Let alpha and k be constants. Prove that the function u(x, t) = e^{-alpha^2 k^2 t} sin(kx) is a...

Question:

Let {eq}\alpha {/eq} and {eq}k {/eq} be constants. Prove that the function {eq}u(x, t) = e^{-\alpha^2 k^2 t} \sin(kx) {/eq} is a solution to the heat equation {eq}u_t = \alpha^2 u_{xx}. {/eq}

Solution of Partial Differential Equation:

In order to verify that a function is a solution of a partial differential equation, we

calculate the partial derivatives of the function and plug them into the equation.

If an identity is achieved, the function is a solution of the differential equation.

Answer and Explanation:

We are given the heat equation

{eq}\displaystyle u_t = \alpha^2 u_{xx}. {/eq}

The partial derivatives of the function

{eq}\displaystyle u(x, t) = e^{-\alpha^2 k^2 t} \sin(kx) {/eq}

that are present in the equation are found as

{eq}\displaystyle u_t(x, t) = -\alpha^2 k^2 e^{-\alpha^2 k^2 t} \sin(kx) \\ \displaystyle u_x(x, t) = ke^{-\alpha^2 k^2 t} \cos(kx) \\ \displaystyle u_{xx}(x, t) = -k^2e^{-\alpha^2 k^2 t} \sin(kx) {/eq}

Upon plugging the partial derivatives in the heat equation we get

{eq}\displaystyle u_t = \alpha^2 u_{xx} \Rightarrow e^{-\alpha^2 k^2 t} \sin(kx) = -\alpha^2 k^2e^{-\alpha^2 k^2 t} \sin(kx) {/eq}

Since the last equation is not an identity, this means that the function is not solution

of the heat equation.


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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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