f : R 2 → R is continuous and f ( t , x ) &lt; 0 holds for t x &gt; 0 and f ( t , x ) &gt; 0 holds for t x &lt; 0I want to show, that x ( t ) = 0 is the only solution of x ′ = f ( t , x )You can say, that there is an solution, because f is continuous. By the intermediate value theorem exists ξ ∈ [ − a , a ], such that f ( t , ξ ) = 0. I only have to show, that x ( t ) = 0 is unique. How can I do that?

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f  :            R      2        →      R   is continuous and   f  (  t  ,  x  )  &amp;lt;  0 holds for   t  x  &amp;gt;  0 and   f  (  t  ,  x  )  &amp;gt;  0 holds for   t  x  &amp;lt;  0I want to show, that   x  (  t  )  =  0 is the only solution of       x    ′    =  f  (  t  ,  x  )You can say, that there is an solution, because f is continuous. By the intermediate value theorem exists   ξ  ∈  [  −  a  ,  a  ], such that   f  (  t  ,  ξ  )  =  0. I only have to show, that   x  (  t  )  =  0 is unique. How can I do that?

mar1nerne · Accepted Answer

As you noticed,   x  (  t  )  =  0 is one solution. If       x    1    (  t  ) is another solution, then for   y  (  t  )  =  x  (  t  )  −      x    1    (  t  ) we have       y    ′    (  t  )  =  0, so   y  (  t  )  =  C. Since   x  (  0  )  =      x    1    (  0  )  =  0, then   C  =  y  (  0  )  =  x  (  0  )  −      x    1    (  0  )  =  0, so   y  (  t  )  =  0 and       x    1    (  t  )  =  x  (  t  )  =  0.

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