1 | initial version |

Hello, @Grim! I am not quite sure about Schrödinger's Equation, but I know how to obtain $k=\frac{n\pi}{a}$. I believe this will result a helpful version of your code:

```
k, a = var('k, a', domain='positive')
psi = function('psi')(x)
eqs = diff(psi, x, 2) + k^(2)*psi == 0
EQS = desolve(eqs, psi, ivar=x)
A, B = var('A, B')
EQS = EQS.subs(_K1=B, _K2=A)
show(LatexExpr(r'\psi(x) = '), EQS)
eq1 = EQS.subs(x=0) == 0 # this is psi(0) = 0
eq2 = EQS.subs(x=a) == 0 # this is psi(a) = 0
show(eq1)
show(eq2)
```

The first line defines the symbolic variables `k`

and `a`

to be positive, so you can avoid using the `assume`

command. The second line defines `psi`

as the unknown function; the third line defines the ordinary differential equation; and the fourth one solves the differential equation.

Now, things get a little complicated. If you write `print(EQS)`

, you will notice that there are a couple of new variables called `_K1`

and `_K2`

. They are integration variables (byproducts of the resolution of the ODE). However, they are not symbolic variables that you have defined, so you won't be able to use them outside of `EQS`

(for example, to use the `solve`

command or any other). So, I replaced them with `A`

and `B`

in the sixth line. The seventh line (the first `show`

command) is there only to check that I have obtained the same definition of $psi$ as you in your code (it will show you $psi(x)=A\cos(kx)+B\sin(kx)$). Then, `eq1`

and `eq2`

represent your boundary conditions $psi(0)=0$ and $psi(a)=0$, respectively.

At this point, there are many ways in which you can obtain the values of $A$, $B$ and $k$. The first method is manually solving equations `eq1`

and `eq2`

, which are shown by the last two lines of my code. You will see that

$$A=0$$ and $$A\cos(ak)+B\sin(ak)=0$$

If you replace the first equation in the second, you obtain $B\sin(ax)=0$. Now, since $B>0$ (according to your code), you must have that $\sin(ak)=0$, which implies that $k=\frac{n\pi}{a}$. Finally, in this case, $B$ can be any positive number (regardless of its value, your ODE and boundary conditions are satisfied.)

The second way to solve this is using Sage itself. You have two equations, namely `eq1`

and `eq2`

, so you will need two variables. You can write

```
sol = solve([eq1,eq2], A,k)
show(sol)
```

just after the previous code. You will get something like

$$\left[\left[A = 0, k = \frac{pi + 2\pi z1879}{a}\right], \left[A = 0, k = \frac{2pi z1900}{a}\right]\right]$$

The variables `z1879`

and `z1900`

are similar to the `_K1`

and `_K2`

that I mentioned above. More precisely, they are integer parameters; just mentally replace them with $m$ and $n$. Then you have two possible solutions. After further analysis, you will notice that these two solutions can be written more simply as $A=0$ and $k=\frac{n\pi}{a}$. Once again, $B$ has no restrictions, besides being positive.

You can also use the `Sympy`

package to solve these equations:

```
sol = solve([eq1,eq2], A,k, algorithm='sympy')
show(sol)
```

You will get

$$\left[\left{A : 0, k : 0\right}, \left{A : 0, k : \frac{pi}{a}\right}\right]$$

These are just two particular solutions that you can easily generalize using reduction formulas and properties of the trigonometric functions.

What happens if you use `solve([eq1, eq2], A, B)`

or `solve([eq1, eq2], B, k)`

? You will get $[[A=0, B=0]]$ and $[]$, respectively. The first is a correct solution to both equations, but not one that you are looking for. The second is Sage telling you that could not find a solution. If you use `Sympy`

, you will get $\left[\left{B: -\frac{A}{\tan(ak)}\right}\right]$, which is a way of admission of defeat, but it is more useful, because it hints you that it is Sage's (actually Sympy's) fault, while the $[]$ before could lead you to think that there is no solution.

