D-s)(D+s)v = 0. $$ The general solution of the second equation is, for $s > 0$, $$ v(x) = Ae^{-sx}+Be^{sx}+E\sin(\pi x)+F\cos(\pi x). $$ Applying $(D-s^{2})$ to the right annihilates $Ae^{-sx}+Be^{sx}$, which leaves \begin{align} -s\sin(\pi x) & = (D^{2}-s^{2})(E\sin(\pi x)+F\cos(\pi x)) \\ & = (-\pi^{2}-s^{2})\{E\sin(\pi x)+F\cos(\pi x)\} \end{align} Hence, $F=0$ and $E=s/(\pi^{2}+s^{2})$. Therefore, $$ v(x) = Ae^{-sx}+Be^{sx}+\frac{s}{\pi^{2}+s^{2}}\sin(\pi x). $$ And $v(0)=0=v(1)$ implies $$ A+B = 0\\ Ae^{-s}+Be^{s} = 0 $$ Unless the equations are dependent, $A=B=0$. They are dependent iff $s=0$. For $s > 0$, $$ v(x) = \frac{s}{\pi^{2}+s^{2}}\sin(\pi x). $$