text: Q:
  
  Solvability of inhomogeneous linear differential equation
  
  In applied mechanics the following equation is common.
  $$w^{(4)}+k^2 w''=f$$ defined on (0,1),subject to homogeneous boundary conditions:
  $$w(0)=w''(0)=w(1)=w''(1)=0$$. It seems that if $ k$ is an eigen value such that the homogeneous version of the above equation $w^{(4)}+k^2 w''=0$ has an eigenfunction $w^*$, then it seems necessary that:
  $$\int_0^1 f w^*=0 $$ to ensure there is a solution to the original inhomogeneous equation. Although this fact is used commonly in applied mechanics analysis, I don't see the way to prove it. Could any one give a proof of it.  
  
  A:
  
  For the integral
  $$\begin{align} 
  &\quad \int_0^1 w^{(4)}w^* + k^2 \int_0^1 w''w^* \\
  &= w'''w^*\Big|_0^1 - \int_0^1 w'''{w^*}' + k^2w'w^*\Big|_0^1 - k^2\int_0^1w'{w^*}' \\
  &= w'''w^*\Big|_0^1 + \int_0^1 w''{w^*}'' + k^2w'w^*\Big|_0^1 + k^2\int_0^1w{w^*}'' \\
  &= w'''w^*\Big|_0^1 + w'{w^*}''\Big|_0^1 - \int_0^1 w' {w^*}''' + k^2w'w^*\Big|_0^1 + k^2\int_0^1w{w^*}'' \\
  &= w'''w^*\Big|_0^1 + w'{w^*}''\Big|_0^1 + \int_0^1 w {w^*}^{(4)} + k^2w'w^*\Big|_0^1 + k^2\int_0^1w{w^*}'' \\
  &= w'''w^*\Big|_0^1 + w'({w^*}'' + k^2w^*)\Big|_0^1 
  \end{align} $$
  without any information on the odd derivatives, this can't be simplified further
  
  For the equation, you can use Green's function to solve it. If $u(x,s)$ satisfies
  $$ u^{(4)}(x,s) + k^2u''(x,s) = \delta(x-s) $$
  Then $$ w(x) = \int_0^1 u(x,s)f(s)\ ds $$
  Let's solve for $u$. For $x \ne s$ the general solution is
  $$ u(x,s) =c_0 + c_1x + c_2\cos(kx) + c_3\sin(kx) = 0 $$
  where $c_i$ are dependent on $s$
  Using the B.C. at $x=0$ for $x<s$ and at $x=1$ for $x > s$ we obtain the piecewise solution
  $$ u(s,x) = \begin{cases} a_1x + a_2\sin (kx), && x < s \\
  b_1(1-x) + b_2\sin\big(k(1-x)\big), && x > s \end{cases} $$
  For continuity, we require
  $$ \begin{align} 
  u(s,s) &= a_1s + a_2\sin(ks) = b_1(1-s) + b_2\sin((k(1-s)) \\ \\ 
  u'(s,s) &= a_1 + ka_2\cos(ks) = -b_1s - kb_2\cos(k(1-s)) \\ \\
  u''(s,s) &= -k^2a_2\sin(ks) = -k^2b_2\sin(k(1-s))
  \end{align}$$
  We also need discontinuity in the third derivative (which you can check by integrating the original equation throughout)
  $$ u'''(s^+,s) - u'''(s^-,s) = k^3a_2\cos(ks) + k^3b_2\cos(k(1-s)) = 1 $$
  Solving the above system gives
  $$ \begin{array}{ll} 
  a_1 = \dfrac{1-s}{k^2(1-s+s^2)}, && b_1 = \dfrac{s}{k^2(1-s+s^2)} \\ 
  a_2 = \dfrac{\sin(k(1-s))}{k^3\sin k}, && b_2 = \dfrac{\sin(ks)}{k^3\sin k}
  \end{array} $$
  
  Note that if $k=0$, the solution takes a different form
  $$ u(x,s) = \begin{cases} a_1x^3 + a_2x, && x < s \\ 
  b_1(1-x)^3 + b_2(1-x), && x > s \end{cases} $$
  with the same conditions as before
  $$ \begin{array}{ll}
  a_1 = -\dfrac{1-s}{6}, && b_1 = -\dfrac{s}{6} \\ 
  a_2 = \dfrac{s(1-s)(2-s)}{6}, && b_2 = \dfrac{s(1-s)(1+s)}{6} 
  \end{array} $$
  
  