Question

Show that the function
i) 3 u(x, t) = A(x + ct)³ is a solution of the one-dimensional wave equation
ii) u( x, t) =exp(-ut) sin x , is a solution of the one-dimensional heat equation

Answer 1

Answer on Question #67205 – Math – Differential Equations

Question

Show that the function

i) $u(x, t) = A(x + ct)^3$ is a solution of the one-dimensional wave equation

ii) $u(x, t) = \exp(-ut)\sin x$ is a solution of the one-dimensional heat equation

Solution

i)

One-dimensional wave equation is

$$\frac{\partial^2 u(x, t)}{\partial t^2} = c^2 \frac{\partial^2 u(x, t)}{\partial x^2}$$

To show that the function

$$u(x, t) = A(x + ct)^3$$

is a solution of the one-dimensional wave equation we first find $\frac{\partial^2 u(x,t)}{\partial t^2}$ and $\frac{\partial^2 u(x,t)}{\partial x^2}$ for the given function:

$$\frac{\partial u(x, t)}{\partial t} = \frac{\partial}{\partial t} [A(x + ct)^3] = 3Ac(x + ct)^2$$ $$\frac{\partial^2 u(x, t)}{\partial t^2} = \frac{\partial}{\partial t} [3Ac(x + ct)^2] = 6Ac^2(x + ct)$$ $$\frac{\partial u(x, t)}{\partial x} = \frac{\partial}{\partial x} [A(x + ct)^3] = 3A(x + ct)^2$$ $$\frac{\partial^2 u(x, t)}{\partial x^2} = \frac{\partial}{\partial x} [3A(x + ct)^2] = 6A(x + ct)$$

Substituting the obtained values into the one-dimensional wave equation one gets the identity

$$6Ac^2(x + ct) = c^2 6A(x + ct)$$

Thus, the function $u(x,t) = A(x + ct)^3$ is a solution of the one-dimensional wave equation.

ii)

One-dimensional heat equation is

$$\frac {\partial u (x , t)}{\partial t} = u \frac {\partial^ {2} u (x , t)}{\partial x ^ {2}},$$

where $u(x,t)$ is the function, $u$ is the constant.

To show that given function is a solution of the one-dimensional heat equation we first find $\frac{\partial u(x,t)}{\partial t}$ and $\frac{\partial^2 u(x,t)}{\partial x^2}$ for the given function

$$\frac {\partial u (x , t)}{\partial t} = \frac {\partial}{\partial t} [ \exp (- u t) \sin x ] = - u \exp (- u t) \sin x$$ $$\frac {\partial u (x , t)}{\partial x} = \frac {\partial}{\partial x} [ \exp (- u t) \sin x ] = \exp (- u t) \cos x$$ $$\frac {\partial^ {2} u (x , t)}{\partial x ^ {2}} = \frac {\partial}{\partial x} [ \exp (- u t) \cos x ] = - \exp (- u t) \sin x$$

Substituting the obtained values into the one-dimensional heat equation one gets the identity

$$- u \exp (- u t) \sin x = - u \exp (- u t) \sin x$$

Thus, the function $u(x,t) = \exp(-ut)\sin x$ is a solution of the one-dimensional heat equation.

Answer:

i) $u(x, t) = A(x + ct)^3$ is a solution of the one-dimensional wave equation;

ii) $u(x, t) = \exp(-ut)\sin x$ is a solution of the one-dimensional heat equation.

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