Consider the wave equation with initial data: u t t ( t , x ) + u x x ( t , x ) = 0 u ( 0 , x ) = u 0 ( x ) u t ( 0 , x ) = u 1 ( x )Hadamard showed that this problem is ill-posed: there exist large solutions with arbitrarily small initial data. For instance, if we take u ( t , x ) = a ω sinh ⁡ ( ω t ) sin ⁡ ( ω x ), then u 0 ( x ) = 0 and u 1 ( x ) = a ω ω sin ⁡ ( ω x ), then we can make u(t,x) grow arbitrarily fast while keeping u 0 and u 1 small.Tweaking this construction, it is not hard to see that for any given k and ϵ &gt; 0, we can construct initial data such that | | u 0 | | ∞ + | | u 0 ( 1 ) | | ∞ + … + | | u 0 ( k ) | | ∞ + | | u 1 | | ∞ + | | u 1 ( 1 ) | | ∞ + … + | | u 1 ( k ) | | ∞ &lt; ϵand | | u ( ϵ , x ) | | ∞ &gt; 1 ϵ This can be interpreted as saying that the problem is ill-posed even in a Holder sense.My question is: can one construct an example of a solution u(t,x) with initial data u 0 ( x ) and u 1 ( x ) such that ∑ i = 0 ∞ | | u 0 ( i ) | | ∞ + | | u 1 ( i ) | | ∞ &lt; ϵ

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Consider the wave equation with initial data:      u          t      t        (  t  ,  x  )  +      u          x      x        (  t  ,  x  )  =  0  u  (  0  ,  x  )  =      u    0    (  x  )      u    t    (  0  ,  x  )  =      u    1    (  x  )Hadamard showed that this problem is ill-posed: there exist large solutions with arbitrarily small initial data. For instance, if we take   u  (  t  ,  x  )  =      a          ω        sinh  ⁡  (  ω  t  )  sin  ⁡  (  ω  x  ), then       u    0    (  x  )  =  0 and       u    1    (  x  )  =      a          ω        ω  sin  ⁡  (  ω  x  ), then we can make u(t,x) grow arbitrarily fast while keeping       u    0   and       u    1   small.Tweaking this construction, it is not hard to see that for any given k and   ϵ  &amp;gt;  0, we can construct initial data such that      |        |        u    0        |              |              ∞        +      |        |        u    0          (      1      )            |              |              ∞        +  …  +      |        |        u    0          (      k      )            |              |              ∞        +      |        |        u    1        |              |              ∞        +      |        |        u    1          (      1      )            |              |              ∞        +  …  +      |        |        u    1          (      k      )            |              |              ∞        &amp;lt;  ϵand       |        |    u  (  ϵ  ,  x  )      |              |              ∞        &amp;gt;      1    ϵ  This can be interpreted as saying that the problem is ill-posed even in a Holder sense.My question is: can one construct an example of a solution u(t,x) with initial data       u    0    (  x  ) and       u    1    (  x  ) such that      ∑          i      =      0              ∞            |        |        u    0          (      i      )            |              |              ∞        +      |        |        u    1          (      i      )            |              |              ∞        &amp;lt;  ϵ

Zayden Wiley · Accepted Answer

I am somewhat doubtful of the claim. Here&#039;s why. Let us suppose the weaker inequality that for any k      |        ∂    x    k        u    0        |    &amp;lt;  ϵ  ,      |        ∂    x    k        u    1        |    &amp;lt;  ϵ  ..Now, let us consider a solution u to the Laplace equation with indicated boundary values. We have that along t=0      |        ∂    x    k    u      |    &amp;lt;  ϵ  ,      |        ∂    x    k        ∂    t    u      |    &amp;lt;  ϵ  .Using the equation we have      |        ∂    x    k        ∂    t    2    u      |    =      |        ∂    x          k      +      2        u      |    &amp;lt;  ϵ  ..And by induction      |        ∂    x    k        ∂    t          2      l        u      |    =      |        ∂    x          k      +      2      l        u      |    &amp;lt;  ϵ  ..and      |        ∂    x    k        ∂    t          2      l      +      1        u      |    =      |        ∂    x          k      +      2      l            ∂    t    u      |    &amp;lt;  ϵ  .This implies that every single derivative, spatial or temporal is bounded by ϵ. The practical implication of this is: let      a          k      l        =      ∂    x    k        ∂    t    l    u  (  0  ,  0  )The formal power series  U  (  t  ,  x  )  =      ∑          k      ,      l      =      0              ∞            1          k      !      l      !            a          k      l            x    k        t    l  has infinite radius of convergence; the function that it defines is a solution, so we can identify U with u.Now, if you look at   t  =  ϵ  ,  x  =  0 of the power series, you see that you have the sum  U  (  ϵ  ,  0  )  =      ∑          l      =      0              ∞            1          l      !            a          0      l            ϵ    l  which we can bound pretty trivially by  (      sup    l        |        a          0      l            |    )  exp  ⁡  (  ϵ  )  &amp;lt;  2  ϵThe above argument is clearly independent of where we do the Taylor series: instead of taking the series relative to the origin, we do it relative to   (  0  ,      x    0    ) and this tells us that the desired inequality is very far from being true.The above is basically a quantitative version of the theorem of Cauchy-Kowalevski, and which can be easily extended to initial data in Gevrey classes; we are using here that if a smooth function is such that all its derivatives are bounded by a fixed constant, the function must be real analytic.

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