# Thread: Sovling a differential equation

1. ## Sovling a differential equation

Hello folks.

Let

$\displaystyle A = ]0,1[^2, \quad f(x) \in C^2(\overline{A})$

Find a solution f(x) for the following problem

$\displaystyle -f''(x) = const =: c > 0, \mbox{ iff } x \in A$

$\displaystyle f(x) = 0, \mbox{ iff } x \in \partial A$

My try:

$\displaystyle -f(x) = \int \int c \ dx dx = \int c x + b dx = 0.5 c x^2 + bx + a$

$\displaystyle f(0) = a = 0$

$\displaystyle f(1) = 0.5 c + b = 0 \Rightarrow c = -2b$

So f(x) should be equal to $\displaystyle 0.5 c x^2 -2b$

Rapha

2. Originally Posted by Rapha
Hello folks.

Let

$\displaystyle A = ]0,1[^2, \quad f(x) \in C^2(\overline{A})$

Find a solution f(x) for the following problem

$\displaystyle -f''(x) = const =: c > 0, \mbox{ iff } x \in A$

$\displaystyle f(x) = 0, \mbox{ iff } x \in \partial A$

My try:

$\displaystyle -f(x) = \int \int c \ dx dx = \int c x + b dx = 0.5 c x^2 + bx + a$

$\displaystyle f(0) = a = 0$

$\displaystyle f(1) = 0.5 c + b = 0 \Rightarrow c = -2b$

So f(x) should be equal to $\displaystyle 0.5 c x^2 -2b$

You just proved that c= -2b so $\displaystyle b= -\frac{1}{2}c$.
$\displaystyle f(x)= \frac{1}{2}cx^2- \frac{1}{2}cx= \frac{1}{2}c(x^2- x)$