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To solve a Partial Differential Equation with solution $ u(x,y) $, I separated $u$ into two variables, $u(x,y) = h(x)g(y)$.

For background: the resulting differential equations were: $ \frac{h''}{h} = -\frac{g''}{g} = \lambda$

My solution got:

For h:

$\lambda = 0 \implies h = c_1$ where $c_1$ is any constant.

$\lambda < 0 \implies -\lambda = (\frac{n\pi}{L})^2$ with corresponding eigenfunction $h_n = \cos{(n \pi x / L)} $ for $ n = 1, 2, 3, ...$

For g:

$\lambda = 0 \implies$ eigenfunction is $ g = y$

$\lambda < 0 \implies$ eigenfunction is $ g_n = \sinh(n \pi y / L)$

How do I now use the principle of superposition to come up with the next step of a solution? I had thought the answer would be:

$u(x,y) = \sum_{n = 1}^{\infty} a_n \cos{(n \pi x / L)} \sinh(n \pi y / L) + \sum_{n = 1}^{\infty} b_n \cos{(n \pi x / L)} y $

But my textbook gives the next step as:

$u(x,y) = c_0 y + \sum_{n = 1}^{\infty} a_n \cos{(n \pi x / L)} \sinh(n \pi y / L) $

Why is this?

Vijay Ramanujan
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1 Answers1

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By looking at $h$ you have found what the eigenvalues are: $$ \lambda_0=0 ,\qquad \lambda_n=-(n\pi/L)^2 \quad (n \ge 1) . $$ To each eigenvalue you have a corresponding function $h_n(x)$.

Then for each eigenvalue you solve $$ g_n''(y)=-\lambda_n g_n(y) ,\qquad g_n(0)=0 , $$ to find $g_n(y$).

If you combine $h_n$ and $g_n$, you get a separated solution to the PDE for each $n \ge 0$: $$ u_n(x,y)=h_n(x) \, g_n(x) . $$

These separated solutions are then put together in a linear combination $$ u(x,y) = \sum_{n \ge 0} c_n \, u_n(x,y) , $$ where the constants $c_n$ are chosen such that $u(x,t)$ satisfies the last boundary condition $u(x,H)=f(x)$.

Hans Lundmark
  • 53,395