(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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].

2. Relevant equations

3. 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Using Laplace transform to solve this (simple) PDE

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**