Understanding the Differential Equation: -u''(x) = X ln(x) on (0,1) with Boundary Conditions
(a) Consider The Differential Equation -u''(x) = X ln(x) on (0,1), with boundary conditions u(0) = u(1) = 0. (i) Show that the solution satisfies (1) == (x - x),
This article delves into solving a specific second-order differential equation with boundary conditions and understanding the properties of its solution. The problem statement involves analyzing the differential equation:
\[
-u''(x) = X \ln(x), \quad x \in (0,1),
\]
with boundary conditions:
\[
u(0) = 0, \quad u(1) = 0.
\]
Our goal is to find the explicit form of the solution and verify certain properties, specifically demonstrating that the solution satisfies a particular relation, denoted as (1) == (x - x). The notation here likely signifies an expected functional form or a specific property of the solution, which we will clarify through the solution process.
Setting Up the Problem and Basic Concepts
Understanding the Differential Equation
The given differential equation is:
\[
-u''(x) = X \ln(x),
\]
which can be rewritten as:
\[
u''(x) = -X \ln(x).
\]
This is a nonhomogeneous second-order ordinary differential equation. The term \(X \ln(x)\) acts as the forcing function.
Boundary Conditions and Their Significance
The boundary conditions:
\[
u(0) = 0, \quad u(1) = 0,
\]
are essential for determining the particular solution. Since \(\ln(x)\) approaches \(-\infty\) as \(x \to 0^+\), special care must be taken to ensure the solution remains finite and well-defined within the domain.
Methodology for Solving the Differential Equation
Step 1: Find the General Solution of the Homogeneous Equation
The associated homogeneous differential equation is:
\[
u''(x) = 0.
\]
Integrating twice:
\[
u''(x) = 0 \implies u'(x) = C1 \implies u(x) = C1 x + C_2,
\]
where \(C1, C2\) are constants determined by boundary conditions.
Step 2: Find a Particular Solution to the Nonhomogeneous Equation
Since the forcing term involves \(\ln(x)\), a suitable method is variation of parameters or integration of the Green's function. Here, we opt for variation of parameters.
The general solution:
\[
u(x) = uh(x) + up(x),
\]
where \(uh(x) = C1 x + C2\), and \(up(x)\) is a particular solution.
To find \(u_p(x)\), recall the general form of variation of parameters for second-order linear ODEs:
\[
up(x) = v1(x) y1(x) + v2(x) y_2(x),
\]
where \(y1(x)\) and \(y2(x)\) are linearly independent solutions of the homogeneous equation. Here, \(y1(x) = x\), and \(y2(x) = 1\).
However, because the right-hand side involves \(\ln(x)\), an alternative approach is to directly integrate twice, considering the structure of the differential equation.
Step 3: Integrate to Find \(u(x)\)
Given:
\[
u''(x) = -X \ln(x),
\]
integrate once:
\[
u'(x) = \int -X \ln(x) dx + D_1,
\]
where \(D_1\) is an integration constant.
Recall:
\[
\int \ln(x) dx = x \ln(x) - x + C,
\]
so:
\[
u'(x) = -X (x \ln(x) - x) + D_1,
\]
or:
\[
u'(x) = -X x \ln(x) + X x + D_1.
\]
Next, integrate \(u'(x)\) to find \(u(x)\):
\[
u(x) = \int u'(x) dx + D_2,
\]
which gives:
\[
u(x) = \int \left(-X x \ln(x) + X x + D1 \right) dx + D2.
\]
Break the integral into parts:
\[
u(x) = -X \int x \ln(x) dx + X \int x dx + D1 \int dx + D2.
\]
Calculate each integral.
Integral 1: \(\int x \ln(x) dx\)
Use integration by parts:
- Let \(u = \ln(x)\), \(dv = x dx\),
- then \(du = \frac{1}{x} dx\), \(v = \frac{x^2}{2}\).
Applying integration by parts:
\[
\int x \ln(x) dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x dx.
\]
Calculate the remaining integral:
\[
\int x dx = \frac{x^2}{2},
\]
so:
\[
\int x \ln(x) dx = \frac{x^2}{2} \ln(x) - \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{2} \ln(x) - \frac{x^2}{4}.
\]
Integral 2: \(\int x dx = \frac{x^2}{2}\)
Integral 3: \(\int dx = x\)
Putting all together:
\[
u(x) = -X \left( \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right) + X \frac{x^2}{2} + D1 x + D2.
\]
Simplify:
\[
u(x) = -\frac{X x^2}{2} \ln(x) + \frac{X x^2}{4} + \frac{X x^2}{2} + D1 x + D2.
\]
Combine similar terms:
\[
u(x) = -\frac{X x^2}{2} \ln(x) + \left( \frac{X x^2}{4} + \frac{X x^2}{2} \right) + D1 x + D2,
\]
\[
u(x) = -\frac{X x^2}{2} \ln(x) + \frac{3 X x^2}{4} + D1 x + D2.
\]
Applying Boundary Conditions to Determine Constants
Recall:
\[
u(0) = 0, \quad u(1) = 0.
\]
However, as \(x \to 0^+\), note that:
\[
u(x) \sim -\frac{X x^2}{2} \ln(x),
\]
and since \(\lim_{x \to 0^+} x^2 \ln(x) = 0\), the first term tends to zero. The remaining parts:
\[
u(0) = 0 + 0 + D1 \cdot 0 + D2 = D_2,
\]
so:
\[
D_2 = 0.
\]
Next, evaluate at \(x=1\):
\[
u(1) = -\frac{X \cdot 1^2}{2} \ln(1) + \frac{3 X \cdot 1^2}{4} + D_1 \cdot 1 + 0,
\]
\[
u(1) = -\frac{X}{2} \cdot 0 + \frac{3X}{4} + D1 = \frac{3X}{4} + D1.
\]
Since \(u(1) = 0\), we have:
\[
D_1 = -\frac{3X}{4}.
\]
Final Solution Expression
Putting all constants back in, the explicit solution is:
\[
u(x) = -\frac{X x^2}{2} \ln(x) + \frac{3 X x^2}{4} - \frac{3X}{4} x.
\]
This expression satisfies the differential equation and boundary conditions.
Verifying the Property: Showing (1) == (x - x)
The problem suggests demonstrating that the solution satisfies a relation akin to \((1) == (x - x)\). Interpreting this,
