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The local solution of an ordinary differential equation (ODE) is the solution you get at a specific point of the function involved in the differential equation. One can Taylor expand the function at this point, turning non-linear ODEs into linear ones, if needed, to find the behavior of the solution around that one specific point. Of course, a local solution tells you very little about the ODE's global solution, but sometimes you don't want to know that anyways.

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Q: What is the local solution of an ordinary differential equation?
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Cauchy problem for first order partial differential equation?

There is a theorem called the Cauchy-Kowalevski theoremwhich deals with the existence of solutions to a system of mdifferential equation in n dimensions when the coefficients are analytic functions. I am guessing this is what you are asking about. A special case of this theorem was proved by Cauchy alone.The theorem talks about the local existence of a solution.Since this is a complicated topic, I will provide a link.


What is the minimum or maximum point called in a quadratic equation?

They are simply referred to as local minimums and maximums. Experience: Algebra 2 Advanced


What is the method of finite differences for linear equations?

Intuitive derivationFinite-difference methods approximate the solutions to differential equations by replacing derivative expressions with approximately equivalent difference quotients. That is, because the first derivative of a function f is, by definition,then a reasonable approximation for that derivative would be to takefor some small value of h. In fact, this is the forward difference equation for the first derivative. Using this and similar formulae to replace derivative expressions in differential equations, one can approximate their solutions without the need for calculus.[edit] Derivation from Taylor's polynomialAssuming the function whose derivatives are to be approximated is properly-behaved, by Taylor's theorem,where n! denotes the factorial of n, and Rn(x) is a remainder term, denoting the difference between the Taylor polynomial of degree n and the original function. Again using the first derivative of the function f as an example, by Taylor's theorem,f(x0 + h) = f(x0) + f'(x0)h + R1(x), which, with some minor algebraic manipulation, is equivalent toso that for R1(x) sufficiently small,[edit] Accuracy and orderThe error in a method's solution is defined as the difference between its approximation and the exact analytical solution. The two sources of error in finite difference methods are round-off error, the loss of precision due to computer rounding of decimal quantities, and truncation error or discretization error, the difference between the exact solution of the finite difference equation and the exact quantity assuming perfect arithmetic (that is, assuming no round-off).The finite difference method relies on discretizing a function on a grid. To use a finite difference method to attempt to solve (or, more generally, approximate the solution to) a problem, one must first discretize the problem's domain. This is usually done by dividing the domain into a uniform grid (see image to the right). Note that this means that finite-difference methods produce sets of discrete numerical approximations to the derivative, often in a "time-stepping" manner.An expression of general interest is the local truncation error of a method. Typically expressed using Big-O notation, local truncation error refers to the error from a single application of a method. That is, it is the quantity f'(xi) &minus; f'i if f'(xi) refers to the exact value and f'i to the numerical approximation. The remainder term of a Taylor polynomial is convenient for analyzing the local truncation error. Using the Lagrange form of the remainder from the Taylor polynomial for f(x0 + h), which is, where x0 < &xi; < x0 + h, the dominant term of the local truncation error can be discovered. For example, again using the forward-difference formula for the first derivative, knowing that f(xi) = f(x0 + ih),and with some algebraic manipulation, this leads toand further noting that the quantity on the left is the approximation from the finite difference method and that the quantity on the right is the exact quantity of interest plus a remainder, clearly that remainder is the local truncation error. A final expression of this example and its order is:This means that, in this case, the local truncation error is proportional to the step size.[edit] Example: ordinary differential equationFor example, consider the ordinary differential equationThe Euler method for solving this equation uses the finite difference quotientto approximate the differential equation by first substituting in for u'(x) and applying a little algebra to getThe last equation is a finite-difference equation, and solving this equation gives an approximate solution to the differential equation.[edit] Example: The heat equationConsider the normalized heat equation in one dimension, with homogeneous Dirichlet boundary conditions(boundary condition) (initial condition) One way to numerically solve this equation is to approximate all the derivatives by finite differences. We partition the domain in space using a mesh x0,...,xJ and in time using a mesh t0,....,tN. We assume a uniform partition both in space and in time, so the difference between two consecutive space points will be h and between two consecutive time points will be k. The pointswill represent the numerical approximation of u(xj,tn).[edit] Explicit methodThe stencil for the most common explicit method for the heat equation. Using a forward difference at time tn and a second-order central difference for the space derivative at position xj ("FTCS") we get the recurrence equation:This is an explicit method for solving the one-dimensional heat equation.We can obtain from the other values this way:where r = k / h2.So, knowing the values at time n you can obtain the corresponding ones at time n+1 using this recurrence relation. and must be replaced by the boundary conditions, in this example they are both 0.This explicit method is known to be numerically stable and convergent whenever . The numerical errors are proportional to the time step and the square of the space step:[edit] Implicit methodThe implicit method stencil. If we use the backward difference at time ti + 1 and a second-order central difference for the space derivative at position xj ("BTCS") we get the recurrence equation:This is an implicit method for solving the one-dimensional heat equation.We can obtain from solving a system of linear equations:The scheme is always numerically stable and convergent but usually more numerically intensive than the explicit method as it requires solving a system of numerical equations on each time step. The errors are linear over the time step and quadratic over the space step.[edit] Crank-Nicolson methodFinally if we use the central difference at time tn + 1 / 2 and a second-order central difference for the space derivative at position xj ("CTCS") we get the recurrence equation:This formula is known as the Crank-Nicolson method.The Crank-Nicolson stencil. We can obtain from solving a system of linear equations:The scheme is always numerically stable and convergent but usually more numerically intensive as it requires solving a system of numerical equations on each time step. The errors are quadratic over the time step and formally are of the fourth degree regarding the space step:However, near the boundaries, the error is often O(h2) instead of O(h4).Usually the Crank-Nicolson scheme is the most accurate scheme for small time steps. The explicit scheme is the least accurate and can be unstable, but is also the easiest to implement and the least numerically intensive. The implicit scheme works the best for large time steps.


What does a local emergency manager coordinate?

local emergency management programs


At a local theater adult tickets cost 8 and student tickets cost 5 At a recent show 500 tickets were sold for a total of 3475 How many adult tickets were sold?

a=# of adult tickets, s=# of student tickets 8a+5s=3475 a+s=500 Solve for s in this equation: s=500-a Now substitute 500-a in for s in the equation 8a+5s=3475: 8a+5(500-a)=3475 8a+2500-5a=3475 3a=975 a=325 ans. There were 325 adult tickets sold.

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Cauchy problem for first order partial differential equation?

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