Using Laplace transform to solve this (simple) PDE

[math2011]

Homework Statement

Solve [itex]\frac{\partial^2 v}{\partial x^2} = \frac{\partial^2 v}{\partial t^2}, x > 0, t > 0[/itex] subject to
[itex]v(x,0) = 0[/itex]
[itex]v_t(x,0) = 0[/itex]
[itex]v(0,t) = f(t)[/itex]
and where [itex]v[/itex] is bounded for all [itex]x > 0[/itex], by taking Laplace transforms with respect to [itex]t[/itex].

Homework Equations

The Attempt at a Solution

Take Laplace transforms with respect to [itex]t[/itex]:
[itex]\frac{\partial^2}{\partial x^2} V = s^2 V(x,s) - s v(x,0) - \frac{dv}{dt}(x,0)[/itex]
[itex]\frac{\partial^2}{\partial x^2} V = s^2 V(x,s)[/itex]
where [itex]V = \mathcal{L} \{ v \}[/itex].

Solve this ODE.
[itex]\lambda^2 - s^2 = 0[/itex]
[itex]\lambda = \pm s[/itex]

Here I start to get confused. By Laplace transform definition [itex]s[/itex] is a complex number. So is this ODE solution correct?

[itex]V(x,s) = A(s) e^{sx} + B(s) e^{-sx}[/itex]

Because [itex]v[/itex] is bounded the transform [itex]V[/itex] is also bounded for all [itex]x > 0[/itex] hence [itex]A(s) = 0[/itex].

[itex]V(x,s) = B(s) e^{-sx}[/itex]
[itex]V(0,s) = F(s) = B(s)[/itex]
[itex]V(x,s) = F(s) e^{-sx}[/itex]

Invert using given formula [itex]\mathcal{L} \{ H(t - c) f(t - x) \} = e^{-cs} F(s), c > 0[/itex] where [itex]H[/itex] is the Heaviside function we get
[itex]v(x,t) = H(t - x)f(t - x)[/itex].

Is this correct?

 

[mighty2000]

Your solution is correct up to the point where you have V(x,s) = F(s) e^{-sx}. However, in order to solve for v(x,t), you need to take the inverse Laplace transform of V(x,s) with respect to s. This will give you v(x,t) = f(t-x)H(t-x), where H is the Heaviside function.

Also, it is important to note that when taking the Laplace transform, s is a complex number, but when taking the inverse Laplace transform, s is replaced by t, which is a real number. So in your final solution, s should be replaced by t in the exponent as well.

Overall, your approach and solution are correct. Keep up the good work!