Prove Existence & Uniqueness for Diff. Eq. w/ Measurable Coeff. & RHS

[bkarpuz]

Dear MHB members,

Suppose that $p,f$ are locally essentially bounded Lebesgue measurable functions and consider the differential equation
$x'(t)=p(t)x(t)+f(t)$ almost for all $t\geq t_{0}$, and $x(t_{0})=x_{0}$.
By a solution of this equation, we mean a function $x$,
which is absolutely continuous in $[t_{0},t_{1}]$ for all $t_{1}\geq t_{0}$,
and satisfies the differential equation almost for all $t\geq t_{0}$ and $x(t_{0})=x_{0}$.

How can I prove existence and uniqueness in the sense of almost everywhere of solutions to this problem?

Thanks.
bkarpuz

 

[Jose27]

I think this works, it's basically the same approach as with classical ODE's IVP: Define the operator $A:L^\infty[t_0,t_1] \to L^\infty[t_0,t_1]$, with $t_1>t_0$ to be defined, as $A(x)(t)=x_0+\int_{t_0}^{t_1} p(s)x(s)+f(s)ds$. It's easy to see $A$ is well defined as a mapping between these spaces and moreover we are looking for a fixed point, ie. $A(x)(t)=x(t)$, for this we use Banach's fixed point theorem: By Hölder's inequality we have

$$\| Ax-Ay\|_\infty \leq \| x-y\|_\infty \int_{t_0}^{t_1} |p(s)|ds$$

so for $t_1$ small enough we get a contraction and thus a solution in $[t_0,t_1]$ (that it's AC is obvious from the definition of $A$). We can apply this procedure again in $[t_1,t_2]$ and the operator $A_1(x)(t)=x_1(t_1)+\int_{t_1}^{t_2} p(s)x(s)+f(s)ds$ where $x_1$ is the solution in $[t_0,t_1]$. Continuing this way we can build a solution for all $t\geq t_0$ (If it had a finite supremum we apply the same argument, contradiction).

For uniqueness take two solution $x,y$, by uniqueness in the fixed point theorem we have $x=y$ in $[t_0,t_1]$, and so they coincide on every extension of these intervals, hence they coincide everywhere.

 

[bkarpuz]

I think this works, it's basically the same approach as with classical ODE's IVP: Define the operator $A:L^\infty[t_0,t_1] \to L^\infty[t_0,t_1]$, with $t_1>t_0$ to be defined, as $A(x)(t)=x_0+\int_{t_0}^{t_1} p(s)x(s)+f(s)ds$. It's easy to see $A$ is well defined as a mapping between these spaces and moreover we are looking for a fixed point, ie. $A(x)(t)=x(t)$, for this we use Banach's fixed point theorem: By Hölder's inequality we have

$$\| Ax-Ay\|_\infty \leq \| x-y\|_\infty \int_{t_0}^{t_1} |p(s)|ds$$

so for $t_1$ small enough we get a contraction and thus a solution in $[t_0,t_1]$ (that it's AC is obvious from the definition of $A$). We can apply this procedure again in $[t_1,t_2]$ and the operator $A_1(x)(t)=x_1(t_1)+\int_{t_1}^{t_2} p(s)x(s)+f(s)ds$ where $x_1$ is the solution in $[t_0,t_1]$. Continuing this way we can build a solution for all $t\geq t_0$ (If it had a finite supremum we apply the same argument, contradiction).

For uniqueness take two solution $x,y$, by uniqueness in the fixed point theorem we have $x=y$ in $[t_0,t_1]$, and so they coincide on every extension of these intervals, hence they coincide everywhere.

Jose27, thank you very much. Here is my approach, please let me know if I am doing wrong.

Existence. Pick some $t_{1}\geq t_{0}$, and define the operator $\Gamma:\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})\to\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$ by $(\Gamma{}x)(t):=x_{0}+\int_{t_{0}}^{t}\big[p(s)x(s)+f(s)\big]\mathrm{d}s$ for $t_{0}\leq{}t\leq{}t_{1}$.
Obviously, $\Gamma\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})\subset\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$.
Let $\{y_{k}\}_{k\in\mathbb{N}_{0}}\subset\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$ to be the sequence of Picard iterates defined by $y_{0}(t):=x_{0}$ for $t_{0}\leq{}t\leq{}t_{1}$ and $y_{k}(t)=(\Gamma{}y_{k-1})(t)$ for $t_{0}\leq{}t\leq{}t_{1}$ and $k\in\mathbb{N}$.

