Partial Differential Equations

Question

Take u_tt = 9 u_xx, with u(0, x) = u_t (0, x) = 0, and u(t, 0) = sin(t), u(t, infinity) = 0 Considering the Laplace Transform of the above PDE.

Find u(t,x)

Answer 1

For your inquisitive mind, I would be absolutely thrilled to solve this wave equation using Laplace transforms.

Given Information

1. PDE: $u_{tt} = 9u_{xx}$
2. Initial conditions: $u(0,x) = 0$, $u_t(0,x) = 0$
3. Boundary conditions: $u(t,0) = \sin(t)$, $u(t,\infty) = 0$

Solution

Step 1: Apply Laplace Transform in time

Let $U(s,x) = \mathcal{L}{u(t,x)}$

Taking the Laplace transform of both sides: L{utt}=9L{uxx}L{utt​}=9L{uxx​}

Using the property $\mathcal{L}{u_{tt}} = s^2U(s,x) - su(0,x) - u_t(0,x)$ and the initial conditions: s2U(s,x)=9∂2U∂x2s2U(s,x)=9∂x2∂2U​

This gives us: ∂2U∂x2−s29U=0∂x2∂2U​−9s2​U=0

Step 2: Solve the ODE in x

The general solution is: U(s,x)=A(s)esx/3+B(s)e−sx/3U(s,x)=A(s)esx/3+B(s)e−sx/3

Step 3: Apply boundary conditions

From $u(t,\infty) = 0$, we need $U(s,\infty) = 0$, which requires $A(s) = 0$.

So: $U(s,x) = B(s)e^{-sx/3}$

From $u(t,0) = \sin(t)$: U(s,0)=L{sin⁡(t)}=1s2+1U(s,0)=L{sin(t)}=s2+11​

Therefore: $B(s) = \frac{1}{s^2+1}$

Thus: U(s,x)=e−sx/3s2+1U(s,x)=s2+1e−sx/3​

Step 4: Inverse Laplace Transform

Using the shifting theorem: $\mathcal{L}^{-1}{e^{-as}F(s)} = f(t-a)H(t-a)$ where $H$ is the Heaviside function.

Since $\mathcal{L}^{-1}\left{\frac{1}{s^2+1}\right} = \sin(t)$:

u(t,x)=sin⁡(t−x3)H(t−x3)u(t,x)=sin(t−3x​)H(t−3x​)

Final Answer

u(t,x)={sin⁡(t−x3)if t≥x30if t<x3u(t,x)={sin(t−3x​)0​if t≥3x​if t<3x​​​

This represents a wave traveling to the right with velocity $c = 3$ (since $c^2 = 9$ in the wave equation). The disturbance at $x = 0$ propagates outward, reaching position $x$ at time $t = x/3$.