desertiev5

Answered

2022-07-03

How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation?

${y}^{\prime}={x}^{3}{y}^{3}-1\phantom{\rule{0ex}{0ex}}$

${y}^{\prime}={x}^{3}{y}^{3}-1\phantom{\rule{0ex}{0ex}}$

Answer & Explanation

trantegisis

Expert

2022-07-04Added 20 answers

Honestly, I don't think the solutions of your ODE can be written in elementary terms.

Actually, any substitution of the type $u={y}^{\alpha}$ won't simplify the ODE, because of that evil constant term -1.

Neverthless, you could look for a power series solution of your ODE using the Frobenius method, that is:

1. assume that a solution of your ODE can be expanded in a power series $\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$

2. evaluate the power series expansion of ${y}^{3}(x)$ and y'(x) and plug them into the ODE,

3. deduce from the ODE a recurrence relation for the coefficients ${a}_{n}$,

4. try to prove that the radius of convergence of $\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$ is $>0$

then the sum $y(x):=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$ will be an analytic solution of your ODE.

It is easy to prove that if $y(x)=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$ then:

1. ${y}^{3}(x)=\sum _{n=0}^{\mathrm{\infty}}{b}_{n}\text{}{x}^{n}$, where ${b}_{n}:=\sum _{k=0}^{n}\sum _{h=0}^{n-k}{a}_{k}\text{}{a}_{h}\text{}{a}_{n-k-h}$ satisfies:

$\{\begin{array}{l}{b}_{0}={a}_{0}^{3}\\ {b}_{n}=\frac{1}{n\text{}{a}_{0}}\text{}\sum _{k=1}^{n}(4k-n){a}_{k}\text{}{b}_{n-k}\end{array}$

2. ${y}^{\mathrm{\prime}}(x)=\sum _{n=0}^{\mathrm{\infty}}(n+1)\text{}{a}_{n+1}\text{}{x}^{n}$,

therefore plugging 1 and 2 into your ODE gives:

${a}_{1}+2{a}_{2}\text{}x+3{a}_{3}\text{}{x}^{2}+\sum _{n=3}^{\mathrm{\infty}}(n+1)\text{}{a}_{n+1}\text{}{x}^{n}=-1+\sum _{n=3}^{\mathrm{\infty}}{b}_{n-3}{x}^{n}\phantom{\rule{thickmathspace}{0ex}}.$

Equating the coefficients of like powers of x, you obtain:

$\{\begin{array}{l}{a}_{1}=-1\\ {a}_{2}=0\\ {a}_{3}=0\\ (n+1)\text{}{a}_{n+1}={b}_{n-3}& \text{, for}n\ge 3\end{array}$

i.e.:

$\begin{array}{}\text{(1)}& \{\begin{array}{l}{a}_{1}=-1\\ {a}_{2}=0\\ {a}_{3}=0\\ {a}_{n+1}=\frac{1}{n+1}\text{}\sum _{k=0}^{n-3}\sum _{h=0}^{n-3-k}{a}_{k}\text{}{a}_{h}\text{}{a}_{n-3-k-h}& \text{, for}n\ge 3.\end{array}\end{array}$

Note that ${a}_{0}$ cannot be determined using (1).

Now there remains to be solved the problem of finding the radius of convergence of the power series whose coefficients are given by (1).

Actually, any substitution of the type $u={y}^{\alpha}$ won't simplify the ODE, because of that evil constant term -1.

Neverthless, you could look for a power series solution of your ODE using the Frobenius method, that is:

1. assume that a solution of your ODE can be expanded in a power series $\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$

2. evaluate the power series expansion of ${y}^{3}(x)$ and y'(x) and plug them into the ODE,

3. deduce from the ODE a recurrence relation for the coefficients ${a}_{n}$,

4. try to prove that the radius of convergence of $\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$ is $>0$

then the sum $y(x):=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$ will be an analytic solution of your ODE.

It is easy to prove that if $y(x)=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}\text{}{x}^{n}$ then:

1. ${y}^{3}(x)=\sum _{n=0}^{\mathrm{\infty}}{b}_{n}\text{}{x}^{n}$, where ${b}_{n}:=\sum _{k=0}^{n}\sum _{h=0}^{n-k}{a}_{k}\text{}{a}_{h}\text{}{a}_{n-k-h}$ satisfies:

$\{\begin{array}{l}{b}_{0}={a}_{0}^{3}\\ {b}_{n}=\frac{1}{n\text{}{a}_{0}}\text{}\sum _{k=1}^{n}(4k-n){a}_{k}\text{}{b}_{n-k}\end{array}$

