## integration – General solution to the Laplace equation in \$ mathbb {R ^ 3} \$

I'm trying to find Green's function for $$nabla ^ 2 phi = S (x)$$ for $$x in mathbb {R ^ 3}$$ and express the general solution to the Laplace equation in $$mathbb {R ^ 3}$$.

To find the function of green, I considered
begin {align} nabla ^ 2G & = delta left ( underline {x} – underline {x & # 39;} right) \ nabla ^ 2G & = 0 left ( text {when} underline {x} neq underline {x & # 39;} right) \ nabla ^ 2G & = frac {1} {r ^ 2} frac {d} {dr} left (r ^ 2 frac {DG} {dr} right) \ G (r) & = – frac {A} {r} + B, A, B in mathbb {R}. end {align}
Then consider a ball centered on $$underline {x & # 39;}$$ with radio $$a$$denoted $$B_a$$. So,
begin {align} iiint_ {B_a} nabla ^ 2G dV & = iiint_ {B_a} delta left ( underline {x} – underline {x & # 39;} right) dV \ iint _ { delta B_a} frac {dG} {dr} bigg | _ {r = a} ds & = 1 text {(divergence theorem)} \ 4 pi A & = 1 \ implies A & = frac {1} {4 pi} end {align}
Therefore (leaving $$B = 0$$ without loss of generality), $$G (r) = – frac {1} {4 pi r}.$$ Now, $$r = | underline {x} – underline {x & # 39;} |$$, meaning $$G ( underline {x}, underline {x & # 39;}) = – frac {1} {4 pi | underline {x} – underline {x & # 39;} |}.$$

Why can the general solution be expressed as $$phi ( underline (x)) = – frac {1} {4 pi} iiint _ { mathbb {R ^ 3}} frac {1} {| underline {x} – underline {x & # 39;} |} S ( underline {x & # 39;}) dV & # 39 ;?$$

## Mathematics: Can anyone figure out the damage equation in this game?

So there's a rather dark turn-based role game called Gods Lands of Infinity. So there is a power of close combat (basically a basic attack statistic) of each character in this game. In this case, the melee power is 132. Each weapon also has a base damage of its own, and in this case, the weapon's damage is 9-11. Now, attacking an enemy that has a defense of -30, the damage output is 11-13. If the damage to the base of the weapon increases to 12-14, the damage output becomes 14-17. Also the attack level is 1 and the defender is also 1, but I do not think the levels are taken into account in the formula.

I tried a lot, but I could not figure out how the damage is

Can anyone figure out the damage equation in this game?

## Physics: evaluation of Hough functions using NDEigensystem in Laplace's tidal equation

I am currently studying the use of Mathematica to solve the classic. tidal equation by M. Laplace:

$$mathcal {F} Theta + gamma Theta = 0$$

whose own functions $$Theta$$ These are the functions of Hough. Here, following the convention in the role of Wang. et al., I let the tidal operator of Laplace $$mathcal {F}$$ be

$$mathcal {F} Theta = frac { mathrm d} { mathrm d mu} left ( frac {1- mu ^ 2} { sigma ^ 2- mu ^ 2} frac { mathrm d Theta} { mathrm d mu} right) – frac1 { sigma ^ 2- mu ^ 2} left ( frac {s} { sigma} frac { sigma ^ 2 + mu ^ 2} { sigma ^ 2- mu ^ 2} + frac {s ^ 2} {1- mu ^ 2} right) Theta$$

where $$s$$ Y $$sigma$$ They are numerical parameters associated with the Hough function.

The role of Wang et al. it uses a pseudospectral approach with the associated Legendre functions normalized or Chebyshev polynomials as a base, but I wanted to see if I can use `NDEigensystem` Instead, since you need less programming from me. Unfortunately, my efforts have so far been in vain, and now I wonder what I am doing wrong.

As a concrete example, let us take the so-called diurnal mode, propagation towards the west, zonal wave number 1 (DW1), corresponding to $$s = 1, sigma = frac12$$, and take three Hough features:

``````s = 1; σ = 1/2; m = 3;
tide = -D[(1 - μ^2)/(σ^2 - μ^2) Θ'[μ], μ]+
1 / (σ ^ 2 - μ ^ 2) (s / σ (σ ^ 2 + μ ^ 2) / (σ ^ 2 - μ ^ 2) + s ^ 2 / (1 - μ ^ 2)) Θ[μ];
limits = DirichletCondition[Θ[Θ[Θ[Θ[μ] == 0, True];
{γ, Θk} = with[{h = Sqrt[\$MachineEpsilon]}
NDEigensystem[{tidal, bounds}, Θ, {μ, -1 + h, 1 - h}, m]];
``````

(I changed the ends of the solution interval to avoid the singularities, but I feel very uncomfortable doing so, so it would be good to see alternatives).