I would like to point out a few interesting teachings derived from your question:

- Do not depend on any software to solve a problem; it's just a tool.
- SageMath is excellent in solving problems, but it is not perfect. Try different approaches (e.g., as we did when we used Sympy), especially if the problem is complex.
- You can use a software to help you, but the more of the resolution you can do manually (using ingenuity), the more sure you will be of your result. In the end, if you can do it manually, why bother using Sage? In the end, there are things that are easy for a human mind, but very difficult for a software.

2 | No.2 Revision |

Hello, @Grim! I am not quite sure about Schrödinger's Equation, but I know how to obtain $k=\frac{n\pi}{a}$. I believe this will result a helpful version of your code:

```
k, a = var('k, a', domain='positive')
psi = function('psi')(x)
eqs = diff(psi, x, 2) + k^(2)*psi == 0
EQS = desolve(eqs, psi, ivar=x)
A, B = var('A, B')
EQS = EQS.subs(_K1=B, _K2=A)
show(LatexExpr(r'\psi(x) = '), EQS)
eq1 = EQS.subs(x=0) == 0 # this is psi(0) = 0
eq2 = EQS.subs(x=a) == 0 # this is psi(a) = 0
show(eq1)
show(eq2)
```

The first line defines the symbolic variables `k`

and `a`

to be positive, so you can avoid using the `assume`

command. The second line defines `psi`

as the unknown function; the third line defines the ordinary differential equation; and the fourth one solves the differential equation.

Now, things get a little complicated. If you write `print(EQS)`

, you will notice that there are a couple of new variables called `_K1`

and `_K2`

. They are integration variables (byproducts of the resolution of the ODE). However, they are not symbolic variables that you have defined, so you won't be able to use them outside of `EQS`

(for example, to use the `solve`

command or any other). So, I replaced them with `A`

and `B`

in the sixth line. The seventh line (the first `show`

command) is there only to check that I have obtained the same definition of $psi$ as you in your code (it will show you $psi(x)=A\cos(kx)+B\sin(kx)$). Then, `eq1`

and `eq2`

represent your boundary conditions $psi(0)=0$ and $psi(a)=0$, respectively.

At this point, there are many ways in which you can obtain the values of $A$, $B$ and $k$. The first method is manually solving equations `eq1`

and `eq2`

, which are shown by the last two lines of my code. You will see that

$$A=0$$ and $$A\cos(ak)+B\sin(ak)=0$$

If you replace the first equation in the second, you obtain $B\sin(ax)=0$. Now, since $B>0$ (according to your code), you must have that $\sin(ak)=0$, which implies that $k=\frac{n\pi}{a}$. Finally, in this case, $B$ can be any positive number (regardless of its value, your ODE and boundary conditions are satisfied.)

The second way to solve this is using Sage itself. You have two equations, namely `eq1`

and `eq2`

, so you will need two variables. You can write

```
sol = solve([eq1,eq2], A,k)
show(sol)
```

just after the previous code. You will get something like

$$\left[\left[A = 0, k = \frac{pi + 2\pi z1879}{a}\right], \left[A = 0, k = \frac{2pi z1900}{a}\right]\right]$$

The variables `z1879`

and `z1900`

are similar to the `_K1`

and `_K2`

that I mentioned above. More precisely, they are integer parameters; just mentally replace them with $m$ and $n$. Then you have two possible solutions. After further analysis, you will notice that these two solutions can be written more simply as $A=0$ and $k=\frac{n\pi}{a}$. Once again, $B$ has no restrictions, besides being positive.

You can also use the Sympy package to solve these equations:`Sympy`

```
sol = solve([eq1,eq2], A,k, algorithm='sympy')
show(sol)
```

You will get

$$\left[\left{A : 0, k : 0\right}, \left{A : 0, k : ~~\frac{pi}{a}\right}\right]$$~~\frac{\pi}{a}\right}\right]$$

These are just two particular solutions that you can easily generalize using reduction formulas and properties of the trigonometric functions.

