# Showing that the solution of the in-homogeneous ODE: $u''(x)-Cu(x)=-f(x)$ is unique

Given the inhomogeneous ODE on the form: $u''(x)-Cu(x)=-f(x)$

where $0 < x < 1, C > 0$ and $f(x)$ is cont. on the interval for x.

With IC: $u(0) = u(1) = 0$

Solving that yields the general solution for the homogeneous part: $u_c = Ae^\sqrt{C} + Be^{-\sqrt{C}}$

Inserting initial conditions yields the equality $\sqrt{C} = -\sqrt{C}$ (which can't be since C is strictly positive, right?) since we find that $A = -B$

Also how would one go on finding a particular solution when the form of f(x) is unknown?

Update

If discretizing this system would that be a valid method to prove this has a unique solution, or would that only be valid for the discrete system?

Analytically i guess we would use Variation of parameters as follows:

A solution would be of the form: $y = y_c + y_p$

An attempt at an solution using variation of parameters yield:

$y_p = y_1u_1+y_2u_2$

$y_1 = e^{(\sqrt{C})x}$

$y_2 = e^{-(\sqrt{C})x}$

$f = f(x)$

$u_1 = \int \dfrac{-y_2 f}{W(y_1,y_2)} dx = -\int \dfrac{e^{-(\sqrt{C})x}f}{1} dx = -\int {e^{-(\sqrt{C})x}f} dx$

$u_2 = \int \dfrac{y_1 f}{W(y_1,y_2)} dx = \int \dfrac{e^{(\sqrt{C})x}f}{1} dx = \int {e^{(\sqrt{C})x}f} dx$

Where (W = Wronskian determinant):

$W(y_1,y_2) = e^{(\sqrt{C}-\sqrt{C})x} = 1$

A solution would be of the form:

$y = y_c + y_p$

$y_c = 0$ so:

$y = y_p = y_1u_1+y_2u_2 = -e^{(\sqrt{C})x} \int {e^{-(\sqrt{C})x}f} dx + e^{-(\sqrt{C})x} \int {e^{(\sqrt{C})x}f} dx$

Inserting for IC yields (all the expontentials cancel out):

$y(0) = y_p = -\int {f(0)} dx + \int {f(0)} dx = 0$

$y(1) = y_p = - \int {f(1)} dx + \int {f(1)} dx = 0$

So $y = 0$ and does this tell us that the solution is unique? But is 0 a valid solution? This solution differs from the Green kernel method in the answer below, so this solution appears to be wrong.

Now this seems problematic, now am I using the wrong technique or using it wrongly?

• In the inhomogeneous equation, the coefficients of the homogeneous part of the solution are again indeterminate. You can write it as $$y =-e^{(\sqrt{C})x} \Bigl(c_1+\int_0^x e^{-(\sqrt{C})s}f(s) ds\Bigr) + e^{-(\sqrt{C})x} \Bigl(c_2+\int_0^x {e^{(\sqrt{C})s}f(s)} ds\Bigr)$$ and $c_1,c_2$ are to be determined by the boundary conditions. – LutzL Nov 6 '15 at 16:03
• The integrals cancel each other out, right, or how does one solve these integrals? And y is a function of x: $y = y(x)$ With $x = 0$ each integral equals 0 – jibo Nov 7 '15 at 15:49
• So then we know at least that $c_1=c_2$ – jibo Nov 7 '15 at 15:55

It should of course be $$u_c(x) = Ae^{\sqrt{C}x} + Be^{-\sqrt{C}x}$$ giving you the conditions $$A+B=0\\ Ae^{\sqrt{C}} + Be^{-\sqrt{C}}=0$$ which indeed tells you that there are no solution except the trivial solution $u_c\equiv 0$.
For the inhomogeneous equation you could either use variation of constants or use the Green kernel to express the same. For the latter, find the solutions with $u(0)=0=v(1)$ and $u'(0)=1=v'(1)$ which are $$u(x)=\frac1{\sqrt C}\sinh(\sqrt Cx),\qquad v(x)=\frac1{\sqrt C}\sinh(\sqrt C(x-1))$$ resulting in a Green function $$G(x,y)=-\frac1C·\sinh(\sqrt C\min(x,y))·\sinh(\sqrt C(1-\max(x,y)))$$ for the particular solution $$u_p(x)=\int_0^1 G(x,y)f(y)\,dy$$