# ODE existence and uniqueness

For $u''(x) + u'(x) = f(x)$, and $u'(0) = u(0) = 1/2\bigl(u'(l) + u(l)\bigr)$, with $f$ a given function, is the solution unique, and why/why not?

Also, does a solution necessarily exist or is there a condition that $f$ must satisfy for existence?

I'm thinking of an integrating factor but that didn't quite get me anywhere.

Thanks

-

(In a comment you asked for the "general theory")

The ODE $$u''(x)+u'(x)=f(x)\qquad ( a<x<b)\ ,\qquad(*)$$ where $f$ is continuous on $\ ]a,b[\$, $\ -\infty\leq a<b\leq\infty$, is an inhomogeneous linear equation of second order, with constant coefficients. The general theory about such equations says the following:

1. All solutions are defined on all of $\ ]a,b[\$.

2. The associated homogeneous ODE $u''+u'=0$ has the solutions $x\mapsto A+Be^{-x}$ with arbitrary constants $A$, $B$. One arrives at these solutions by studying the characteristic polynomial $\chi(\lambda):=\lambda^2+\lambda$ of the given ODE.

3. The general solution of the given ODE $(*)$ is of the form $$u(x)=u_p(x) + A +Be^{-x}\ ,$$ where the so-called particular solution $u_p(\cdot)$ can be any single solution of $(*)$ found by whatever means (guesswork, "Ansatz", "variation of constants", Laplace transform, etc.).

4. When $f$ is a function of the form $x\mapsto x^n e^{\gamma x}$, or a linear combination of such functions, then a $u_p$ can be found by means of linear algebra.

In your example you can substitute $u'(x):=y(x)e^{-x}$ and will then get $y'(x)=f(x)e^x$, or $y(x)=y_0+\int_0^x f(t)e^t\ dt$. This leads to $u'(x)=y_0e^{-x}+\int_0^x f(t)e^{t-x}\ dt$. Integrating once more we have $$u(x)=y_1-y_0e^{-x}+\int_0^x\ \int_0^{x'}f(t)e^{t-x'}\ dt\ dx'=y_1-y_0e^{-x}+\int_0^x f(t)\bigl(1-e^{t-x}\bigr)\ dt\ .$$ Note that the terms $y_1-y_0e^{-x}$ correspond to $A +Be^{-x}$ in the above general setup.

But we are not done yet: The given problem may still have $0$, $1$, or an infinity of solutions. This has to do with the somewhat nonstandard boundary conditions. I don't think there is a "general theory" covering exactly your data. Therefore we have to proceed ad hoc.

One of your conditions is $u(0)=u'(0)$. Now $u(0)=y_1-y_0$ and $u'(0)=y_0$. Therefore necessarily $y_1=2y_0$. To satisfy the boundary condition at $L$ one has to collect the $f$-integrals for $u$ and $u'$, and then check whether $y_0$ can be chosen suitably.

If my calculations are correct, the answer is the following: When $\int_0^Lf(t)\ dt=0$ then the problem has infinitely many solutions, otherwise none.

-

We can try a direct approach, without using general theorems.

The equation can be written as $e^x(u''(x)+u'(x))=e^xf(x)$, that is $\frac d{dx}(e^xu'(x))=e^xf(x)$. We get $e^xu'(x)=\int_0^xe^tf(t)dt+C_1$, hence $u'(x)=e^{-x}\int_0^xe^tf(t)dt+C_1e^{-x}$. Integrating, this will give a second constant $C_2$, and you can determine $C_1$ and $C_2$ thanks to the initial conditions.

-
Can you also explain how we can use the general theorems? That may be more beneficial for me. Thanks! –  mary Sep 11 '12 at 9:42
You can transform the equation as a first order equation (but you will work in dimension $2$) and check the conditions of Cauchy-Lipschitz theorem. –  Davide Giraudo Sep 11 '12 at 9:46

One may rewrite the ODE $$\tag{1} u''(x)+u'(x)=f(x)$$ as $$\tag{2} w'(x)=Aw(x)+f(x)e_2,$$ where $w={u\choose u'}$, and $A=\left[\begin{array}{cc}0&1\\0&-1\end{array}\right]$, with $e_1, e_2$ the standard basis of $\mathbb{R}^2$. Solving (2) yields $$w(x)=e^{xA}w(0)+\int_0^xf(s)e^{(x-s)A}e_2ds.$$ Since \begin{eqnarray} e^{xA}&=&I_2+\sum_{k=1}^\infty\frac{x^{2k}}{(2k)!}A^2+\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}A\cr &=&I_2+(\cosh(x)-1)A^2+\sinh(x)A\cr &=&\left[\begin{array}{cc}1&1-e^{-x}\cr0&e^{-x}\end{array}\right], \end{eqnarray} it follows that $$u(x)=u(0)+u(0)(1-e^{-x})+\int_0^x(1-e^{x-s})f(s)ds.$$ Because $$2u(0)=u(l)+u'(l)=2u(0)+\int_0^l(1-2e^{l-s})f(s)ds,$$ $f$ and $l$ must satisfy the condition $$\tag{F} \int_0^l(1-2e^{l-s})f(s)ds=0.$$

Note For $f(x)=a \ne 0$ for every $x$, and $l \ne 0$ the condition (F) reads $$l+2-2e^l=0.$$ Clearly (F) is not satisfied since $s+2-2e^s \ne 0$ for every $s \ne 0$. Thus the condition (F) is not always satisfied!

-