Integral-form change of variable in differential equation

In summary, the conversation discusses a differential equation known as the general Sturm-Liouville problem and a change of variable that can be used to express the equation in terms of a new variable, y. The speaker also presents an approach for solving the equation by expressing it in terms of the new variable.
  • #1
Jaime_mc2
8
9
I have the following differential equation, which is the general Sturm-Liouville problem,
$$
\dfrac{d}{dx} \left[ p(x) \dfrac{d\varphi}{dx} \right] + \left[ \lambda w(x) - q(x) \right] \varphi(x) = 0\ ,
$$
and I want to perform the change of variable
$$
x \rightarrow y = \int_a^x \sqrt{\lambda \dfrac{w(t)}{p(t)}}\, dt\ .
$$

I used the fundamental theorem of calculus to get
$$
\dfrac{dy}{dx} = \sqrt{\lambda\dfrac{w(x)}{p(x)}}\ ,
$$
and get the differentiation operator with respect to the new variable as
$$
\dfrac{d}{dx} = \dfrac{dy}{dx}\dfrac{d}{dy} = \sqrt{\lambda\dfrac{w(x)}{p(x)}} \dfrac{d}{dy}\ .
$$

My question is about how do I get the differential equation to be expressed in terms of ##y##. My first thought was realising that ##\varphi(x)## could just be expressed as ##\varphi(y)## if there exists an inversion such that ##x = x(y)##. Then I just extended this idea to ##w##, ##p## and ##q##, and I arrived to the equation
$$
\dfrac{d}{dy} \left[ \sqrt{w(y)p(y)} \dfrac{d\varphi}{dy} \right] + \left[ \sqrt{w(y)p(y)} - \dfrac{q(y)}{\lambda} \sqrt{\dfrac{p(y)}{w(y)}} \right] \varphi(y) = 0\ .
$$

Is this approach correct?
 
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  • #2
I observe if you choose w as w=q, the equation becomes easy enough to solve.
 

FAQ: Integral-form change of variable in differential equation

1. What is the integral-form change of variable in a differential equation?

The integral-form change of variable is a technique used in solving differential equations. It involves replacing the independent variable in the differential equation with a new variable, which is usually a function of the original variable. This allows for easier integration and solution of the differential equation.

2. Why is the integral-form change of variable useful?

The integral-form change of variable is useful because it can simplify the process of solving differential equations. It can transform a complex differential equation into a simpler form that is easier to integrate and solve. It also allows for the use of standard integration techniques, making the solution process more straightforward.

3. How do you perform an integral-form change of variable?

To perform an integral-form change of variable, you first need to identify the independent variable in the differential equation. Then, you choose a new variable that is a function of the original variable. You substitute this new variable into the differential equation, and then integrate both sides of the equation. Finally, you solve for the new variable and then substitute it back into the original equation to obtain the solution.

4. What are some common examples of integral-form change of variable?

Some common examples of integral-form change of variable include substitution of trigonometric functions, exponential functions, and logarithmic functions. For example, in solving a differential equation with a trigonometric function, you can use the substitution u = sin(x) or u = cos(x) to simplify the equation and make it easier to solve.

5. Are there any limitations to using the integral-form change of variable?

Yes, there are limitations to using the integral-form change of variable. It is not always possible to find a suitable substitution that will simplify the differential equation. In some cases, the substitution may lead to a more complex equation or may not work at all. Additionally, the integral-form change of variable may not be applicable to all types of differential equations, such as those with non-constant coefficients.

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