What happens if you use `solve([eq1, eq2], A, B)`

or `solve([eq1, eq2], B, k)`

? You will get $[[A=0, B=0]]$ and $[]$, respectively. The first is a correct solution to both equations, but not one that you are looking for. The second is Sage telling you that could not find a solution. If you use `Sympy`

, you will get $\left[\left{B: -\frac{A}{\tan(ak)}\right}\right]$, which is a way of admission of defeat, but it is more useful, because it hints you that it is Sage's (actually Sympy's) fault, while the $[]$ before could lead you to think that there is no solution.

I would like to point out a few interesting teachings derived from your question:

- Do not depend on any software to solve a problem; it's just a tool.
- SageMath is excellent in solving problems, but it is not perfect. Try different approaches (e.g., as we did when we used Sympy), especially if the problem is complex.
- You can use a software to help you, but the more of the resolution you can do manually (using ingenuity), the more sure you will be of your result. In the end, if you can do it manually, why bother using Sage? In the end, there are things that are easy for a human mind, but very difficult for a software.

3 | No.3 Revision |

Hello, @Grim! I am not quite sure about Schrödinger's Equation, but I know how to obtain $k=\frac{n\pi}{a}$. I believe this will result a helpful version of your code:

```
k, a = var('k, a', domain='positive')
psi = function('psi')(x)
eqs = diff(psi, x, 2) + k^(2)*psi == 0
EQS = desolve(eqs, psi, ivar=x)
A, B = var('A, B')
EQS = EQS.subs(_K1=B, _K2=A)
show(LatexExpr(r'\psi(x) = '), EQS)
eq1 = EQS.subs(x=0) == 0 # this is psi(0) = 0
eq2 = EQS.subs(x=a) == 0 # this is psi(a) = 0
show(eq1)
show(eq2)
```

The first line defines the symbolic variables `k`

and `a`

to be positive, so you can avoid using the `assume`

command. The second line defines `psi`

as the unknown function; the third line defines the ordinary differential equation; and the fourth one solves the differential equation.

Now, things get a little complicated. If you write `print(EQS)`

, you will notice that there are a couple of new variables called `_K1`

and `_K2`

. They are integration variables (byproducts of the resolution of the ODE). However, they are not symbolic variables that you have defined, so you won't be able to use them outside of `EQS`

(for example, to use the `solve`

command or any other). So, I replaced them with `A`

and `B`

in the sixth line. The seventh line (the first `show`

command) is there only to check that I have obtained the same definition of $psi$ as you in your code (it will show you $psi(x)=A\cos(kx)+B\sin(kx)$). Then, `eq1`

and `eq2`

represent your boundary conditions $psi(0)=0$ and $psi(a)=0$, respectively.

At this point, there are many ways in which you can obtain the values of $A$, $B$ and $k$. The first method is manually solving equations `eq1`

and `eq2`

, which are shown by the last two lines of my code. You will see that

$$A=0$$ and $$A\cos(ak)+B\sin(ak)=0$$

If you replace the first equation in the second, you obtain $B\sin(ax)=0$. Now, since $B>0$ (according to your code), you must have that $\sin(ak)=0$, which implies that $k=\frac{n\pi}{a}$. Finally, in this case, $B$ can be any positive number (regardless of its value, your ODE and boundary conditions are satisfied.)

The second way to solve this is using Sage itself. You have two equations, namely `eq1`

and `eq2`

, so you will need two variables. You can write

```
sol = solve([eq1,eq2], A,k)
show(sol)
```

just after the previous code. You will get something like

$$\left[\left[A = 0, k = \frac{pi + 2\pi z1879}{a}\right], \left[A = 0, k = \frac{2pi z1900}{a}\right]\right]$$

The variables `z1879`

and `z1900`

are similar to the `_K1`

and `_K2`

that I mentioned above. More precisely, they are integer parameters; just mentally replace them with $m$ and $n$. Then you have two possible solutions. After further analysis, you will notice that these two solutions can be written more simply as $A=0$ and $k=\frac{n\pi}{a}$. Once again, $B$ has no restrictions, besides being positive.