2. ${y}^{\mathrm{\prime}}(x)=\sum _{n=0}^{\mathrm{\infty}}(n+1)\text{}{a}_{n+1}\text{}{x}^{n}$,

therefore plugging 1 and 2 into your ODE gives:

${a}_{1}+2{a}_{2}\text{}x+3{a}_{3}\text{}{x}^{2}+\sum _{n=3}^{\mathrm{\infty}}(n+1)\text{}{a}_{n+1}\text{}{x}^{n}=-1+\sum _{n=3}^{\mathrm{\infty}}{b}_{n-3}{x}^{n}\phantom{\rule{thickmathspace}{0ex}}.$

Equating the coefficients of like powers of x, you obtain:

$\{\begin{array}{l}{a}_{1}=-1\\ {a}_{2}=0\\ {a}_{3}=0\\ (n+1)\text{}{a}_{n+1}={b}_{n-3}& \text{, for}n\ge 3\end{array}$

i.e.:

$\begin{array}{}\text{(1)}& \{\begin{array}{l}{a}_{1}=-1\\ {a}_{2}=0\\ {a}_{3}=0\\ {a}_{n+1}=\frac{1}{n+1}\text{}\sum _{k=0}^{n-3}\sum _{h=0}^{n-3-k}{a}_{k}\text{}{a}_{h}\text{}{a}_{n-3-k-h}& \text{, for}n\ge 3.\end{array}\end{array}$

Note that ${a}_{0}$ cannot be determined using (1).

Now there remains to be solved the problem of finding the radius of convergence of the power series whose coefficients are given by (1).

Jovany Clayton

Expert

2022-07-05Added 4 answers

You should highly notice that Bernoulli differental equation is of the form ${y}^{\prime}=f(x){y}^{n}+g(x)y$ rather than of the form ${y}^{\prime}=f(x){y}^{n}+g(x)$

Approach 1:

In fact ${y}^{\prime}={x}^{3}{y}^{3}-1$ belongs to an Abel equation of the first kind.

Approach 2:

Let u=xy,

Then $y={\displaystyle \frac{u}{x}}$

$\frac{dy}{dx}}={\displaystyle \frac{1}{x}}{\displaystyle \frac{du}{dx}}-{\displaystyle \frac{u}{{x}^{2}}$

$\therefore {\displaystyle \frac{1}{x}}{\displaystyle \frac{du}{dx}}-{\displaystyle \frac{u}{{x}^{2}}}={u}^{3}-1$

$\frac{1}{x}}{\displaystyle \frac{du}{dx}}={\displaystyle \frac{u}{{x}^{2}}}+{u}^{3}-1$

Let $v={\displaystyle \frac{1}{{x}^{2}}}$

Then $\frac{du}{dx}}={\displaystyle \frac{du}{dv}}{\displaystyle \frac{dv}{dx}}=-{\displaystyle \frac{2}{{x}^{3}}}{\displaystyle \frac{du}{dv}$

$\therefore -{\displaystyle \frac{2}{{x}^{4}}}{\displaystyle \frac{du}{dv}}={\displaystyle \frac{u}{{x}^{2}}}+{u}^{3}-1$

$(uv+{u}^{3}-1){\displaystyle \frac{dv}{du}}=-2{v}^{2}$

This belongs to an Abel equation of the second kind.

Approach 1:

In fact ${y}^{\prime}={x}^{3}{y}^{3}-1$ belongs to an Abel equation of the first kind.

Approach 2:

Let u=xy,

Then $y={\displaystyle \frac{u}{x}}$

$\frac{dy}{dx}}={\displaystyle \frac{1}{x}}{\displaystyle \frac{du}{dx}}-{\displaystyle \frac{u}{{x}^{2}}$

$\therefore {\displaystyle \frac{1}{x}}{\displaystyle \frac{du}{dx}}-{\displaystyle \frac{u}{{x}^{2}}}={u}^{3}-1$

$\frac{1}{x}}{\displaystyle \frac{du}{dx}}={\displaystyle \frac{u}{{x}^{2}}}+{u}^{3}-1$

Let $v={\displaystyle \frac{1}{{x}^{2}}}$

Then $\frac{du}{dx}}={\displaystyle \frac{du}{dv}}{\displaystyle \frac{dv}{dx}}=-{\displaystyle \frac{2}{{x}^{3}}}{\displaystyle \frac{du}{dv}$

$\therefore -{\displaystyle \frac{2}{{x}^{4}}}{\displaystyle \frac{du}{dv}}={\displaystyle \frac{u}{{x}^{2}}}+{u}^{3}-1$

$(uv+{u}^{3}-1){\displaystyle \frac{dv}{du}}=-2{v}^{2}$

This belongs to an Abel equation of the second kind.

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