Unfortunately, the eigenvalues ​​that I obtain from this are markedly different from the eigenvalues ​​I obtained with the reimplementation of the pseudospectral approach (whose code I am ashamed to share due to all the For loops that it has); in particular, I can not reproduce figure 1 in Wang et al. paper.

Therefore, I am interested to know if I have used the Mathematica functionality incorrectly or if Mathematica is not equipped to handle the Laplace tidal equation. Ideas and other forms are, of course, welcome.

## using NDEigensystem to solve Mathieu's equation

In order to apply Mathematica's differential equation capabilities to my graduate thesis, I am trying to apply NDEigensystem to a problem of my own whose solution I know of, but I have some problems to do so.

As a test problem, I'm using an algebraic version of Mathieu's equation,

$$(1- zeta ^ {2}) w ^ { prime prime} – zeta w ^ { prime} + left (a + 2q-4q zeta ^ {2} right) w = 0$$

For this example I put myself $$q = 4/3$$ and take only the first three eigen pairs:

``````m = 3; q = 4/3;
op = - (1 - ζ ^ 2) u & # 39; & # 39;[ζ] + ζ u & # 39;[ζ] + 2 q (2 ζ ^ 2 - 1) u[ζ];
bc = DirichletCondition[or[or[u[u[ζ] == 0, True];
{λ, fl} = NDEigensystem[{op, bc}, u, {ζ, 0, 1}, m];
``````

I chose the Mathieu equation as a non-trivial example, since Mathematica already has a function for its evaluation:

``````λt = Table[MathieuCharacteristicB[2 k, q], {k, m}];
flt = Table[With[{j = j},
MathieuS[MathieuCharacteristicB[2 k, q], q, ArcCos[#]]&], {k, m}];
``````

The problem is that I do not get the expected eigenvalues!

``````λ
(* {4.0708, 17.3259, 39.1877} *)
north[λt]
(* {3.85298, 16.0581, 36.0254} *)
``````

And, of course, the plot shows that the equation is not satisfied at all:

``````With[{u = fl[[1]], b = λ[[1]]}
Plot[(1-ζ^2)u&#39;&#39;[(1-ζ^2)u''[(1-ζ^2)u''[(1-ζ^2)u''[ζ] - ζ u & # 39;[ζ] + (b + 2 q - 4 q ζ ^ 2) u[ζ], {ζ, 0, 1}]]With[{u=flt[{u=flt[{u=flt[{u=flt[[1]], b = λt[[1]]}
Plot[(1-ζ^2)u&#39;&#39;[(1-ζ^2)u''[(1-ζ^2)u''[(1-ζ^2)u''[ζ] - ζ u & # 39;[ζ] + (b + 2 q - 4 q ζ ^ 2) u[ζ], {ζ, 0, 1}]]
``````

What was wrong with my attempt? If I can make this example work, I should be able to apply it to my current, more complicated problem, so any good idea would be welcome.

## Keynote equation: is it possible to predefine LaTeX macros?

Is it possible to add to the LaTeX preamble for the Keynote equation tool? For example, the default setting is that vec {x} generates an arrow over the letter `X`, but I would like this to produce a bold `X`.

## It was not possible to solve this equation with Resolver and you can also obtain an almost correct plot with wolfram alpha

As suggested by Chip, you need DSolve. But if you want to trace it, you also need an initial condition. This is the reason why Wolfram Alpha's graph shows multiple curves … they correspond to different initial conditions. This is what I have to work:

``````result[x_] =
Y[x]/.NDSolver[{and&#39;[{Y'[{y'[{y'[x]== - (224.8 ((and[x]^ 2-y[x]/ 10 ^ 11
- (0.000010385 x ^ 3) / E ^ (2 x)) / x ^ 2)), and[1] == 1}, and[x], {x, 0, 8}]The[[[1]]Plot[result[result[resultado[result[x], {x, 1,8}];
``````

## Equation of wave with medium term porous.

The classic equation of porous media is

$$u_t + Delta (u ^ m) = 0 m> 1.$$

It has the wave equation (degenerate)
$$u_ {tt} + Delta (u ^ m) = 0$$
Have you been the subject of studies?

