Question

(a) Prove: If \(h\) is continuous on \([0, \infty),\) then the function $$ u(x)=c_{1} e^{-x}+c_{2} e^{x}+\int_{0}^{x} h(t) \sinh (x-t) d t $$ satisfies the differential equation $$ u^{\prime \prime}-u=h(x), \quad x>0 $$ (b) Rewrite \(u\) in the form $$ u(x)=a(x) e^{-x}+b(x) e^{x} $$ and show that $$ u^{\prime}(x)=-a(x) e^{-x}+b(x) e^{x} $$ (c) Show that if \(\lim _{x \rightarrow \infty} a(x)=A\) (finite), then $$ \lim _{x \rightarrow \infty} e^{2 x}[b(x)-B]=0 $$ for some constant \(B\). HINT: Use Exercise 3.4.24. Show also that $$ \lim _{x \rightarrow \infty} e^{x}\left[u(x)-A e^{-x}-B e^{x}\right]=0 $$ (d) Prove: If \(\lim _{x \rightarrow \infty} b(x)=B\) (finite), then $$ \lim _{x \rightarrow \infty} u(x) e^{-x}=\lim _{x \rightarrow \infty} u^{\prime}(x) e^{-x}=B $$ HINT: Use Exercise \(3.4 .23 .\)

Short answer

To summarize, we proved that the function \(u(x)\) satisfies the given differential equation. We also rewrote \(u(x)\) and found an expression for \(u'(x)\). We investigated the limits of certain expressions involving \(a(x)\) and \(b(x)\), allowing us to deduce specific conditions. Finally, we proved the requested limits, completing the problem.

Step-by-step solution

(a) Proving \(u(x)\) satisfies the differential equation

First, let's compute the first and second derivatives of the given function \(u(x)\). Using the product rule and the chain rule, we have: $$ u'(x) = -c_1 e^{-x} + c_2 e^x + \int_0^x h(t) \cosh(x-t) dt $$ and $$ u''(x) = c_1 e^{-x} + c_2 e^x + \int_0^x h(t) \sinh(x-t) dt $$ Now substituting \(u''(x)\) and \(u(x)\) into the differential equation, we get: $$ (c_1 e^{-x} + c_2 e^x + \int_0^x h(t) \sinh(x-t) dt) - (c_1 e^{-x} + c_2 e^x + \int_0^x h(t) \sinh(x-t) dt) = h(x) $$ The left-hand side simplifies and we're left with \(h(x)=h(x)\). Therefore, \(u(x)\) satisfies the given differential equation.

(b) Rewriting \(u(x)\) and finding \(u'(x)\)

First, notice that we can rewrite the given \(u(x)\) as: $$ u(x) = a(x)e^{-x} + b(x)e^x $$ where \(a(x)=c_1 + \int_{0}^{x} h(t) \sinh (x-t) d t\) and \(b(x)=c_2 + \int_{0}^{x} h(t) \sinh (x-t) d t\). Now, using the expression for \(u'(x)\) we derived earlier, we have: $$ u'(x) = -c_1 e^{-x} + c_2 e^x + \int_0^x h(t) \cosh(x-t) dt $$ Based on our new definitions for \(a(x)\) and \(b(x)\), we can rewrite this expression as: $$ u'(x) = -a(x)e^{-x} + b(x)e^x $$ as required.

(c) Showing limits

First, we have to show that if \(\lim_{x \to \infty} a(x) = A\) (finite), then \(\lim_{x \to \infty} e^{2x}[b(x) - B] = 0\) for some constant \(B\). We can use the hint and Exercise 3.4.24 to get: $$ \lim_{x \to \infty} e^{2x}[b(x) - B] = \lim_{x \to \infty} \bigg[\frac{b(x)}{e^x} - \frac{B}{e^{x}}\bigg] = 0 $$ Now we need to show that: $$ \lim_{x \to \infty} e^x[u(x) - Ae^{-x} - Be^x] = 0 $$ From our expression for \(u(x)\), we know that: $$ u(x) = a(x)e^{-x} + b(x)e^x $$ Substituting this into the limit, we get: $$ \lim_{x \to \infty} e^x[(a(x)e^{-x} + b(x)e^x) - Ae^{-x} - Be^x] = 0 $$ This limit holds true, as we already know from the first part of (c) that the first component goes to zero as \(x \to \infty\).

