Transformation not reproducible












2












$begingroup$



Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
For
$$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
$mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.




The only advice given is to introduce the function
$$
F(X):=frac{X}{sqrt{1+|X|^2}}
$$
and use a green formula for
$$
F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
$$



Attempt:
When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get



$$
mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
= 0
= mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
$$



Here I am stuck as I don't find a suiting Green formula.






real-analysis integration pde







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
    For
    $$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
    let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
    $mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.




    The only advice given is to introduce the function
    $$
    F(X):=frac{X}{sqrt{1+|X|^2}}
    $$
    and use a green formula for
    $$
    F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
    $$



    Attempt:
    When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get



    $$
    mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
    = 0
    = mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
    $$



    Here I am stuck as I don't find a suiting Green formula.






    real-analysis integration pde







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
      For
      $$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
      let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
      $mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.




      The only advice given is to introduce the function
      $$
      F(X):=frac{X}{sqrt{1+|X|^2}}
      $$
      and use a green formula for
      $$
      F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
      $$



      Attempt:
      When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get



      $$
      mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
      = 0
      = mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
      $$



      Here I am stuck as I don't find a suiting Green formula.






      real-analysis integration pde







      share|cite|improve this question











      $endgroup$





      Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
      For
      $$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
      let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
      $mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.




      The only advice given is to introduce the function
      $$
      F(X):=frac{X}{sqrt{1+|X|^2}}
      $$
      and use a green formula for
      $$
      F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
      $$



      Attempt:
      When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get



      $$
      mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
      = 0
      = mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
      $$



      Here I am stuck as I don't find a suiting Green formula.






      real-analysis integration pde




      real-analysis integration pde






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 12 at 14:09









      Viktor Glombik

      790527




      790527










      asked Jan 12 at 14:03









      EpsilonDeltaEpsilonDelta

      6611615




      6611615






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Now you need to compute. First note that
          $$
          partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
          $$
          which in particular means that
          $$
          partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
          $$



          Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
          $$
          0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
          $$
          and we get the desired result by setting
          $$
          A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
          $$
          To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
          $$
          partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
          $$
          which we can rewrite as
          $$
          DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
          $$
          We can then plug in above to get more information about $A$.



          EDIT: As requested, here is some more information about $A$.



          It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
          $$
          DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
          $$
          where the last inequality follows from Cauchy-Schwarz. Next note that
          $$
          |t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
          $$
          which means that
          $$
          (1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
          $$
          Then
          $$
          A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
          $$
          and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
            $endgroup$
            – EpsilonDelta
            Jan 12 at 16:30












          • $begingroup$
            @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
            $endgroup$
            – Glitch
            Jan 12 at 17:15











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          1












          $begingroup$

          Now you need to compute. First note that
          $$
          partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
          $$
          which in particular means that
          $$
          partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
          $$



          Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
          $$
          0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
          $$
          and we get the desired result by setting
          $$
          A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
          $$
          To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
          $$
          partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
          $$
          which we can rewrite as
          $$
          DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
          $$
          We can then plug in above to get more information about $A$.



          EDIT: As requested, here is some more information about $A$.



          It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
          $$
          DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
          $$
          where the last inequality follows from Cauchy-Schwarz. Next note that
          $$
          |t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
          $$
          which means that
          $$
          (1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
          $$
          Then
          $$
          A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
          $$
          and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
            $endgroup$
            – EpsilonDelta
            Jan 12 at 16:30












          • $begingroup$
            @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
            $endgroup$
            – Glitch
            Jan 12 at 17:15
















          1












          $begingroup$

          Now you need to compute. First note that
          $$
          partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
          $$
          which in particular means that
          $$
          partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
          $$



          Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
          $$
          0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
          $$
          and we get the desired result by setting
          $$
          A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
          $$
          To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
          $$
          partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
          $$
          which we can rewrite as
          $$
          DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
          $$
          We can then plug in above to get more information about $A$.



          EDIT: As requested, here is some more information about $A$.



          It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
          $$
          DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
          $$
          where the last inequality follows from Cauchy-Schwarz. Next note that
          $$
          |t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
          $$
          which means that
          $$
          (1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
          $$
          Then
          $$
          A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
          $$
          and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
            $endgroup$
            – EpsilonDelta
            Jan 12 at 16:30












          • $begingroup$
            @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
            $endgroup$
            – Glitch
            Jan 12 at 17:15














          1












          1








          1





          $begingroup$

          Now you need to compute. First note that
          $$
          partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
          $$
          which in particular means that
          $$
          partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
          $$



          Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
          $$
          0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
          $$
          and we get the desired result by setting
          $$
          A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
          $$
          To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
          $$
          partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
          $$
          which we can rewrite as
          $$
          DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
          $$
          We can then plug in above to get more information about $A$.



          EDIT: As requested, here is some more information about $A$.



          It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
          $$
          DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
          $$
          where the last inequality follows from Cauchy-Schwarz. Next note that
          $$
          |t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
          $$
          which means that
          $$
          (1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
          $$
          Then
          $$
          A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
          $$
          and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.






          share|cite|improve this answer











          $endgroup$



          Now you need to compute. First note that
          $$
          partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
          $$
          which in particular means that
          $$
          partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
          $$



          Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
          $$
          0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
          $$
          and we get the desired result by setting
          $$
          A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
          $$
          To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
          $$
          partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
          $$
          which we can rewrite as
          $$
          DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
          $$
          We can then plug in above to get more information about $A$.



          EDIT: As requested, here is some more information about $A$.



          It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
          $$
          DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
          $$
          where the last inequality follows from Cauchy-Schwarz. Next note that
          $$
          |t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
          $$
          which means that
          $$
          (1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
          $$
          Then
          $$
          A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
          $$
          and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 17:15

























          answered Jan 12 at 15:40









          GlitchGlitch

          5,5881030




          5,5881030












          • $begingroup$
            Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
            $endgroup$
            – EpsilonDelta
            Jan 12 at 16:30












          • $begingroup$
            @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
            $endgroup$
            – Glitch
            Jan 12 at 17:15


















          • $begingroup$
            Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
            $endgroup$
            – EpsilonDelta
            Jan 12 at 16:30












          • $begingroup$
            @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
            $endgroup$
            – Glitch
            Jan 12 at 17:15
















          $begingroup$
          Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
          $endgroup$
          – EpsilonDelta
          Jan 12 at 16:30






          $begingroup$
          Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
          $endgroup$
          – EpsilonDelta
          Jan 12 at 16:30














          $begingroup$
          @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
          $endgroup$
          – Glitch
          Jan 12 at 17:15




          $begingroup$
          @EpsilonDelta I'm glad you like it. See the edit for ellipticity.
          $endgroup$
          – Glitch
          Jan 12 at 17:15


















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