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The function u(x,t)=ei(x+ct)𝑢(𝑥,𝑡)=𝑒𝑖(𝑥+𝑐𝑡) is another solution to the wave equation∂2u∂t2−c2∂2u∂x2=0∂2𝑢∂𝑡2−𝑐2∂2𝑢∂𝑥2=0  Let's investiage the case when c=2+5i𝑐=2+5𝑖 .  ThenThe real part of u𝑢 is    The imaginary part of u𝑢 is   .So the function oscillates in the complex plane, but these oscillations are dampened by the exponential term. As time goes onlimt→∞u(x,t)=lim𝑡→∞𝑢(𝑥,𝑡)=     .

Question

The function u(x,t)=ei(x+ct)𝑢(𝑥,𝑡)=𝑒𝑖(𝑥+𝑐𝑡) is another solution to the wave equation∂2u∂t2−c2∂2u∂x2=0∂2𝑢∂𝑡2−𝑐2∂2𝑢∂𝑥2=0  Let's investiage the case when c=2+5i𝑐=2+5𝑖 .  ThenThe real part of u𝑢 is    The imaginary part of u𝑢 is   .So the function oscillates in the complex plane, but these oscillations are dampened by the exponential term. As time goes onlimt→∞u(x,t)=lim𝑡→∞𝑢(𝑥,𝑡)=     .

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Solution

Primero, consideremos la función u(x,t)=ei(x+ct) u(x,t) = e^{i(x+ct)} con c=2+5i c = 2 + 5i . Sustituyendo c c en la función, obtenemos:

u(x,t)=ei(x+(2+5i)t)=ei(x+2t+5it) u(x,t) = e^{i(x + (2 + 5i)t)} = e^{i(x + 2t + 5it)}

Podemos separar la parte real y la parte imaginaria de la expresión en el exponente:

u(x,t)=ei(x+2t)−5t u(x,t) = e^{i(x + 2t) - 5t}

Usando la identidad de Euler eiθ=cos⁡(θ)+isin⁡(θ) e^{i\theta} = \cos(\theta) + i\sin(\theta) , tenemos:

u(x,t)=e−5t⋅ei(x+2t)=e−5t(cos⁡(x+2t)+isin⁡(x+2t)) u(x,t) = e^{-5t} \cdot e^{i(x + 2t)} = e^{-5t} \left( \cos(x + 2t) + i\sin(x + 2t) \right)

Por lo tanto, la parte real de u u es:

Re(u)=e−5tcos⁡(x+2t) \text{Re}(u) = e^{-5t} \cos(x + 2t)

Y la parte imaginaria de u u es:

Im(u)=e−5tsin⁡(x+2t) \text{Im}(u) = e^{-5t} \sin(x + 2t)

Observamos que la función oscila en el plano complejo, pero estas oscilaciones están amortiguadas por el término exponencial e−5t e^{-5t} . A medida que el tiempo avanza, es decir, cuando t→∞ t \to \infty :

lim⁡t→∞u(x,t)=lim⁡t→∞e−5t(cos⁡(x+2t)+isin⁡(x+2t))=0 \lim_{t \to \infty} u(x,t) = \lim_{t \to \infty} e^{-5t} \left( \cos(x + 2t) + i\sin(x + 2t) \right) = 0

Por lo tanto, la función tiende a cero a medida que el tiempo tiende a infinito.

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