You can also use the Sympy package to solve these equations:

```
sol = solve([eq1,eq2], A,k, algorithm='sympy')
show(sol)
```

You will get

$$\left[\left{A : 0, k : 0\right}, \left{A : 0, k : \frac{\pi}{a}\right}\right]$$

These are just two particular solutions that you can easily generalize using reduction formulas and properties of the trigonometric functions.

What happens if you use `solve([eq1, eq2], A, B)`

or `solve([eq1, eq2], B, k)`

? You will get $[[A=0, B=0]]$ and $[]$, respectively. The first is a correct solution to both equations, but not one that you are looking for. The second is Sage telling you that could not find a solution. If you use `Sympy`

, you will get $\left[\left{B: -\frac{A}{\tan(ak)}\right}\right]$, which is a way of admission of defeat, but it is more useful, because it hints you that it is Sage's (actually Sympy's) fault, while the $[]$ before could lead you to think that there is no solution.

I would like to point out a few interesting teachings derived from your question:

- Do not depend on any software to solve a problem; it's just a tool.
- SageMath is excellent in solving problems, but it is not perfect. Try different approaches (e.g., as we did when we used Sympy), especially if the problem is complex.
- You can use a software to help you, but the more of the resolution you can do manually (using ingenuity), the more sure you will be of your result. In the end, if you can do it manually, why bother using Sage? In the end, there are things that are easy for a human mind, but very difficult for a software.

4 | No.4 Revision |

```
k, a = var('k, a', domain='positive')
psi = function('psi')(x)
eqs = diff(psi, x, 2) + k^(2)*psi == 0
EQS = desolve(eqs, psi, ivar=x)
A, B = var('A, B')
EQS = EQS.subs(_K1=B, _K2=A)
show(LatexExpr(r'\psi(x) = '), EQS)
eq1 = EQS.subs(x=0) == 0 # this is psi(0) = 0
eq2 = EQS.subs(x=a) == 0 # this is psi(a) = 0
show(eq1)
show(eq2)
```

`k`

and `a`

to be positive, so you can avoid using the `assume`

command. The second line defines `psi`

as the unknown function; the third line defines the ordinary differential equation; and the fourth one solves the differential equation.

`print(EQS)`

, you will notice that there are a couple of new variables called `_K1`

and `_K2`

. They are integration variables (byproducts of the resolution of the ODE). However, they are not symbolic variables that you have defined, so you won't be able to use them outside of `EQS`

(for example, to use the `solve`

command or any other). So, I replaced them with `A`

and `B`

in the sixth line. The seventh line (the first `show`

command) is there only to check that I have obtained the same definition of $psi$ as you in your code (it will show you $psi(x)=A\cos(kx)+B\sin(kx)$). Then, `eq1`

and `eq2`

represent your boundary conditions $psi(0)=0$ and $psi(a)=0$, respectively.

`eq1`

and `eq2`

, which are shown by the last two lines of my code. You will see that

$$A=0$$ and $$A\cos(ak)+B\sin(ak)=0$$

`eq1`

and `eq2`

, so you will need two variables. You can write

```
sol = solve([eq1,eq2], A,k)
show(sol)
```

just after the previous code. You will get something like

`z1879`

and `z1900`

are similar to the `_K1`

and `_K2`

that I mentioned above. More precisely, they are integer parameters; just mentally replace them with $m$ and $n$. Then you have two possible solutions. After further analysis, you will notice that these two solutions can be written more simply as $A=0$ and $k=\frac{n\pi}{a}$. Once again, $B$ has no restrictions, besides being positive.

