Question

show that function u(x,t)=(exp^-(mu)t)sin x is a solution of of one dimensional heat equation

Answer 1

ANSWER on Question #75734 – Math – Differential Equations

Show that function

$$u(x, t) = e^{-\mu t} \sin x$$

is a solution of one dimensional heat equation.

SOLUTION

By the definition, the one dimensional heat equation has form

$$\frac{\partial u}{\partial t} = \mu \frac{\partial^2 u}{\partial x^2}$$

(More information: https://en.wikipedia.org/wiki/Heat_equation)

Then, we calculate the above partial derivatives of the proposed solution

$$u(x, t) = e^{-\mu t} \sin x \rightarrow \frac{\partial u}{\partial t} = \frac{\partial (e^{-\mu t} \sin x)}{\partial t} = \frac{d (e^{-\mu t})}{dt} \sin x = -\mu e^{-\mu t} \sin x$$ $$\boxed{\frac{\partial u}{\partial t} = -\mu e^{-\mu t} \sin x}$$ $$u(x, t) = e^{-\mu t} \sin x \rightarrow \frac{\partial u}{\partial x} = \frac{\partial (e^{-\mu t} \sin x)}{\partial x} = e^{-\mu t} \frac{d (\sin x)}{dx} = e^{-\mu t} \cos x \rightarrow$$ $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial x}\right) = \frac{\partial}{\partial x} (e^{-\mu t} \cos x) = e^{-\mu t} \frac{d (\cos x)}{dx} = -e^{-\mu t} \sin x$$ $$\boxed{\frac{\partial^2 u}{\partial x^2} = -e^{-\mu t} \sin x}$$

Conclusion,

$$u(x, t) = e^{-\mu t} \sin x \rightarrow \begin{cases} \frac{\partial u}{\partial t} = -\mu e^{-\mu t} \sin x \\ \frac{\partial^2 u}{\partial x^2} = -e^{-\mu t} \sin x \end{cases}$$

Then,

$$\frac {\partial u}{\partial t} = \mu \frac {\partial^ {2} u}{\partial x ^ {2}} \leftrightarrow - \mu e ^ {- \mu t} \sin x = \mu \cdot (- e ^ {- \mu t} \sin x)$$

This means that this function $u(x,t)$ is indeed a solution of the one-dimensional heat equation.