Problem 5 - Applied Partial Differential Equations

Question

Consider $$ L=\frac{d^{2}}{d x^{2}}+6 \frac{d}{d x}+9 $$ (a) Show that \(L\left(e^{r x}\right)=(r+3)^{2} e^{r \pi}\). (b) Use part (a) to obtain solutions of \(L(y)-0\) (a second order constant coefficient differential equation). (c) If \(z\) depends on \(x\) and a parameter \(r\), show that $$ \frac{\partial}{\partial r} L(z)=L\left(\frac{\partial z}{\partial r}\right) \text {. } $$ (d) Using part (c), evaluate \(L(\partial z / \partial r)\) if \(z-e^{r x}\). (e) Obtain a second solution of \(L(y)=0\), using part (d).

Step-by-Step Solution

Answer

The individual solutions to the differential equation \(L(y) = 0\) are \(y = e^{-3x}\) and \(y = -3xe^{-3x}\).

Step 1: Prove the given relationship

First, consider the function \(f(x) = e^{rx}\). The derivatives of \(f(x)\) are \(f'(x) = re^{rx}\) and \(f''(x) = r^2e^{rx}\). We can substitute these values into \(L(f(x))\), giving us \(L(f(x)) = r^2e^{rx} + 6re^{rx} + 9e^{rx} = e^{rx}(r^2 + 6r + 9) = (r + 3)^2e^{rx}\). Therefore, \(L\left(e^{r x}\right) = (r + 3)^{2} e^{r \pi}\).

Step 2: Obtain solutions to differential equation

Given that \(L(y) = 0\), we have from part (a) that \((r + 3)^2e^{rx} = 0\). This is only true when \(r = -3\). Thus, a solution to \(L(y) = 0\) is \(y = e^{-3x}\).

Step 3: Prove the partial derivative interaction with the operator

We know that the differential operator \(L\) acts linearly. This means that \(L(af + bg) = aL(f) + bL(g)\) for any functions \(f\) and \(g\) and constants \(a\) and \(b\). Therefore, we have the chain rule at our disposal. Let \(z\) depend on \(x\) and a parameter \(r\). By differentiating \(L(z)\) with respect to \(r\) we get \(\frac{\partial}{\partial r}L(z) = L\left(\frac{\partial z}{\partial r}\right)\).

Step 4: Evaluate the partial derivative of z

We have \(z = e^{rx}\), so applying the chain rule to differentiate with respect to \(r\) gives \(\frac{\partial z}{\partial r} = xe^{rx}\). Then, applying the operator \(L\) to \(\frac{\partial z}{\partial r}\) gives \(L\left(\frac{\partial z}{\partial r}\right) = 0\).

Step 5: Find the second solution

From Step 4, we find that \(\frac{\partial z}{\partial r} = xe^{rx}\) satisfies \(L\left(\frac{\partial z}{\partial r}\right) = 0\), indicating that \(y = xe^{rx}\) would be a second solution to the differential equation \(L(y) = 0\). However, we know that \(r = -3\), so the second solution to the differential equation would be \(y = -3xe^{-3x}\).

Key Concepts

Second Order Constant Coefficient Differential Equation

In mathematics, a second order constant coefficient differential equation is a common type of linear differential equation. It involves a second derivative and has constant coefficients—it doesn’t change as the dependent variable changes. An equation like this generally looks like:\[ L = \frac{d^2}{dx^2} + a\frac{d}{dx} + b \]where \(a\) and \(b\) are constants.
For students, mastering such equations is crucial because they appear in physical phenomena descriptions frequently.Differential equations of this form are significant because they often describe systems with simple harmonic motion, circuits, or vibrations. The solutions to these equations are usually exponential functions or polynomials, depending on the nature of the roots. To find these solutions, we can apply the characteristic equation method, which involves assuming a solution of exponential form \(e^{r x}\) and substituting it back into the differential equation to find the value of \(r\).
For example, the process involves:

• Setting up the characteristic quadratic
• Solving for \(r\) to find distinct real roots, repeated roots, or complex roots
• Construct the general solution based on the root type

Using the example in the exercise, we found that \(r = -3\) is a solution, giving us a particular solution of \(y = e^{-3x}\).
Expanding upon initial solutions with other legitimate functions based on constants is often necessary to fully describe the solution space.

Exponential Functions in PDEs

Exponential functions play a crucial role in solving differential equations. They often appear as solutions in linear differential equations, especially those with constant coefficients.In the context of partial differential equations (PDEs), exponential solutions are particularly helpful due to their convenient properties when differentiating. For example, if we have an equation like \(L\left(e^{r x}\right)\), the differential operations turn into simple algebraic manipulations.
We use the property that the derivative of an exponential is again an exponential, multiplied by the factor of derivation. So for \(f(x) = e^{rx}\):

• The first derivative \( f'(x) = re^{rx} \)
• The second derivative \(f''(x) = r^2e^{rx} \)

This means that when inserting \(e^{rx}\) into a differential operator with constant coefficients, the exponent \(r\) controls all aspects of the output style. For the equation \(L = \frac{d^2}{dx^2} + 6\frac{d}{dx} + 9\), the characteristic polynomial approach was used:
\(L(f(x)) = (r + 3)^2e^{rx}\). This indicates that manipulating the value \(r\) changes the form of solutions significantly, which helps in building a complete set of solutions to the differential equation.

Partial Derivative Interaction with Operators

Partial derivatives and operators, like the ones used in PDEs, work hand in hand to solve complex mathematical problems. Understanding this interaction will simplify many calculus problems.When we have a function \(z(x, r)\) depending on both \(x\) and a parameter \(r\), partial derivatives help us see how changes in \(r\) affect \(z\).
The linearity of differential operators like \(L\) ensures that:\[\frac{\partial}{\partial r} L(z) = L\left(\frac{\partial z}{\partial r}\right) \]This powerful property lets us differentiate before or after applying the operator without changing the outcome.Consider a function \(z = e^{rx}\). By differentiating with respect to \(r\), you obtain:

• \(\frac{\partial z}{\partial r} = xe^{rx} \)

Applying the differential operator \(L\) to this, we find that it simplifies problems significantly and helps find solutions without tedious calculations. As seen in the example, substituting and evaluating gives another critical solution, allowing the construction of a robust solution set like \(y = -3xe^{-3x}\).
Learning how to handle the interactions between operators and partial derivatives hence provides the ability to elegantly solve and explore complex differential equations.