You can also use the Sympy package to solve these equations:

```
sol = solve([eq1,eq2], A,k, algorithm='sympy')
show(sol)
```

You will get

~~$$\left[\left{A ~~$$\left[\left\{A : 0, k : ~~0\right}, \left{A ~~0\right\}, \left\{A : 0, k : ~~\frac{\pi}{a}\right}\right]$$~~\frac{\pi}{a}\right\}\right]$$

What happens if you use `solve([eq1, eq2], A, B)`

or `solve([eq1, eq2], B, k)`

? You will get $[[A=0, B=0]]$ and $[]$, respectively. The first is a correct solution to both equations, but not one that you are looking for. The second is Sage telling you that could not find a solution. If you use `Sympy`

, you will get ~~$\left[\left{B: -\frac{A}{\tan(ak)}\right}\right]$, ~~$\left[\left\{B: -\frac{A}{\tan(ak)}\right\}\right]$, which is a way of admission of defeat, but it is more useful, because it hints you that it is Sage's (actually Sympy's) fault, while the $[]$ before could lead you to think that there is no solution.

I would like to point out a few interesting teachings derived from your question:

- Do not depend on any software to solve a problem; it's just a tool.

5 | No.5 Revision |

```
k, a = var('k, a', domain='positive')
psi = function('psi')(x)
eqs = diff(psi, x, 2) + k^(2)*psi == 0
EQS = desolve(eqs, psi, ivar=x)
A, B = var('A, B')
EQS = EQS.subs(_K1=B, _K2=A)
show(LatexExpr(r'\psi(x) = '), EQS)
eq1 = EQS.subs(x=0) == 0 # this is psi(0) = 0
eq2 = EQS.subs(x=a) == 0 # this is psi(a) = 0
show(eq1)
show(eq2)
```

`k`

and `a`

to be positive, so you can avoid using the `assume`

command. The second line defines `psi`

as the unknown function; the third line defines the ordinary differential equation; and the fourth one solves the differential equation.

Now, things get a little complicated. If you write `print(EQS)`

, you will notice that there are a couple of new variables called `_K1`

and `_K2`

. They are integration variables (byproducts of the resolution of the ODE). However, they are not symbolic variables that you have defined, so you won't be able to use them outside of `EQS`

(for example, to use the `solve`

command or any other). So, I replaced them with `A`

and `B`

in the sixth line. The seventh line (the first `show`

command) is there only to check that I have obtained the same definition of ~~$psi$ ~~$\psi$ as you in your code (it will show you ~~$psi(x)=A\cos(kx)+B\sin(kx)$). ~~$\psi(x)=A\cos(kx)+B\sin(kx)$). Then, `eq1`

and `eq2`

represent your boundary conditions ~~$psi(0)=0$ and $psi(a)=0$, ~~$\psi(0)=0$ and $\psi(a)=0$, respectively.

`eq1`

and `eq2`

, which are shown by the last two lines of my code. You will see that

$$A=0$$ and $$A\cos(ak)+B\sin(ak)=0$$

`eq1`

and `eq2`

, so you will need two variables. You can write

```
sol = solve([eq1,eq2], A,k)
show(sol)
```

just after the previous code. You will get something like

`z1879`

and `z1900`

are similar to the `_K1`

and `_K2`

that I mentioned above. More precisely, they are integer parameters; just mentally replace them with $m$ and $n$. Then you have two possible solutions. After further analysis, you will notice that these two solutions can be written more simply as $A=0$ and $k=\frac{n\pi}{a}$. Once again, $B$ has no restrictions, besides being positive.

You can also use the Sympy package to solve these equations:

```
sol = solve([eq1,eq2], A,k, algorithm='sympy')
show(sol)
```

You will get

$$\left[\left\{A : 0, k : 0\right\}, \left\{A : 0, k : \frac{\pi}{a}\right\}\right]$$

What happens if you use `solve([eq1, eq2], A, B)`

or `solve([eq1, eq2], B, k)`

? You will get $[[A=0, B=0]]$ and $[]$, respectively. The first is a correct solution to both equations, but not one that you are looking for. The second is Sage telling you that could not find a solution. If you use `Sympy`

, you will get $\left[\left\{B: -\frac{A}{\tan(ak)}\right\}\right]$, which is a way of admission of defeat, but it is more useful, because it hints you that it is Sage's (actually Sympy's) fault, while the $[]$ before could lead you to think that there is no solution.

