Solution to the parabolic cylinder equation












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In the Gradshteyn & Ryzhik (7th ed.) the differential equation (9.255) leading to parabolic cylinder functions is $$frac{d^2u}{dz^2}+(p+frac{1}{2}-frac{z^2}{4})u=0.$$ The solutions are $u=D_p(z),D_p(-z),D_{-p-1}(iz),D_{-p-1}(-iz)$, where $D_p(z)$ is the parabolic cylinder function. These four solutions are linearly dependent. My question is why is there four solutions to the second order ODE? In my case $p$ is complex, and Mathematica gives solution in the form $C_1 D_p(z)+C_2 D_{-p-1}(iz)$.










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    In the Gradshteyn & Ryzhik (7th ed.) the differential equation (9.255) leading to parabolic cylinder functions is $$frac{d^2u}{dz^2}+(p+frac{1}{2}-frac{z^2}{4})u=0.$$ The solutions are $u=D_p(z),D_p(-z),D_{-p-1}(iz),D_{-p-1}(-iz)$, where $D_p(z)$ is the parabolic cylinder function. These four solutions are linearly dependent. My question is why is there four solutions to the second order ODE? In my case $p$ is complex, and Mathematica gives solution in the form $C_1 D_p(z)+C_2 D_{-p-1}(iz)$.










    share|cite|improve this question

























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      In the Gradshteyn & Ryzhik (7th ed.) the differential equation (9.255) leading to parabolic cylinder functions is $$frac{d^2u}{dz^2}+(p+frac{1}{2}-frac{z^2}{4})u=0.$$ The solutions are $u=D_p(z),D_p(-z),D_{-p-1}(iz),D_{-p-1}(-iz)$, where $D_p(z)$ is the parabolic cylinder function. These four solutions are linearly dependent. My question is why is there four solutions to the second order ODE? In my case $p$ is complex, and Mathematica gives solution in the form $C_1 D_p(z)+C_2 D_{-p-1}(iz)$.










      share|cite|improve this question













      In the Gradshteyn & Ryzhik (7th ed.) the differential equation (9.255) leading to parabolic cylinder functions is $$frac{d^2u}{dz^2}+(p+frac{1}{2}-frac{z^2}{4})u=0.$$ The solutions are $u=D_p(z),D_p(-z),D_{-p-1}(iz),D_{-p-1}(-iz)$, where $D_p(z)$ is the parabolic cylinder function. These four solutions are linearly dependent. My question is why is there four solutions to the second order ODE? In my case $p$ is complex, and Mathematica gives solution in the form $C_1 D_p(z)+C_2 D_{-p-1}(iz)$.







      differential-equations special-functions






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      Galkina

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          This second order ODE has not four solutions as you wrote, but has an infinity of solutions.



          Don't write



          << The solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1}(iz),u_4=D_{-p-1}(-iz)$ >> ,



          better write



          << Some solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1 }(iz),u_4=D_{-p-1}(-iz)$ >>.



          For example $u_5=3u_1-7u_3$ is also a solution of the ODE. You will not say "Why is there five solutions to the second order ODE ? ".



          In fact, among the infinity of solutions we can only find COUPLES of linearly INDEPENDENT solutions. Not TRIPLET.



          All solutions can be defined on the form of a linear combination of any couple of independent solution.



          Thus the general solution expressed on the form $$u(z)=C_1 D_p(z)+C_2 D_{-p-1}(iz)$$ is equivalent to $$u(z)=C_3 D_p(-z)+C_4 D_{-p-1}(iz)$$ or equivalent to $$u(z)=C_5 D_p(-z)+C_6 D_{-p-1}(-iz)$$
          Etc.



          Of course the coefficients are generally not the same: For example $C_2neq C_6$.






          share|cite|improve this answer























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            1 Answer
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            1 Answer
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            active

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            active

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            active

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            1














            This second order ODE has not four solutions as you wrote, but has an infinity of solutions.



            Don't write



            << The solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1}(iz),u_4=D_{-p-1}(-iz)$ >> ,



            better write



            << Some solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1 }(iz),u_4=D_{-p-1}(-iz)$ >>.



            For example $u_5=3u_1-7u_3$ is also a solution of the ODE. You will not say "Why is there five solutions to the second order ODE ? ".



