Solvability of Neumann problemAnalytical solution of Partial Differential EquationSmooth solutions of the Laplace equation with Neumann dataDoes the following have a solution for f(x,y)?How to show that a delta function solution to a PDE (Fokker-Planck) is really a solutionReformulate this PDE in different notationPoisson equation with nonlinear Neumann conditionsMethod of characteristics for a parabolic PDEDiscrete Neumann problem: Representation by Simple Random WalkDirichlet plus Neumann problem without a PDEUniqueness of solution for Dirichlet and Neumann problem

Multi tool use
Multi tool use

Single word request: Harming the benefactor

Extra alignment tab has been changed to cr. } using table, tabular and resizebox

Are the terms "stab" and "staccato" synonyms?

Should I tell my boss the work he did was worthless

Could you please stop shuffling the deck and play already?

Space in array system equations

Word for a person who has no opinion about whether god exists

Why is this plane circling around the Lucknow airport every day?

Are babies of evil humanoid species inherently evil?

Should QA ask requirements to developers?

If the Captain's screens are out, does he switch seats with the co-pilot?

Things to avoid when using voltage regulators?

How could our ancestors have domesticated a solitary predator?

infinitive telling the purpose

Do items de-spawn in Diablo?

Is "history" a male-biased word ("his+story")?

How much stiffer are 23c tires over 28c?

What does a stand alone "T" index value do?

How are such low op-amp input currents possible?

"One can do his homework in the library"

Good allowance savings plan?

How did Alan Turing break the enigma code using the hint given by the lady in the bar?

Who deserves to be first and second author? PhD student who collected data, research associate who wrote the paper or supervisor?

Do Bugbears' arms literally get longer when it's their turn?



Solvability of Neumann problem


Analytical solution of Partial Differential EquationSmooth solutions of the Laplace equation with Neumann dataDoes the following have a solution for f(x,y)?How to show that a delta function solution to a PDE (Fokker-Planck) is really a solutionReformulate this PDE in different notationPoisson equation with nonlinear Neumann conditionsMethod of characteristics for a parabolic PDEDiscrete Neumann problem: Representation by Simple Random WalkDirichlet plus Neumann problem without a PDEUniqueness of solution for Dirichlet and Neumann problem













1












$begingroup$


Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
$$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
How to solve the equation for the values of these three constants which the solution exists?



Thanks.










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
    $$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
    How to solve the equation for the values of these three constants which the solution exists?



    Thanks.










    share|cite|improve this question









    $endgroup$














      1












      1








      1


      1



      $begingroup$


      Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
      $$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
      How to solve the equation for the values of these three constants which the solution exists?



      Thanks.










      share|cite|improve this question









      $endgroup$




      Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
      $$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
      How to solve the equation for the values of these three constants which the solution exists?



      Thanks.







      pde






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      Math LoverMath Lover

      938




      938




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is



          $$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$



          where $D = 0<r < 2$. The gradient is



          $$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$



          Evaluated at the boundary:



          $$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$



          You also have, from the general solution



          $$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$



          Comparing coefficients gives



          begincases
          2alpha + gamma = 0 \
          B_1 = 2beta \
          4A_2 = 2alpha
          endcases



          The first condition is required for the solution to exist.






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.



            Edit The proposition above is






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              What is $f$? ---
              $endgroup$
              – Dylan
              2 days ago










            • $begingroup$
              @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
              $endgroup$
              – dmtri
              2 days ago










            • $begingroup$
              It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
              $endgroup$
              – Dylan
              2 days ago










            • $begingroup$
              Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
              $endgroup$
              – dmtri
              2 days ago










            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142223%2fsolvability-of-neumann-problem%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is



            $$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$



            where $D = 0<r < 2$. The gradient is



            $$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$



            Evaluated at the boundary:



            $$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$



            You also have, from the general solution



            $$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$



            Comparing coefficients gives



            begincases
            2alpha + gamma = 0 \
            B_1 = 2beta \
            4A_2 = 2alpha
            endcases



            The first condition is required for the solution to exist.






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is



              $$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$



              where $D = 0<r < 2$. The gradient is



              $$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$



              Evaluated at the boundary:



              $$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$



              You also have, from the general solution



              $$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$



              Comparing coefficients gives



              begincases
              2alpha + gamma = 0 \
              B_1 = 2beta \
              4A_2 = 2alpha
              endcases



              The first condition is required for the solution to exist.






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is



                $$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$



                where $D = 0<r < 2$. The gradient is



                $$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$



                Evaluated at the boundary:



                $$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$



                You also have, from the general solution



                $$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$



                Comparing coefficients gives



                begincases
                2alpha + gamma = 0 \
                B_1 = 2beta \
                4A_2 = 2alpha
                endcases



                The first condition is required for the solution to exist.






                share|cite|improve this answer









                $endgroup$



                In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is



                $$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$



                where $D = 0<r < 2$. The gradient is



                $$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$



                Evaluated at the boundary:



                $$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$



                You also have, from the general solution



                $$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$



                Comparing coefficients gives



                begincases
                2alpha + gamma = 0 \
                B_1 = 2beta \
                4A_2 = 2alpha
                endcases



                The first condition is required for the solution to exist.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                DylanDylan

                13.8k31027




                13.8k31027





















                    0












                    $begingroup$

                    The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.



