Difference between revisions of "Initial Value Problem"
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| + | == Definition == | ||
| + | An '''initial value problem''' is a differential equation | ||
| + | :<math>y'(t) = f(t, y(t))</math> with <math>f\colon \Omega \subset \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n</math> where <math>\Omega</math> is an open set of <math>\mathbb{R} \times \mathbb{R}^n</math>, | ||
| + | together with a point in the domain of <math>f</math> | ||
| + | :<math>(t_0, y_0) \in \Omega,</math> | ||
| + | called the initial condition. | ||
| + | A '''solution''' to an initial value problem is a function <math>y</math> that is a solution to the differential equation and satisfies | ||
| + | :<math>y(t_0) = y_0.</math> | ||
| + | |||
| + | In higher dimensions, the differential equation is replaced with a family of equations <math>y_i'(t)=f_i(t, y_1(t), y_2(t), \dotsc)</math>, and <math>y(t)</math> is viewed as the vector <math>(y_1(t), \dotsc, y_n(t))</math>, most commonly associated with the position in space. More generally, the unknown function <math>y</math> 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. <math>y''(t)=f(t,y(t),y'(t))</math>. | ||
| + | |||
| + | ===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. | ||
| + | |||
| + | A simple example is to solve <math>y'(t) = 0.85 y(t)</math> and <math>y(0) = 19</math>. We are trying to find a formula for <math>y(t)</math> that satisfies these two equations. | ||
| + | |||
| + | Rearrange the equation so that <math>y</math> is on the left hand side | ||
| + | |||
| + | : <math>\frac{y'(t)}{y(t)} = 0.85</math> | ||
| + | |||
| + | Now integrate both sides with respect to <math>t</math> (this introduces an unknown constant <math>B</math>). | ||
| + | |||
| + | : <math>\int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt</math> | ||
| + | : <math>\ln |y(t)| = 0.85t + B </math> | ||
| + | |||
| + | Eliminate the logarithm with exponentiation on both sides | ||
| + | |||
| + | : <math> | y(t) | = e^Be^{0.85t} </math> | ||
| + | |||
| + | Let <math>C</math> be a new unknown constant, <math>C = \pm e^B</math>, so | ||
| + | |||
| + | : <math> y(t) = Ce^{0.85t} </math> | ||
| + | |||
| + | Now we need to find a value for <math>C</math>. Use <math>y(0) = 19</math> as given at the start and substitute 0 for <math>t</math> and 19 for <math>y</math> | ||
| + | |||
| + | : <math> 19 = C e^{0.85 \cdot 0}</math> | ||
| + | : <math> C = 19 </math> | ||
| + | |||
| + | this gives the final solution of <math> y(t) = 19e^{0.85t}</math>. | ||
| + | |||
| + | ;Second example | ||
| + | The solution of | ||
| + | |||
| + | : <math>y'+3y=6t+5,\qquad y(0)=3</math> | ||
| + | |||
| + | can be found to be | ||
| + | |||
| + | : <math>y(t)=2e^{-3t}+2t+1. \,</math> | ||
| + | |||
| + | Indeed, | ||
| + | |||
| + | : <math> \begin{align} | ||
| + | y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\ | ||
| + | &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\ | ||
| + | &= 6t+5. | ||
| + | \end{align} </math> | ||
| + | |||
| + | Some more simple examples of initial value problems: | ||
| + | * <math> y' = 2x </math>, <math> y(2) = 0 </math>. With this point and the general solution <math>y = x^2 + C</math>, we can calculate the constant C to be -4. Thus the particular solution is <math>y = x^2 - 4</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>. | ||
| + | |||
| + | == Licensing == | ||
| + | Content obtained and/or adapted from: | ||
| + | * [https://en.wikipedia.org/wiki/Initial_value_problem Initial value problem, Wikipedia] under a CC BY-SA license | ||
Latest revision as of 21:40, 5 November 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
called the initial condition.
A solution to an initial value problem is a function that is a solution to the differential equation and satisfies
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.
A simple example is to solve 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) = 0.85 y(t)} 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(0) = 19} . We are trying to find a formula for 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)} that satisfies these two equations.
Rearrange the equation so that 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} is on the left hand side
- 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 \frac{y'(t)}{y(t)} = 0.85}
Now integrate both sides with respect to 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} (this introduces an unknown constant 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 B} ).
- 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 \int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt}
- 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 \ln |y(t)| = 0.85t + B }
Eliminate the logarithm with exponentiation on both sides
- 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) | = e^Be^{0.85t} }
Let 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 C} be a new unknown constant, 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 C = \pm e^B} , 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 y(t) = Ce^{0.85t} }
Now we need to find a value for 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 C} . Use 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) = 19} as given at the start and substitute 0 for 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} and 19 for 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}
- 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 19 = C e^{0.85 \cdot 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 C = 19 }
this gives the final solution 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 y(t) = 19e^{0.85t}} .
- Second example
The solution 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 y'+3y=6t+5,\qquad y(0)=3}
can be found to be
- 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)=2e^{-3t}+2t+1. \,}
Indeed,
- 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 \begin{align} y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\ &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\ &= 6t+5. \end{align} }
Some more simple examples of initial value problems:
- 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 } , . 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} } .
Licensing
Content obtained and/or adapted from:
- Initial value problem, Wikipedia under a CC BY-SA license
