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how we can resolve the problem of Chauchy $$ \begin{cases} y'=y^{3/4}\\ y(x_0)=0 \end{cases} $$ I try to write the equation as equation of separable variables, but the problem is that we don't know if y=0 or $\exists x_i: y(x_i)=0$. And theorem Cauchy Lipschitsz of existence and unicity give unicity of only fonction y to take his values in $]0,+\infty[$. So how we can resolve this problem? Please.

2 Answers 2

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Hint: $y=0$ is a solution to the problem.

Now, just integrate $dy/y^{3/4}=dx \implies \frac{1}{-3/4+1}y^{-3/4+1}=x+c_1 \implies 4y^{1/4}=x+c_1 \implies y^{1/4}=x/4+c_2$

$\implies y =(x/4+c_2)^4$. For $c_2=-x_0/4$ the condition $y(x_0)=0$ is also fulfilled.

EDIT: We had to assume that $y\neq 0$ in order to derive this solution, but it turns out that we don't need this additional constraint, as the solution $y=(x/4-x_0/4)^4$ is a solution to the ODE (check this by inserting it into the ODE).

What we see is, that we found two solutions, hence the solution is not unique.

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    but, what about the case where $y$ isn't identiquely vanish, but $\exists x_i: y(x_i) =0$? – 2017-02-15
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    Then, the second solution will be the right solution. Also read my EDIT in the answer. – 2017-02-15
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    Sorry, but i don't understand. How you devide on $y^{-3/4}$, and there is a possibility to have $\exists x_i: y(x_i)=0$ and we don't know hwo's theses points $x_i$? – 2017-02-15
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    I know what you mean, but as I said the way of getting the solution is not important. The solution that we obtained does not have any problems for $y(x_0)=0$, and it solves the ODE with your given condition. We only would have to consider this problematic, if $y(x_0)=0$ would lead to a singularity of your function, but this is not the case as you can easily verify. – 2017-02-15
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    Thank you very much for the answer. – 2017-02-16
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This is a separable equation:

$$\text{y}'\left(x\right)=\text{y}\left(x\right)^\text{n}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(x\right)}{\text{y}\left(x\right)^\text{n}}\space\text{d}x=\int1\space\text{d}x\tag1$$

For some constant $\text{n}$, where $\text{n}\ne1$

Now, we can use:

  • Substitute $\text{u}=\text{y}\left(x\right)$: $$\int\frac{\text{y}'\left(x\right)}{\text{y}\left(x\right)^\text{n}}\space\text{d}x=\int\frac{1}{\text{u}^\text{n}}\space\text{d}\text{u}=\frac{\text{u}^{1-\text{n}}}{1-\text{n}}+\text{C}_1=\frac{\text{y}\left(x\right)^{1-\text{n}}}{1-\text{n}}+\text{C}_1\tag2$$
  • $$\int1\space\text{d}x=x+\text{C}_2\tag3$$

So, we get:

$$\frac{\text{y}\left(x\right)^{1-\text{n}}}{1-\text{n}}=x+\text{C}\space\Longleftrightarrow\space\text{y}\left(x\right)^{1-\text{n}}=x\left(1-\text{n}\right)+\text{C}\tag4$$

Now, when we have an initial condition: $\text{y}\left(\alpha\right)=\beta$, we get:

$$\beta^{1-\text{n}}=\alpha\left(1-\text{n}\right)+\text{C}\space\Longleftrightarrow\space\text{C}=\beta^{1-\text{n}}-\alpha\left(1-\text{n}\right)\tag5$$

So, we end up with:

$$\text{y}\left(x\right)^{1-\text{n}}=x\left(1-\text{n}\right)+\beta^{1-\text{n}}-\alpha\left(1-\text{n}\right)\tag6$$

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    Thank you very much for the answer. – 2017-02-16
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    @jiji You're welcome, I'm glad that I could help you! – 2017-02-16