## Weapons – Can anyone figure out the damage equation in this game?

So there's a rather dark turn-based role game called Gods Lands of Infinity. So there is a melee power of each character in this game. In this case, the melee power is 132. Each weapon also has its own damage, and in this case, the weapon's damage is 9-11. Now, attacking an enemy that has a defense of -30, the damage output is 11-13. If the damage to the base of the weapon increases to 12-14, the damage output becomes 14-17.

Can anyone figure out the damage equation in this game?

## Solve an equation for a specific unknown?

I have this equation:

``````((a * (1 - t) ^ 2 + 2 * d * (1 - t) * t + g * t ^ 2) - j) ^ 2 + ((b * (1 - t) ^ 2 +
2 * e * (1 - t) * t + h * t ^ 2) - k) ^ 2 +
((c * (1 - t) ^ 2 + 2 * f * (1 - t) * t + i * t ^ 2) - l) ^ 2 == r ^ 2
``````

It is long and very badly edited, but I count on Wolfram to simplify it. But here I have a lot of unknowns, while in real all the variables except $$t$$ are known, I plan to use it in a video game, so I even know that I do not know $$a$$, $$b$$, $$c$$, … These variables will be known in the calculation.
So I want to tell Wolfram, just $$t$$ It is unknown and the result is obtained. $$t = a + b ^ 2 …$$.

Currently I have tried to solve [my equation] for $$t$$ And I looked at the equation page but I did not find my solution.

## Solving an equation with 3 free parameters in a way to satisfy 1 condition

I have an equation in terms of $$q$$, $$c_2$$, $$c_3$$ Y $$r$$. I only know that $$q$$ Y $$r$$ are positive (there are no restrictions on $$c_2$$ Y $$c_3$$).

``````eq1 = (10 q ^ 2) / r ^ 3 - 2 (3 c3 + c2 r)
``````

Solving this equation in terms of $$r$$ With the mentioned conditions it gives:

``````sol1 = r /. Solve[eq1 == 0 && q > 0 && r > 0, r, Reals]
``````

{Conditional expression[Root[-5 q^2 + 3 c3 #1^3 + c2 #1^4 &, 1],
0 <q <27/16 Sqrt[3/5] Sqrt[-(c3^4/c2^3)] && c2 < 0 && c3 > 0],
Conditional expression[
Root[-5 q^2 + 3 c3 #1^3 + c2 #1^4 &,
2], (c2> 0 && c3> 0 &&
q> 0) || (0 <q <27/16 Sqrt[3/5] Sqrt[-(c3^4/c2^3)] && c2 < 0 && c3 > 0) || (c3 < 0 && c2 > 0 && q> 0)]}

I want those solutions of this equation with 2 roots that make eq3 positive:

``````eq3 = (3 c3 + 2 c2 r) / (8 [Pi] r ^ 3);
``````

I could not find the analytical way to do it, but here is my numeric code to find the parameters for which the condition is met. It takes too much time and is not efficient at all (until now I did not find a single set of parameters to satisfy the condition).

``````ansnum = flat[
ParallelTable[{q, c2, c3, sol1}, {q, 0.001, 1, 0.01}, {c2, -20,
0.001, 0.1}, {c3, 0.001, 20, 0.1}], two];

ansnum = ansnum /. Indefinite -> -100;

j = 0;

Table[

If[ansnum[[i, 4, 1]]! = -100 && Length[ansnum[[i, 4]]]== 2,
j = j + 1;
eq3fin[j] = {ansnum[[i, 1]]Ansum[[i, 2]],
Ansum[[i, 3]], {ansnum[[i, 4, 1]],
Ansum[[i, 4,
2]]}, {eq3 /. {q -> ansnum[[i, 1]], c2 -> ansnum[[i, 2]],
c3 -> ansnum[[i, 3]], r -> ansnum[[i, 4, 1]]}
eq3 /. {q -> ansnum[[i, 1]], c2 -> ansnum[[i, 2]],
c3 -> ansnum[[i, 3]], r -> ansnum[[i, 4, 2]]}}}
], {i, 1, length[ansnum]}
]/. Indefinite -> sequence[];

Table

The
Yes[eq3fin[eq3fin[eq3fin[eq3fin[j][[5, 1]]> 0; eq3fin[j][[5, 2]]> 0,
Print[eq3fin[eq3fin[eq3fin[eq3fin[j]]], {j, 1, j}];
``````

My question is, is there any analytical way to do that? If not, how can I make my numerical code efficient?

Thank you,