I would like to point out a few interesting teachings derived from your question:

- Do not depend on any software to solve a problem; it's just a tool.

6 | No.6 Revision |

```
k, a = var('k, a', domain='positive')
psi = function('psi')(x)
eqs = diff(psi, x, 2) + k^(2)*psi == 0
EQS = desolve(eqs, psi, ivar=x)
A, B = var('A, B')
EQS = EQS.subs(_K1=B, _K2=A)
show(LatexExpr(r'\psi(x) = '), EQS)
eq1 = EQS.subs(x=0) == 0 # this is psi(0) = 0
eq2 = EQS.subs(x=a) == 0 # this is psi(a) = 0
show(eq1)
show(eq2)
```

`k`

and `a`

to be positive, so you can avoid using the `assume`

command. The second line defines `psi`

as the unknown function; the third line defines the ordinary differential equation; and the fourth one solves the differential equation.

Now, things get a little complicated. If you write `print(EQS)`

, you will notice that there are a couple of new variables called `_K1`

and `_K2`

. They are integration variables (byproducts of the resolution of the ODE). However, they are not symbolic variables that you have defined, so you won't be able to use them outside of `EQS`

(for example, to use the `solve`

command or any other). So, I replaced them with `A`

and `B`

in the sixth line. The seventh line (the first `show`

command) is there only to check that I have obtained the same definition of $\psi$ as you in your code (it will show you $\psi(x)=A\cos(kx)+B\sin(kx)$). Then, `eq1`

and `eq2`

represent your boundary conditions $\psi(0)=0$ and $\psi(a)=0$, respectively.

`eq1`

and `eq2`

, which are shown by the last two lines of my code. You will see that

$$A=0$$ and $$A\cos(ak)+B\sin(ak)=0$$

`eq1`

and `eq2`

, so you will need two variables. You can write

`sol = `~~solve([eq1,eq2], A,k)
~~solve([eq1, eq2], A, k)
show(sol)

just after the previous code. You will get something like

$$\left[\left[A = 0, k = ~~\frac{pi ~~\frac{\pi + 2\pi ~~z1879}{a}\right], ~~z_{1879}}{a}\right], \left[A = 0, k = ~~\frac{2pi z1900}{a}\right]\right]$$~~\frac{2 \pi z_{1900}}{a}\right]\right]$$

`z1879`

and `z1900`

are similar to the `_K1`

and `_K2`

that I mentioned above. More precisely, they are integer parameters; just mentally replace them with $m$ and $n$. Then you have two possible solutions. After further analysis, you will notice that these two solutions can be written more simply as $A=0$ and $k=\frac{n\pi}{a}$. Once again, $B$ has no restrictions, besides being positive.

You can also use the Sympy package to solve these equations:

`sol = `~~solve([eq1,eq2], A,k, ~~solve([eq1, eq2], A, k, algorithm='sympy')
show(sol)

You will get

$$\left[\left\{A : 0, k : 0\right\}, \left\{A : 0, k : \frac{\pi}{a}\right\}\right]$$

What happens if you use `solve([eq1, eq2], A, B)`

or `solve([eq1, eq2], B, k)`

? You will get $[[A=0, B=0]]$ and $[]$, respectively. The first is a correct solution to both equations, but not one that you are looking for. The second is Sage telling you that could not find a solution. If you use SymPy, you will get $\left[\left\{B: -\frac{A}{\tan(ak)}\right\}\right]$, which is a way of admission of defeat, but it is more useful, because it hints you that it is Sage's (actually `Sympy`

, ~~Sympy's) ~~SymPy's) fault, while the $[]$ before could lead you to think that there is no solution.

I would like to point out a few interesting teachings derived from your question:

- Do not depend on any software to solve a problem; it's just a tool.
- SageMath is excellent in solving problems, but it is not perfect. Try different approaches (e.g., as we did when we used
~~Sympy),~~SymPy), especially if the problem is complex.

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