- #1
sigmund
- 23
- 0
We have the first order ODE
[tex] y'=4t \sqrt y,~y(0)=1, [/tex]
for which i have found the exact solution, namely a fourth order polynomial.
I want a numerical method to solve the problem exactly. This method has to be a fourth order method, since this implies that the local error vanishes.
Now we change the problem so it becomes
[tex] y'=4t \sqrt y - \lambda(y-(1+t^2)^2),~y(0)=a, [/tex]
and the question is: for which values of [itex]\lambda[/itex] and [itex]a[/itex] does a method that has the above mentioned property solve the new problem exactly.
Of course, the obvious case is [itex]\lambda=0[/itex] and [itex]a=1[/itex], because in this case the new problem reduces to the first problem.
My idea is that the solution must be a fourth order polynomial, since a fourth order numerical method has to solve the new problem exactly.
Although I want your view on this and a strategy to find the values of [itex]\lambda[/itex] and [itex]a[/itex] for which the new problem is solved exactly by a fourth order numerical method.
[tex] y'=4t \sqrt y,~y(0)=1, [/tex]
for which i have found the exact solution, namely a fourth order polynomial.
I want a numerical method to solve the problem exactly. This method has to be a fourth order method, since this implies that the local error vanishes.
Now we change the problem so it becomes
[tex] y'=4t \sqrt y - \lambda(y-(1+t^2)^2),~y(0)=a, [/tex]
and the question is: for which values of [itex]\lambda[/itex] and [itex]a[/itex] does a method that has the above mentioned property solve the new problem exactly.
Of course, the obvious case is [itex]\lambda=0[/itex] and [itex]a=1[/itex], because in this case the new problem reduces to the first problem.
My idea is that the solution must be a fourth order polynomial, since a fourth order numerical method has to solve the new problem exactly.
Although I want your view on this and a strategy to find the values of [itex]\lambda[/itex] and [itex]a[/itex] for which the new problem is solved exactly by a fourth order numerical method.