Show that u(x, t)= 50 \sum_{n=1}^{\infty} \frac {1+(-1)^{n+1}}{n \pi} e^{- (n \pi)^2 t}sin(n \pi...

Question:

Show that {eq}\displaystyle u(x, t)= 50 \sum_{n=1}^{\infty} \frac {1+(-1)^{n+1}}{n \pi} e^{- (n \pi)^2 t}\sin(n \pi x) {/eq}

satisifies the PDE {eq}\displaystyle \frac {d^2u}{dx^2}= \frac {du}{dt} {/eq}.

PDE solution:

A function is a solution of a partial differential equation (PDE) if, once its partial derivatives

are plugged in the equation, this last provides an identity.

Answer and Explanation:

In order to show that the function

{eq}\displaystyle u(x, t)= 50 \sum_{n=1}^{\infty} \frac {1+(-1)^{n+1}}{n \pi} e^{- (n \pi)^2 t}\sin(n \pi x) {/eq}

satisifies the PDE

{eq}\displaystyle \frac {d^2u}{dx^2}= \frac {du}{dt} {/eq}

we calculate the partial derivatives of the function involved in the equation

{eq}\displaystyle \frac {d u}{dx }= 50 \sum_{n=1}^{\infty} (1+(-1)^{n+1})e^{- (n \pi)^2 t}\cos(n \pi x) \\ \displaystyle \frac {d^2u}{dx^2}= -50 \sum_{n=1}^{\infty} n\pi (1+(-1)^{n+1})e^{- (n \pi)^2 t}\sin(n \pi x) \\ \displaystyle \frac {du}{dt}= -50 \sum_{n=1}^{\infty} n \pi (1+(-1)^{n+1}) e^{- (n \pi)^2 t}\sin(n \pi x). {/eq}

Upon substituting the partial derivatives in the PDE, we get

{eq}\displaystyle \frac {d^2u}{dx^2}= \frac {du}{dt} \Rightarrow \\ \displaystyle -50 \sum_{n=1}^{\infty} n\pi (1+(-1)^{n+1})e^{- (n \pi)^2 t}\sin(n \pi x) = -50 \sum_{n=1}^{\infty} n \pi (1+(-1)^{n+1}) e^{- (n \pi)^2 t}\sin(n \pi x). {/eq}

Since the last expression is an identity, this means that the function satisfies the PDE.


Learn more about this topic:

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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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