At school, or in a first-year course on DEs, we learn (perhaps in less abstract language) that if you have a linear nth-order differential equation L y = fthen the general solution is something of the form y = a 1 y 1 + . . . + a n y n + gwhere the y i are independent and satisfy L y i = 0, and g satisfies L g = f. Then we receive lots of training in how to find the y i and g.Obviously any choice of the a i will give us a solution to L y = f. But how do you know that there aren't any more solutions?We justify this by making an analogy with systems of linear equations A x = b, saying something along the lines of 'the space of solutions has the same dimension as the kernel of A'. But that works in finite dimensions - how do we know that the same is true with linear operators?

Question

At school, or in a first-year course on DEs, we learn (perhaps in less abstract language) that if you have a linear   nth-order differential equation  L  y  =  fthen the general solution is something of the form  y  =      a    1        y    1    +  .  .  .  +      a    n        y    n    +  gwhere the       y    i   are independent and satisfy   L      y    i    =  0, and   g satisfies   L  g  =  f. Then we receive lots of training in how to find the       y    i   and   g.Obviously any choice of the       a    i   will give us a solution to   L  y  =  f. But how do you know that there aren&#039;t any more solutions?We justify this by making an analogy with systems of linear equations   A  x  =  b, saying something along the lines of &#039;the space of solutions has the same dimension as the kernel of   A&#039;. But that works in finite dimensions - how do we know that the same is true with linear operators?

delalbaef · Accepted Answer

Theorem [Existence and Uniqueness]: Let      a          0        (  x  )  ,      a          1        (  x  )  ,  ⋯  ,      a          n      −      1        (  x  ) be continuous functions on   [  a  ,  b  ] and let constants       y          0        ,      y          1        ,  ⋯  ,      y          n      −      1       be given. Then there exists one and only one n-times continuously differentiable solution   y  (  x  ) on   [  a  ,  b  ] of the differential equation      y          (      n      )        (  x  )  +      a          n      −      1            y          (      n      −      1      )        (  x  )  +  ⋯  +      a          1            y          (      1      )        (  x  )  +      a          0        y  (  x  )  =  0with initial endpoint values  y  (  a  )  =      y          0        ,        y    ′    (  a  )  =      y          1        ,        y    ″    (  a  )  =      y          2        ,  ⋯  ,      y          (      n      −      1      )        (  a  )  =      y          n      −      1        .In other words, the solution of the homogenous differential equation   y is uniquely determined once you know the values of   y  (  a  )  ,      y    ′    (  a  )  ,      y          (      2      )        (  a  )  ,  ⋯  ,      y          (      n      −      1      )        (  a  ). So the following map is a linear bijection between the solution space of the homogeneous equation and             R              n      :  y  ↦      [                            y          (          a          )                                                  y                          (              1              )                                (          a          )                                      ⋮                                                  y                          (              n              −              1              )                                (          a          )                      ]  If you choose       y          j       to be the unique solution whose image under this map is the j-th standard basis vector for             R              n      , then every solution of the homogeneous equation can be written uniquely as       α          1            y          1        (  x  )  +  ⋯  +      α          n            y          n        (  x  ) for real constants       α          j       (why?). The difference of any two solutions solutions of the inhomogeneous equation is a solution of the homogeneous equation and, hence, must be such a linear combination.

At school, or in a first-year course on DEs, we learn (perhaps in less abstract language) that if yo

Answered question

Answer & Explanation

New Questions in Linear algebra