I am studying numeric solutions of differential equations, and part of my reading is found in Simmonds' book, Differential Equations with Applications and Historical Notes. Although the chapter on numerical methods is written by John S. Robertson.In the treatment of Euler's method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?For some context, the differential equations under consideration are those of the form y'=f(x,y), defined on some interval [a,b], with some initial value y ( a ) = αThese are then transformed into the integral equation y ( x 1 ) − y ( x 0 ) = ∫ x 0 x 1 f ( x , y ) d x .Then, using Taylor's theorem and substituting back into that formula, the error is found to be h 2 2 y ″ ( ξ ) ,which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y''(x) is bounded on the entire interval, and so, by extension, is y ″ ( ξ ). I don't understand why this is necessarily true.

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I am studying numeric solutions of differential equations, and part of my reading is found in Simmonds&#039; book, Differential Equations with Applications and Historical Notes. Although the chapter on numerical methods is written by John S. Robertson.In the treatment of Euler&#039;s method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?For some context, the differential equations under consideration are those of the form y&#039;=f(x,y), defined on some interval [a,b], with some initial value   y  (  a  )  =  αThese are then transformed into the integral equation  y  (      x    1    )  −  y  (      x    0    )  =      ∫                  x        0                            x        1              f  (  x  ,  y  )    d  x  .Then, using Taylor&#039;s theorem and substituting back into that formula, the error is found to be            h      2        2        y    ″    (  ξ  )  ,which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y&#039;&#039;(x) is bounded on the entire interval, and so, by extension, is       y    ″    (  ξ  ). I don&#039;t understand why this is necessarily true.

stigliy0 · Accepted Answer

You need some condition to be satisfied (e.g. f(t,y) is continuous, and Lipschitz in the y variable) to be sure that the solution exists, and is continuous on the interval.Then since y&#039;=f(t,y), it follows that y&#039; exists, and is continuous on the interval.Then see that       y    ″    =            ∂      f              ∂      t        (  t  ,  y  )  +      y    ′              ∂      f              ∂      y        (  t  ,  y  ). If we know that the partial derivatives of f are continuous, we can see that y&#039;&#039; is continuous on the interval.Any continuous function on a closed interval is bounded.You can iterate this method to get higher derivatives as well if needed.

I am studying numeric solutions of differential equations, and part of my reading is found in Simmon

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