We may find two positive constants $M_{1}$ and $M_{2}$ such that $\|y_{1}-x_{0}\|_{\mathrm{ess}}\leq{}M_{1}$ and $\|p\|_{\mathrm{ess}}\leq{}M_{2}$ and show by induction that $\|y_{k}-y_{k-1}\|_{\mathrm{ess}}\leq{}M_{1}M_{2}^{k-1}\frac{(t-t_{0})^{k-1}}{(k-1)!}$ for all $k\in\mathbb{N}$.

Since the majorant series $\sum_{\ell=0}^{\infty}M_{1}M_{2}^{\ell}\frac{(t-t_{0})^{\ell}}{\ell!}$ converges to $M_{1}\mathrm{e}^{M_{2}(t-t_{0})}$, which is bounded above by $M_{1}\mathrm{e}^{M_{2}(t_{1}-t_{0})}$, we see that the sequence $\big\{y_{k}=x_{0}+\sum_{\ell=0}^{k-1}[y_{\ell+1}-y_{\ell}]\big\}_{k\in\mathbb{N}_{0}}$ converges uniformly due to Weierstrass $M$-test.
Let $y:=\lim_{k\to\infty}y_{k}$ (do we know here that $y\in\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$?), which implies that the fixed point of $\Gamma$ is $y$, i.e., $y=\Gamma{}y$ on $[t_{0},t_{1}]$, which shows that $x'(t)=p(t)x(t)+f(t)$ almost for all $t_{0}\leq{}t\leq{}t_{1}$ (actually, I need some clarification here, i.e., how the solution is absolutely continuous).

Uniqueness. Assume that there exist two solutions $x$ and $y$, define $z(t):=\mathrm{ess\,sup}_{t_{0}\leq{}s\leq{}t}|x(s)-y(s)|$ for $t_{0}\leq{}t\leq{}t_{1}$. Note that $z$ is nonnegative and monotone.
Then, we have $\|x(t)-y(t)\|\leq{}M_{2}\int_{t_{1}}^{t}p(s)z(s)\mathrm{d}s$ for all$t_{0}\leq{}t\leq{}t_{1}$, which yields $z(t)\leq{}M_{2}\mathrm{ess\,sup}_{t_{0}\leq s\leq t}\int_{t_{1}}^{s}p(r)z(r)\mathrm{d}r\leq M_{2}\int_{t_{1}}^{t}p(s)z(s)\mathrm{d}s$ for all $t_{0}\leq{}t\leq{}t_{1}$.
By an application of the Grönwall's inequality, we see that $z(t)\leq0$ for all $t_{0}\leq{}t\leq{}t_{1}$, i.e., $x=y$ almost everywhere in $[t_{0},t_{1}]$.

Since $t_{1}$ is arbitrary, we may let $t_{1}\to\infty$ to complete the proof.

Thanks.
bkarpuz

 

[bkarpuz]

Is there a reference rather than Coddington & Levinson - Theory of Ordinary Differential Equations, McGraw Hill, 1955, which presents Carathéodory's existence theorem?

Thanks.
bkarpuz

 

[bkarpuz]

Dear MHB members,

Suppose that $p,f$ are locally essentially bounded Lebesgue measurable functions and consider the differential equation
$x'(t)=p(t)x(t)+f(t)$ almost for all $t\geq t_{0}$, and $x(t_{0})=x_{0}$.
By a solution of this equation, we mean a function $x$,
which is absolutely continuous in $[t_{0},t_{1}]$ for all $t_{1}\geq t_{0}$,
and satisfies the differential equation almost for all $t\geq t_{0}$ and $x(t_{0})=x_{0}$.

How can I prove existence and uniqueness in the sense of almost everywhere of solutions to this problem?

Thanks.
bkarpuz

Here is the complete proof.

