**Solving the Heat Equation Boundary Value Problem****Objective**: Solve completely to the final form of the solution as a Fourier Series the heat equation boundary value problem. The problem comprises a homogeneous Partial Differential Equation (PDE) with Dirichlet non-homogeneous but constant Boundary Conditions (B.C.).**Problem Setup**:**Domain and Parameters**:- $ x \in [0, L] $- $ L = 4 $- $ k = 3 $- $ t \geq 0 $**Equation**:\[\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}\]**Boundary Conditions (B.C.)**:- $ u(0, t) = 5 $- $ u(L, t) = -5 $**Initial Condition (I.C.)**:- $ u(x, 0) = x $Our goal is to use Fourier Series to find the solution to this boundary value problem while taking into account both the initial and boundary conditions provided.

Given Heat equation is the following :Let, so we get By substitution ,we get

Answered: B.C. IC Solve. Completely to the final…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Solving the Heat Equation Boundary Value Problem**

**Objective**:
Solve completely to the final form of the solution as a Fourier Series the heat equation boundary value problem. The problem comprises a homogeneous Partial Differential Equation (PDE) with Dirichlet non-homogeneous but constant Boundary Conditions (B.C.).

**Problem Setup**:

**Domain and Parameters**:
- $ x \in [0, L] $
- $ L = 4 $
- $ k = 3 $
- $ t \geq 0 $

**Equation**:
\[
\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}
\]

**Boundary Conditions (B.C.)**:
- $ u(0, t) = 5 $
- $ u(L, t) = -5 $

**Initial Condition (I.C.)**:
- $ u(x, 0) = x $

Our goal is to use Fourier Series to find the solution to this boundary value problem while taking into account both the initial and boundary conditions provided.

Expert Solution

Step 1: Heat equation solve

Given Heat equation is the following :

$fraction numerator partial differential u over denominator partial differential t end fraction equals 3 fraction numerator partial differential squared u over denominator partial differential x squared end fraction comma 0 less or equal than x less or equal than 4 comma t greater than 0$

$u left parenthesis 0 comma t right parenthesis equals 5 comma u left parenthesis 4 comma t right parenthesis equals negative 5 comma t greater than 0 space a n d space$

$u left parenthesis x comma 0 right parenthesis equals x comma 0 less or equal than x less or equal than 4$

Let, $v left parenthesis x comma t right parenthesis equals u left parenthesis x comma t right parenthesis minus open curly brackets 5 plus x over 4 open parentheses negative 5 minus 5 close parentheses close curly brackets equals u left parenthesis x comma t right parenthesis plus 5 over 2 x minus 5$

so we get $v subscript t equals u subscript t space a n d space v subscript x x end subscript equals u subscript x x end subscript$

By substitution ,we get $v subscript t equals 3 v subscript x x end subscript$