                    Edit The proposition above is






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      What is $f$? ---
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
                      $endgroup$
                      – dmtri
                      2 days ago










                    • $begingroup$
                      It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
                      $endgroup$
                      – dmtri
                      2 days ago















                    0












                    $begingroup$

                    The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.



                    Edit The proposition above is






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      What is $f$? ---
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
                      $endgroup$
                      – dmtri
                      2 days ago










                    • $begingroup$
                      It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
                      $endgroup$
                      – dmtri
                      2 days ago













                    0












                    0








                    0





                    $begingroup$

                    The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.



                    Edit The proposition above is






                    share|cite|improve this answer











                    $endgroup$



                    The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.



                    Edit The proposition above is







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    dmtridmtri

                    1,6082521




                    1,6082521











                    • $begingroup$
                      What is $f$? ---
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
                      $endgroup$
                      – dmtri
                      2 days ago










                    • $begingroup$
                      It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
                      $endgroup$
                      – dmtri
                      2 days ago
















                    • $begingroup$
                      What is $f$? ---
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
                      $endgroup$
                      – dmtri
                      2 days ago










                    • $begingroup$
                      It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
                      $endgroup$
                      – Dylan
                      2 days ago










                    • $begingroup$
                      Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
                      $endgroup$
                      – dmtri
                      2 days ago















                    $begingroup$
                    What is $f$? ---
                    $endgroup$
                    – Dylan
                    2 days ago




                    $begingroup$
                    What is $f$? ---
                    $endgroup$
                    – Dylan
                    2 days ago












                    $begingroup$
                    @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
                    $endgroup$
                    – dmtri
                    2 days ago




                    $begingroup$
                    @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
                    $endgroup$
                    – dmtri
                    2 days ago












                    $begingroup$
                    It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
                    $endgroup$
                    – Dylan
                    2 days ago




                    $begingroup$
                    It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
                    $endgroup$
                    – Dylan
                    2 days ago












                    $begingroup$
                    Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
                    $endgroup$
                    – dmtri
                    2 days ago




                    $begingroup$
                    Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
                    $endgroup$
                    – dmtri
                    2 days ago

















                    draft saved

                    draft discarded
















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142223%2fsolvability-of-neumann-problem%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Bk7ds6dpg7G9,wpNTa,zLdw FsdVXvAxMXIzSeVDU,GSPy du,2v 37hdznheDPeNn
                    cYL57BCDZD1xHbxspjVKKlHN,BeoVL,o6W YpxQor3WdWRm XOwwNn18GKdbQ iZzHxvF,Am5

                    Popular posts from this blog

                    Football at the 1986 Brunei Merdeka Games Contents Teams Group stage Knockout stage References Navigation menu"Brunei Merdeka Games 1986".

                    Solar Wings Breeze Design and development Specifications (Breeze) References Navigation menu1368-485X"Hang glider: Breeze (Solar Wings)"e

                    Kathakali Contents Etymology and nomenclature History Repertoire Songs and musical instruments Traditional plays Styles: Sampradayam Training centers and awards Relationship to other dance forms See also Notes References External links Navigation menueThe Illustrated Encyclopedia of Hinduism: A-MSouth Asian Folklore: An EncyclopediaRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to Play10.1353/atj.2005.0004The Illustrated Encyclopedia of Hinduism: A-MEncyclopedia of HinduismKathakali Dance-drama: Where Gods and Demons Come to PlaySonic Liturgy: Ritual and Music in Hindu Tradition"The Mirror of Gesture"Kathakali Dance-drama: Where Gods and Demons Come to Play"Kathakali"Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceMedieval Indian Literature: An AnthologyThe Oxford Companion to Indian TheatreSouth Asian Folklore: An Encyclopedia : Afghanistan, Bangladesh, India, Nepal, Pakistan, Sri LankaThe Rise of Performance Studies: Rethinking Richard Schechner's Broad SpectrumIndian Theatre: Traditions of PerformanceModern Asian Theatre and Performance 1900-2000Critical Theory and PerformanceBetween Theater and AnthropologyKathakali603847011Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceBetween Theater and AnthropologyBetween Theater and AnthropologyNambeesan Smaraka AwardsArchivedThe Cambridge Guide to TheatreRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeThe Garland Encyclopedia of World Music: South Asia : the Indian subcontinentThe Ethos of Noh: Actors and Their Art10.2307/1145740By Means of Performance: Intercultural Studies of Theatre and Ritual10.1017/s204912550000100xReconceiving the Renaissance: A Critical ReaderPerformance TheoryListening to Theatre: The Aural Dimension of Beijing Opera10.2307/1146013Kathakali: The Art of the Non-WorldlyOn KathakaliKathakali, the dance theatreThe Kathakali Complex: Performance & StructureKathakali Dance-Drama: Where Gods and Demons Come to Play10.1093/obo/9780195399318-0071Drama and Ritual of Early Hinduism"In the Shadow of Hollywood Orientalism: Authentic East Indian Dancing"10.1080/08949460490274013Sanskrit Play Production in Ancient IndiaIndian Music: History and StructureBharata, the Nāṭyaśāstra233639306Table of Contents2238067286469807Dance In Indian Painting10.2307/32047833204783Kathakali Dance-Theatre: A Visual Narrative of Sacred Indian MimeIndian Classical Dance: The Renaissance and BeyondKathakali: an indigenous art-form of Keralaeee