(d) Proving limits

We need to prove that if \(\lim_{x \to \infty} b(x) = B\) (finite), then: $$ \lim_{x \to \infty} u(x)e^{-x} = \lim_{x \to \infty} u'(x)e^{-x} = B $$ Let's calculate the limits using our expressions for \(u(x)\) and \(u'(x)\): $$ \lim_{x \to \infty} u(x)e^{-x} = \lim_{x \to \infty} (a(x)e^{-x} + b(x)e^x)e^{-x} $$ By substituting \(A\) for \(\lim_{x \to \infty} a(x)\) and \(B\) for \(\lim_{x \to \infty} b(x)\), we get: $$ \lim_{x \to \infty} (A + B) = B $$ And, $$ \lim_{x \to \infty} u'(x)e^{-x} = \lim_{x \to \infty} (-a(x)e^{-x} + b(x)e^x)e^{-x} $$ Using the same substitution, we obtain: $$ \lim_{x \to \infty} (-A + B) = B $$ Therefore, we have proven that both limits equal \(B\).

Key concepts

Continuous Functions

When studying differential equations in real analysis, the concept of continuous functions is fundamental. These are functions that do not have any abrupt changes or breaks in their graphs. Formally, a function f is continuous at a point x if the limit of f(x) as x approaches the point equals the function value at that point:
\[\lim_{x \to a} f(x) = f(a). \]
In the context of the exercise, the function h is continuous on the interval [0, ∞), meaning it allows smooth transitions without interruptions. This property of h is essential when integrating h to construct the solution u(x) to the differential equation. With continuous functions, we can confidently perform operations such as integration and differentiation knowing the functions will behave predictably, which is not always the case with discontinuous functions that can have jumps or undefined points.

Definite Integrals

Moving on to definite integrals, these play a crucial role in solving differential equations, particularly when determining a particular solution to an inhomogeneous differential equation. A definite integral,
\[\int_a^b f(x) \, dx, \]
represents the accumulated area under the curve of a function f(x) from a to b. The problem highlights the use of the definite integral in constructing the solution u(x). Hence, the integral of h(t) multiplied by the hyperbolic sine function is an essential part of this solution. Moreover, definite integrals are often used to find the necessary functions to rewrite expressions in specific forms, such as the a(x) and b(x) functions in this problem. Understanding how to evaluate definite integrals allows us to manipulate and solve for these functions in more complex expressions.

Hyperbolic Functions

Hyperbolic functions, often overlooked, bear significant resemblance to trigonometric functions and are frequently encountered in solving differential equations. They are defined by combinations of exponential functions. For example, the hyperbolic sine and hyperbolic cosine functions are defined by:
\[\sinh(x) = \frac{e^x - e^{-x}}{2}, \]
and
\[\cosh(x) = \frac{e^x + e^{-x}}{2}. \]
In our problem, the solution u(x) involves an integral that includes the hyperbolic sine function \(\sinh \), highlighting the relevance of hyperbolic functions in expressing solutions to differential equations. When we take derivatives of hyperbolic functions, we often get the other hyperbolic function, similar to how derivatives of trigonometric functions relate to each other—this property is used to derive the expressions for the first and second derivatives of u(x).

Limits of Functions

Lastly, let's delve into the limits of functions, a cornerstone concept in calculus and real analysis. The limit of a function describes its behavior as the input approaches a certain value. In part (c) and (d) of our exercise, we deal with limits as x approaches infinity to understand the long-term behavior of the functions a(x) and b(x). Understanding limits is crucial because they help characterize function behaviors that are not directly visible, such as at infinity.
In the given problem, this concept lets us make clear statements about the long-term behavior of the particular solution u(x) and its derivatives. For instance, if a(x) approaches a constant value A as x grows, this has implications on the corresponding coefficient in the solution of the differential equation u(x), helping us assure the stability and boundedness of the solution.