Question

Show that
u=sin(wx)e^(-w^2c^2t)
is a solution of the one-dimensional heat equation.
note : here w stands for omega

Answer 1

Answer on Question#38998 - Math - Differential Calculus|Equation

Question: Show that

$$u(x, t) = \sin(\omega x) e^{-\omega^2 c^2 t}$$

is a solution of the one-dimensional heat equation.

Solution. Recall the one-dimensional heat equation:

$$u_t = c^2 u_{xx}.$$

We need to find the above derivatives $(u_t, u_{xx})$ of the given function and substitute them into the equation.

$$\begin{aligned} u_t(x, t) &= \sin(\omega x) \cdot e^{-\omega^2 c^2 t} \cdot (-\omega^2 c^2) = -\omega^2 c^2 \sin(\omega x) e^{-\omega^2 c^2 t}, \\ u_x(x, t) &= \cos(\omega x) \cdot \omega \cdot e^{-\omega^2 c^2 t} = \omega \cos(\omega x) e^{-\omega^2 c^2 t}, \\ u_{xx}(x, t) &= -\sin(\omega x) \cdot \omega^2 \cdot e^{-\omega^2 c^2 t} = -\omega^2 \sin(\omega x) e^{-\omega^2 c^2 t}. \end{aligned}$$

Therefore, we see that

$$\begin{aligned} u_t(x, t) &= \sin(\omega x) \cdot e^{-\omega^2 c^2 t} \cdot (-\omega^2 c^2) = -\omega^2 c^2 \sin(\omega x) e^{-\omega^2 c^2 t} = c^2 \cdot \left(-\omega^2 \sin(\omega x) e^{-\omega^2 c^2 t}\right) \\ &= c^2 \cdot u_{xx}(x, t), \end{aligned}$$

or

$$u_t = c^2 u_{xx},$$

so the function $u(x, t) = \sin(\omega x) e^{-\omega^2 c^2 t}$ is indeed a solution of the one-dimensional heat equation.