            In fact, among the infinity of solutions we can only find COUPLES of linearly INDEPENDENT solutions. Not TRIPLET.



            All solutions can be defined on the form of a linear combination of any couple of independent solution.



            Thus the general solution expressed on the form $$u(z)=C_1 D_p(z)+C_2 D_{-p-1}(iz)$$ is equivalent to $$u(z)=C_3 D_p(-z)+C_4 D_{-p-1}(iz)$$ or equivalent to $$u(z)=C_5 D_p(-z)+C_6 D_{-p-1}(-iz)$$
            Etc.



            Of course the coefficients are generally not the same: For example $C_2neq C_6$.






            share|cite|improve this answer




























              1














              This second order ODE has not four solutions as you wrote, but has an infinity of solutions.



              Don't write



              << The solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1}(iz),u_4=D_{-p-1}(-iz)$ >> ,



              better write



              << Some solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1 }(iz),u_4=D_{-p-1}(-iz)$ >>.



              For example $u_5=3u_1-7u_3$ is also a solution of the ODE. You will not say "Why is there five solutions to the second order ODE ? ".



              In fact, among the infinity of solutions we can only find COUPLES of linearly INDEPENDENT solutions. Not TRIPLET.



              All solutions can be defined on the form of a linear combination of any couple of independent solution.



              Thus the general solution expressed on the form $$u(z)=C_1 D_p(z)+C_2 D_{-p-1}(iz)$$ is equivalent to $$u(z)=C_3 D_p(-z)+C_4 D_{-p-1}(iz)$$ or equivalent to $$u(z)=C_5 D_p(-z)+C_6 D_{-p-1}(-iz)$$
              Etc.



              Of course the coefficients are generally not the same: For example $C_2neq C_6$.






              share|cite|improve this answer


























                1












                1








                1






                This second order ODE has not four solutions as you wrote, but has an infinity of solutions.



                Don't write



                << The solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1}(iz),u_4=D_{-p-1}(-iz)$ >> ,



                better write



                << Some solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1 }(iz),u_4=D_{-p-1}(-iz)$ >>.



                For example $u_5=3u_1-7u_3$ is also a solution of the ODE. You will not say "Why is there five solutions to the second order ODE ? ".



                In fact, among the infinity of solutions we can only find COUPLES of linearly INDEPENDENT solutions. Not TRIPLET.



                All solutions can be defined on the form of a linear combination of any couple of independent solution.



                Thus the general solution expressed on the form $$u(z)=C_1 D_p(z)+C_2 D_{-p-1}(iz)$$ is equivalent to $$u(z)=C_3 D_p(-z)+C_4 D_{-p-1}(iz)$$ or equivalent to $$u(z)=C_5 D_p(-z)+C_6 D_{-p-1}(-iz)$$
                Etc.



                Of course the coefficients are generally not the same: For example $C_2neq C_6$.






                share|cite|improve this answer














                This second order ODE has not four solutions as you wrote, but has an infinity of solutions.



                Don't write



                << The solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1}(iz),u_4=D_{-p-1}(-iz)$ >> ,



                better write



                << Some solutions are $u_1=D_p(z),u_2=D_p(-z),u_3=D_{-p-1 }(iz),u_4=D_{-p-1}(-iz)$ >>.



                For example $u_5=3u_1-7u_3$ is also a solution of the ODE. You will not say "Why is there five solutions to the second order ODE ? ".



                In fact, among the infinity of solutions we can only find COUPLES of linearly INDEPENDENT solutions. Not TRIPLET.



                All solutions can be defined on the form of a linear combination of any couple of independent solution.



                Thus the general solution expressed on the form $$u(z)=C_1 D_p(z)+C_2 D_{-p-1}(iz)$$ is equivalent to $$u(z)=C_3 D_p(-z)+C_4 D_{-p-1}(iz)$$ or equivalent to $$u(z)=C_5 D_p(-z)+C_6 D_{-p-1}(-iz)$$
                Etc.



                Of course the coefficients are generally not the same: For example $C_2neq C_6$.







                share|cite|improve this answer














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                edited yesterday

























                answered yesterday









                JJacquelin

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                42.6k21750






























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