- Thread starter Kayle
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Cauchy posed and solved many problems.I want to know ,What is theto the Cauchy problem?solution

Please define Cauchy's Problem.

Please share your thoughts/work regarding this assignment.

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Perhaps you mean solving a partial differential equation for, say, f, given values of f on some domain (that's what Wikipedia gives as the "Cauchy problem"). You have actually asked two different things! Your title says "What is the meaning of solution to Cauchy problem" but in your post you said "What is the solution to the Cauchy problem" which implies that you are asking for a solution or a method of solution.

If you actually want the "meaning" of "solution", the solution to a Cauchy problem is a function that satisfies the given differential equation and also takes on the given values. An example of a Cauchy problem is to find a function, f(x,y), that satisfies \(\displaystyle \nabla^2 f= \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial^2 f}{\partial y^2}= 0\), with "boundary conditions" f(0, y)= y, f(1, y)= 1- y, f(x, 0)= x, f(x, 1)= 1- x.

A problem where you are given a partial differential equation and values of, say, the partial derivatives on some set is NOT a "Cauchy problem. An example of a problem that is NOT a "Cauchy problem" would be given a wire pulled tight between to points, say (0,0) and (L, 0) on an xy-coordinate system, the wire being stretched into the shape y= f(x), and then released at t= 0. The differential equation is the "wave equation" \(\displaystyle \frac{\partial^20 y}{\partial x^2}= \frac{1}{c^2}\frac{\partial^2 y}{\partial t^2}\) (c is the "wave speed" in the wire and depends on the linear density of the wire and the tenson in the wire. The boundary conditions are, since the wire is fixed at (0, 0) and (L, 0), y(0, t)= 0, y(L, t)= 0. The initial conditions are \(\displaystyle y(x, 0)= f(x)\) and \(\displaystyle y_t(x, 0)= 0\). That is NOT a Cauchy problem because part of the conditions is a value of the derivative, not y itself.

If you are asking how to solve a Cauchy problem, that depends strongly on both the equation and the geometry of the set on which the value of the function is given. It is relatively easy to see that the solution to the example Cauchy problem I gave above is \(\displaystyle f(x,y)= x(1- y)+ y(1- x)= x+ y- 2xy\).

If you actually want the "meaning" of "solution", the solution to a Cauchy problem is a function that satisfies the given differential equation and also takes on the given values. An example of a Cauchy problem is to find a function, f(x,y), that satisfies \(\displaystyle \nabla^2 f= \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial^2 f}{\partial y^2}= 0\), with "boundary conditions" f(0, y)= y, f(1, y)= 1- y, f(x, 0)= x, f(x, 1)= 1- x.

A problem where you are given a partial differential equation and values of, say, the partial derivatives on some set is NOT a "Cauchy problem. An example of a problem that is NOT a "Cauchy problem" would be given a wire pulled tight between to points, say (0,0) and (L, 0) on an xy-coordinate system, the wire being stretched into the shape y= f(x), and then released at t= 0. The differential equation is the "wave equation" \(\displaystyle \frac{\partial^20 y}{\partial x^2}= \frac{1}{c^2}\frac{\partial^2 y}{\partial t^2}\) (c is the "wave speed" in the wire and depends on the linear density of the wire and the tenson in the wire. The boundary conditions are, since the wire is fixed at (0, 0) and (L, 0), y(0, t)= 0, y(L, t)= 0. The initial conditions are \(\displaystyle y(x, 0)= f(x)\) and \(\displaystyle y_t(x, 0)= 0\). That is NOT a Cauchy problem because part of the conditions is a value of the derivative, not y itself.

If you are asking how to solve a Cauchy problem, that depends strongly on both the equation and the geometry of the set on which the value of the function is given. It is relatively easy to see that the solution to the example Cauchy problem I gave above is \(\displaystyle f(x,y)= x(1- y)+ y(1- x)= x+ y- 2xy\).

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