Question

Let $h$ be continuous on $[0,L]$ and let $u_0,u_L,$ and $a$ be constants, with $a>0.$ Show that it's always possible to find a function $q$ that satisfies (a), (b), or (c), but that this isn't so for (d). (a) $a^2{q}^{\prime \prime }+h=0,q(0)=u_0,q(L)=u_L$ (b) $a^2{q}^{\prime \prime }+h=0,q^{\prime }(0)=u_0,q(L)=u_L$ (c) $a^2{q}^{\prime \prime }+h=0,q(0)=u_0,q^{\prime }(L)=u_L$ (d) $a^2{q}^{\prime \prime }+h=0,q^{\prime }(0)=u_0,q^{\prime }(L)=u_L$

Short answer

Short Answer: We can find functions q(x) that satisfy the given conditions for cases (a), (b), and (c) by following a step-by-step process involving the integration of the given differential equation and the application of the boundary conditions. However, for case (d), since the condition $u_L=u_0$ restricts the possible values for the function q(x), we cannot find a function q(x) that satisfies the conditions for this case in general.

Step-by-step solution

Analyze the given conditions

We are given the following conditions: $a^2{q}^{\prime \prime }+h=0,q(0)=u_0,q(L)=u_L$ Our goal is to find a function q that satisfies these conditions.

Integrate the differential equation

Integrate the given equation twice with respect to x: First integration: $a^2{q}^{\prime }+h(x)=C_1$ Second integration: $a^{2}q(x)+{\int }_0^{x}h(t)dt=C_{1}x+C_2$

Apply the boundary conditions

Apply the given boundary conditions: At $x=0$, we have $q(0)=u_0$: $0+C_2=u_0⟹C_2=u_0$ At $x=L$, we have $q(L)=u_L$: $a^2{u}_L+{\int }_0^{L}h(t)dt=C_{1}L+u_0⟹C_1=\frac{a^2(u_L-u_0)-{\int }_0^{L}h(t)dt}{L}$

Write the function q

The function q that satisfies the given conditions (a) can be written as: $q(x)=u_0+\frac{1}{a^2}({\int }_0^{x}h(t)dt+\frac{a^2(u_L-u_0)-{\int }_0^{L}h(t)dt}{L}(x))$ Now let's move on to case (b): Case (b):

Analyze the given conditions

We are given the following conditions: $a^2{q}^{\prime \prime }+h=0,q^{\prime }(0)=u_0,q(L)=u_L$ Our goal is to find a function q that satisfies these conditions.

Integrate the differential equation

Integrate the given equation twice with respect to x (same as in Case (a)): First integration: $a^2{q}^{\prime }+h(x)=C_1$ Second integration: $a^{2}q(x)+{\int }_0^{x}h(t)dt=C_{1}x+C_2$

Apply the boundary conditions

Apply the given boundary conditions: At $x=0$, we have $q^{\prime }(0)=u_0$: $C_1=u_0$ At $x=L$, we have $q(L)=u_L$: $a^2{u}_L+{\int }_0^{L}h(t)dt=u_{0}L+C_2⟹C_2=u_L-u_{0}L$

Write the function q

The function q that satisfies the given conditions (b) can be written as: $q(x)=(u_L-u_{0}L)+\frac{1}{a^2}(u_{0}x+{\int }_0^{x}h(t)dt)$ Now let's move on to case (c): Case (c):

Analyze the given conditions

We are given the following conditions: $a^2{q}^{\prime \prime }+h=0,q(0)=u_0,q^{\prime }(L)=u_L$ Our goal is to find a function q that satisfies these conditions.

Integrate the differential equation

Integrate the given equation twice with respect to x (same as in Case (a)): First integration: $a^2{q}^{\prime }+h(x)=C_1$ Second integration: $a^{2}q(x)+{\int }_0^{x}h(t)dt=C_{1}x+C_2$

Apply the boundary conditions

Apply the given boundary conditions: At $x=0$, we have $q(0)=u_0$: $0+C_2=u_0⟹C_2=u_0$ At $x=L$, we have $q^{\prime }(L)=u_L$: $C_1=u_L-\frac{1}{a^2}{\int }_0^{L}h(t)dt$

Write the function q

The function q that satisfies the given conditions (c) can be written as: $q(x)=u_0+\frac{1}{a^2}((u_L-\frac{1}{a^2}{\int }_0^{L}h(t)dt)(x)+{\int }_0^{x}h(t)dt)$ Finally, let's move on to case (d): Case (d):

Analyze the given conditions

We are given the following conditions: $a^2{q}^{\prime \prime }+h=0,q^{\prime }(0)=u_0,q^{\prime }(L)=u_L$ Our goal is to find a function q that satisfies these conditions.

Integrate the differential equation

Integrate the given equation once with respect to x: $a^2{q}^{\prime }+h(x)=C_1$

Apply the boundary conditions

Apply the given boundary conditions: At $x=0$, we have $q^{\prime }(0)=u_0$: $C_1=u_0$ At $x=L$, we have $q^{\prime }(L)=u_L$: $C_1=u_L⟹u_L=u_0$

Discuss the possibility of finding q

The condition $u_L=u_0$ for case (d) implies that it is not possible to find a function q that satisfies the conditions in general, as $u_L$ and $u_0$ are constants that can have any value. In conclusion, we can find a function q that satisfies the conditions for cases (a), (b), and (c), but not for case (d).

Key concepts

Boundary Value Problems

Boundary value problems are an important class of differential equations where there are conditions (known as boundary conditions) specified at the boundaries of the interval in which the solution is desired. For example, in the exercise, conditions are given at points 0 and $L$. These conditions can be values of the function itself (Dirichlet conditions), derivatives of the function (Neumann conditions), or a mix of the two. Boundary value problems come up in many areas of physics and engineering, such as heat conduction, wave propagation, and fluid dynamics. Solving these problems involves finding a function that satisfies the differential equation together with the boundary conditions. In many cases, these problems don't have simple solutions and require numerical methods for solving.

Second Order Differential Equations

Second order differential equations involve the second derivative of the unknown function, like $a^2{q}^{″}+h=0$ in the exercise. These equations describe many physical phenomena, such as harmonic oscillators, vibrations of strings, and more. Solving these equations typically involves integrating the differential equation and applying boundary conditions, as shown in the step-by-step solution. The general solution to a second order differential equation includes constants of integration, which are determined using the boundary conditions. In this way, a specific solution meeting given criteria is found.

Mathematical Proofs

Mathematical proofs provide a logical argument that demonstrates the truth of a statement based on previously established axioms and propositions. In this exercise, we used a series of logical steps to determine that it is possible to find functions $q$ for conditions (a), (b), and (c), but generally not for (d). Proofs typically involve breaking down complex problems into simpler ones, applying known theorems, and using logical deduction. In the case provided, each step of the solution is supported by appropriate mathematical reasoning. Mathematical proofs require clarity and precision to ensure that each inference logically follows from the previous one.

Continuity of Functions

The concept of continuity is fundamental in calculus and mathematical analysis. A continuous function is one where small changes in the input result in small changes in the output. This idea is crucial when dealing with differential equations and boundary value problems. In the exercise, the function $h$ is continuous on the interval $[0,L]$, making it easier to integrate and work with. This property helps guarantee that solutions to the differential equation behave well and smoothly across the interval. Continuity also ensures that the integrals involved in solving the differential equations are well-defined.