Proof. Existence. Pick some $t_{1}\in[t_{0},\infty)$, and consider the differential equation
$\begin{cases}
x^{\prime}(t)=p(t)x(t)+f(t)\quad\text{almost for all}\ t\in[t_{0},t_{1}]\\
x(t_{0})=x_{0}.
\end{cases}$____________________________(1)
Now, define the corresponding integral operator $\Gamma:\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})\to\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$ by
$(\Gamma{}x)(t):=x_{0}+\int_{t_{0}}^{t}\big[p(\eta)x(\eta)+f(\eta)\big]\mathrm{d}\eta$ for $t\in[t_{0},t_{1}]$.
Obviously, $\Gamma\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})\subset\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$.
Let $\{y_{k}\}_{k\in\mathbb{N}_{0}}\subset\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$ to be the sequence of Picard iterates defined by
$y_{k}(t):=
\begin{cases}
x_{0},&k=0\\
(\Gamma{}y_{k-1})(t),&k\in\mathbb{N}_{0}
\end{cases}\quad\text{for}\ t\in[t_{0},t_{1}].$___________________________(2)
We may find $M_{1},M_{2}\in\mathbb{R}^{+}$ such that $\|y_{1}-x_{0}\|_{\mathrm{ess}}\leq{}M_{1}$ and $\|p\|_{\mathrm{ess}}\leq{}M_{2}$ and show by induction that
$|y_{k}(t)-y_{k-1}(t)|\leq{}M_{1}M_{2}^{k-1}\frac{(t-t_{0})^{k-1}}{(k-1)!}$ for all $k\in\mathbb{N}$.
Since
$\sum_{\ell=0}^{\infty}M_{1}M_{2}^{\ell}\frac{(t-t_{0})^{\ell}}{\ell!}=M_{1}\mathrm{e}^{M_{2}(t-t_{0})}\leq{}M_{1}\mathrm{e}^{M_{2}(t_{1}-t_{0})}$ for all $t\in[t_{0},t_{1}]$,
we see that the sequence $\big\{y_{k}=x_{0}+\sum_{\ell=0}^{k-1}[y_{\ell+1}-y_{\ell}]\big\}_{k\in\mathbb{N}_{0}}$ converges uniformly due to Weierstrass $M$-test.
Let $y:=\lim_{k\to\infty}y_{k}$, we have $y\in\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$ since $\mathcal{L}^{\infty}([t_{0},t_{1}],\mathbb{R})$ is a complete Banach space.
Obviously, $y(t_{0})=x_{0}$. Letting $k\to\infty$ in (2) implies that the fixed point of $\Gamma$ is $y$, i.e., $y=\Gamma{}y$ on $[t_{0},t_{1}]$.
Due to the Fundamental theorem of calculus for the Lebesgue integral,
we see that $y\in\mathrm{AC}([t_{0},t_{1}],\mathbb{R})$ and $y^{\prime}(t)=p(t)y(t)+f(t)$ almost for all $t\in[t_{0},t_{1}]$.
The proof of existence of a solution to (1) is therefore completed.

Uniqueness. Assume that there exist two solutions $x,y\in\mathrm{AC}([t_{0},t_{1}],\mathbb{R})$, define $z\in\mathrm{C}([t_{0},t_{1}],\mathbb{R}_{0}^{+})$ by
$z(t):=\sup_{\xi\in[t_{0},t]}|x(\xi)-y(\xi)|$ for $t\in[t_{0},t_{1}].$
Note that $z$ is monotone.
Then, we have
$|x(t)-y(t)|\leq{}M_{2}\int_{t_{0}}^{t}p(\eta)z(\eta)\rm{d}\eta$ for all $t\in[t_{0},t_{1}],$
which yields
$z(t)\leq M_{2}\sup_{\xi\in[t_{0},t]}\bigg\{\int_{t_{0}}^{\xi}p(\eta)z(\eta)\rm{d}\eta\bigg\}=0+M_{2}\int_{t_{0}}^{t}p(\eta)z(\eta)\rm{d}\eta$ for all $t\in[t_{0},t_{1}]$.
By an application of the Grönwall's inequality, we see that
$z(t)\leq0\cdot\mathrm{e}^{M_{2}(t-t_{0})}=0$ for all $t\in[t_{0},t_{1}]$
showing that $x=y$ on $[t_{0},t_{1}]$.
Hence, the uniqueness of solutions to (1) is proved.

Since $t_{1}$ is arbitrary, we may let $t_{1}\to\infty$ to complete the proof of existence and uniqueness of solutions to
$\begin{cases}
x^{\prime}(t)=p(t)x(t)+f(t)\quad\text{almost for all}\ t\in[t_{0},\infty)\\
x(t_{0})=x_{0}.
\end{cases}$