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Study the Existence and Uniqueness of the IVP : $$y'(x) = (x-y(x))^{4/5}, \space y(3)=3$$

Regarding existence, I'd first assume $f$ the function such :

$$f(x,y)=(x-y)^{4/5}$$

We observe that $f(x,y)$ is not continuous in $ \mathbb R^2$ since we need $x-y\geq0$.

Letting $D$ be a domain such :

$$D=\{(x,y)\in \mathbb R^2: |x-3| \leq ε_1, \space |y-3|\leq ε_2 \}$$

we can say that there exists $ε_1,ε_2 \in \mathbb R$ such that $f$ is continuous in $D$, which means that the IVP has a solution in $D$.

Question : Is the above statement correct ? Shall I go into more details explaining why $f$ can be continuous in such a domain ?

Finally, regarding uniqueness, I know that showing that $f(x,y)=(x-y)^{4/5}$ is a Lipschitz function is enough. :

$$|f(x,y_2) -f(x,y_1)|=|(x-y_2)^{4/5}-(x-y_1)^{4/5}|$$

The function $g(y)=(x-y)^{4/5}$ is continuous in the interval $[y_1,y_2]$ with $y_1,y_2 \in D_f$ and $g$ is also differentiable in $(y_1,y_2)$ , which means that we can apply the Mean Value Theorem and yield :

$$g'(ξ)=\frac{g(y_2)-g(y_1)}{y_2-y_1}\Rightarrow \bigg|\frac{4}{5(x-ξ)^{1/5}}\bigg||y_2-y_1|=|(x-y_2)^{4/5}-(x-y_1)^{4/5}|$$

Here, we can see that there isn't a bound for the expression $\bigg|\frac{4}{5(x-ξ)^{1/5}}\bigg|$, thus meaning that the solution is not unique.

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The function $f(x,y) := (x-y)^{4/5}$ is continuous is $\mathbb{R}^2$, being the composition of the continuous functions $\phi(t) := t^{4/5} = \sqrt[5]{t^4}$, $t\in\mathbb{R}$ and $g(x,y) := x-y$, $(x,y)\in\mathbb{R}^2$.

By Peano's existence theorem, the Cauchy problems associated with $f$ do have solutions.

The function $f$ is not (locally) Lipschitz continuous, but this is only a sufficient condition in order to get uniqueness of solution of Cauchy problems.

It is easy to see that $y(x)$ is a solution of $y'=f(x,y)$ if and only if $z(x) := y(x) - x$ is a solution of $z' = h(x,z)$, with $h(x,z) := z^{4/5} - 1$. In particular, $y(x)$ is a solution of the given Cauchy problem if and only if $z(x) := y(x) - x$ is a solution of the Cauchy problem $$ \begin{cases} z' = z^{4/5} - 1,\\ z(3) = 0. \end{cases} $$ This Cauchy problem admits, locally, a unique solution, implicitly defined by $$ \int_0^z \frac{1}{s^{4/5}-1}\, ds = x-3. $$ It can be proved (but this requires a simple qualitative study) that this unique local solution can be prolonged to a unique global solution.

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  • $\begingroup$ but what about $y=x$ what I am missing? $\endgroup$ – daulomb Dec 10 '17 at 14:13
  • $\begingroup$ It is not a solution: indeed $y' = 1$, so the equation is not satisfied. $\endgroup$ – Rigel Dec 10 '17 at 14:14
  • $\begingroup$ The function $f(x,y) = (x-y)^{4/5}$ is not continuous in $\mathbb R^2$ since the domain of it is : $D_f = \{(x,y) \in \mathbb R^2 : x \geq y\}$. $\endgroup$ – Rebellos Dec 10 '17 at 14:15
  • $\begingroup$ ahh you re right my crelesness thanks for clarification $\endgroup$ – daulomb Dec 10 '17 at 14:15
  • $\begingroup$ @Rebellos: that function is continuous in $\mathbb{R}^2$ since, by definition, is $\sqrt[5]{(x-y)^4}$. $\endgroup$ – Rigel Dec 10 '17 at 14:17
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The function $f$ is not defined in your set $D$ since $D$ contains points where $x<y$. So your statement is not correct.

Also for uniqueness, the theorem says that if $f$ is Lipschitz in $y$, then the solution is unique, but the converse is not true. So you cannot say that since $f$ is not Lipschitz, there is no uniqueness.

It is true that one could define $t^{4/5}=(t^4)^{1/5}$ but then you lose important properties like $t^{ab}=(t^a)^b$. In general it is better not to define $t^a$ if $a$ is a rational and $t<0$. A way out in this case would be to extend $f$ to be zero in the region $x<y$. So now the problem $y'(x)=f(x,y)$, $y(3)=3$ has a solution since $f$ is continuous in $\mathbb R^2$. Then you can follow Riegel's proof to get existence and show that the solution that you get stays in the region $x-y\ge 0$ so it is a solution of your original equation.

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  • $\begingroup$ The definition of $t^{4/5}$ can be a matter of taste. Anyway, in my personal experience, if it is given a power with a fixed rational exponent (i.e., I am not referring to functions like $x^x$), then the most common interpretation is the one that I have given. Anyway, one can proceed as you have shown. $\endgroup$ – Rigel Dec 10 '17 at 14:34
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    $\begingroup$ Agree. If you decide to define $t^{4/5}$ for $t$ negative then yours is the only possible definition. My point is that it is better not to define it, since it creates too many mistakes. Students are so used to properties like $(t^a)^b=t^{ab}$ or $t^a=e^{\log t^a}=e^{a\log t}$, which clearly fail if you extend the definition of $t^{4/5}$ to negative numbers. When I teach analysis, I do not define it. But yeah, it is a matter of taste. $\endgroup$ – Gio67 Dec 10 '17 at 14:49
  • $\begingroup$ @Gio67: very good explanation. so these equalities are not true $t^{4/5}= (t^4)^{1/5}=(t^{1/5})^4$ for negative $t$ values. $\endgroup$ – daulomb Dec 10 '17 at 15:06
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    $\begingroup$ The elaboration and explanation of Gio67 is exactly the reason for which I argued about the continuity (and probably the reason WA shows the property I elaborated in my question form). $\endgroup$ – Rebellos Dec 10 '17 at 15:16
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    $\begingroup$ The typical example of what could go wrong is $i=(-1)^{1/2}=(-1)^{2/4}=((-1)^{2})^{1/4}=1^{1/4}=1$. $\endgroup$ – Gio67 Dec 10 '17 at 15:57

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