Difference between revisions of "Initial Value Problem"
| Line 13: | Line 13: | ||
Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. <math>y''(t)=f(t,y(t),y'(t))</math>. | Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. <math>y''(t)=f(t,y(t),y'(t))</math>. | ||
| − | ===Examples== | + | ===Examples=== |
With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on. | With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point <math>(x, y(x))</math>, for a second order equation we need 2 points (typically <math>(x_1, y(x_1))</math> and either <math>(x_2, y(x_2))</math> or <math>(x_2, y'(x_2))</math>), and so on. | ||
| Line 21: | Line 21: | ||
* <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>. | * <math> y' - y = 0 </math>, <math> y(0) = 3 </math>. <math> 3 = Ce^{0} \implies C = 3</math>, so the particular solution is <math> y = 3e^{x} </math>. | ||
* <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>. | * <math> y'' + y' - 2y = 0 </math>, <math> y(0) = 2 </math>, <math> y'(0) = -1 </math>. So, <math> 2 = Ce^{0} + De^{0} = C + D </math> and <math> -1 = Ce^{0} - 2De^{0} = C - 2D</math>. Thus C = 1 and D = 1, and the particular solution is <math> y = e^{x} + e^{-2x} </math>. | ||
| + | |||
| + | ==Resources== | ||
| + | * [https://en.wikipedia.org/wiki/Initial_value_problem Initial value problem], Wikipedia | ||
Revision as of 16:00, 14 October 2021
Definition
An initial value problem is a differential equation
- with where is an open set of ,
together with a point in the domain of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (t_0, y_0) \in \Omega,}
called the initial condition.
A solution to an initial value problem is a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} that is a solution to the differential equation and satisfies
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t_0) = y_0.}
In higher dimensions, the differential equation is replaced with a family of equations Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_i'(t)=f_i(t, y_1(t), y_2(t), \dotsc)} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(t)} is viewed as the vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (y_1(t), \dotsc, y_n(t))} , most commonly associated with the position in space. More generally, the unknown function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} can take values on infinite dimensional spaces, such as Banach spaces or spaces of distributions.
Initial value problems are extended to higher orders by treating the derivatives in the same way as an independent function, e.g. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y''(t)=f(t,y(t),y'(t))} .
Examples
With initial value problems, we are given a differential equation, and one or more points (depending on the order of the equation) to solve the constants in the general solution. For a first order differential equation we need 1 point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x, y(x))} , for a second order equation we need 2 points (typically Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_1, y(x_1))} and either Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_2, y(x_2))} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_2, y'(x_2))} ), and so on.
Examples:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' = 2x } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(2) = 0 } . With this point and the general solution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^2 + C} , we can calculate the constant C to be -4. Thus the particular solution is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^2 - 4} .
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y' - y = 0 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(0) = 3 } . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3 = Ce^{0} \implies C = 3} , so the particular solution is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = 3e^{x} } .
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'' + y' - 2y = 0 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y(0) = 2 } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'(0) = -1 } . So, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 = Ce^{0} + De^{0} = C + D } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1 = Ce^{0} - 2De^{0} = C - 2D} . Thus C = 1 and D = 1, and the particular solution is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = e^{x} + e^{-2x} } .
Resources
- Initial value problem, Wikipedia
