Solution to the parabolic cylinder equation
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
asked Jan 3 at 19:37
Galkina
412
add a comment |
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
asked Jan 3 at 19:37
Galkina
412
add a comment |
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
asked Jan 3 at 19:37
Galkina
412
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
differential-equations special-functions
asked Jan 3 at 19:37
Galkina
412
asked Jan 3 at 19:37
Galkina
412
asked Jan 3 at 19:37
Galkina
412
asked Jan 3 at 19:37
Galkina
412
asked Jan 3 at 19:37
Galkina
412
412
add a comment |
add a comment |
1 Answer
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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 $C_2neq C_6$.
answered yesterday
JJacquelin
42.7k21750
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1 Answer
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1 Answer
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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 $C_2neq C_6$.
answered yesterday
JJacquelin
42.7k21750
add a comment |
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 $C_2neq C_6$.
answered yesterday
JJacquelin
42.7k21750
add a comment |
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 $C_2neq C_6$.
answered yesterday
JJacquelin
42.7k21750
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 $C_2neq C_6$.
answered yesterday
JJacquelin
42.7k21750
edited yesterday
answered yesterday
JJacquelin
42.7k21750
answered yesterday
JJacquelin
42.7k21750
answered yesterday
JJacquelin
42.7k21750
42.7k21750
add a comment |
add a comment |
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