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Consider the second-order linear PDE $$ y_t(x,t) = y_{xx}(x,t) - a^2y(x,t) $$ where $a > 0$ in all cases and the equation is restricted to the domain $x = [0, X]$. If we have some way of expressing $y(x,t)$ as e.g. $$ y(x,t) = f(x)g(t) $$ where both $f(x)$ and $g(t)$ are known, and given boundary conditions \begin{align} y(0, t) &= y_0(t) \\ y_t(0,t) &= y'_0(t) \\ y_x(X, t) &= 0 \end{align}

is there an analytical or approximate solution to the PDE?

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  • $\begingroup$ This is a PDE not an ODE and the answer is yes, see en.wikipedia.org/wiki/Separation_of_variables $\endgroup$
    – Gregory
    Nov 30, 2017 at 19:41
  • $\begingroup$ Do you mean $\partial y / \partial y$ and $\partial^2 y / \partial x^2$? $\endgroup$
    – jacques
    Nov 30, 2017 at 19:42
  • $\begingroup$ Thank you @Jakob, I fixed my notation. $\endgroup$
    – mikeck
    Nov 30, 2017 at 20:11
  • $\begingroup$ Based on your edit, it appears you are stuck at solving ODEs, have you solved second order ODEs before? Particularly ones with constant coefficients? Try the solution $f(x) = e^{rt}$ and find the $r$ which satisfies the resulting characteristic equation. $\endgroup$
    – Gregory
    Dec 1, 2017 at 0:06

2 Answers 2

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You can solve it with the reference @Gregory has provided you. Also, by substituting $y(t,x) = e^{-a^2t}z(t,x)$ you can transform your equation into $$\frac{\partial z}{\partial t} = \frac{\partial ^2z}{\partial x^2}$$ a cannonical heat equation which has a ready solution.

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  • $\begingroup$ Thanks @Hans, could you elaborate? I updated my question to include my attempt to use separation of variables but am curious about your solution. $\endgroup$
    – mikeck
    Nov 30, 2017 at 21:30
  • $\begingroup$ Have you tried the substitution I mentioned in the answer? See if you get the same PDE I wrote in my answer. Then check the heat equation link. By the way, the function in your boundary condition is confusing, as there is only one variable in the functions and it is not clear which variable the $'$ is differentiating. $\endgroup$
    – Hans
    Nov 30, 2017 at 23:24
  • $\begingroup$ You need one more boundary condition, e.g. at x=0, to guarantee the uniqueness of the solution unless you do not need the uniqueness. $\endgroup$
    – Hans
    Dec 4, 2017 at 21:59
  • $\begingroup$ Thanks @Hans, I have worked through it and I think knowing $y_t(0,t)$ as a third boundary condition is sufficient? $\endgroup$
    – mikeck
    Dec 4, 2017 at 22:11
  • $\begingroup$ What do you think of the relationship between your first and second boundary condition is? How many boundary conditions do you think you actually have now? $\endgroup$
    – Hans
    Dec 5, 2017 at 6:07
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This is directly solvable with separation of variables as commented by @Gregory---the substitution in the accepted answer is not actually required and leads to the same solution as below.

General solution

Given the PDE

$$ y_t(x,t) = y_{xx}(x,t) - a y(x,t) $$ use separation of variables to express $y(x,t)$ as \begin{equation} y(x,t) = f(x)g(t) \end{equation} Substituting this into the first equation yields \begin{equation} \begin{split} & \partial_t \left( f(x)g(t)\right) = \partial_{xx} \left( f(x) g(t) \right) - a f(x)g(t) \\ \\ \rightarrow~~~& f(x)g_t(t) = f_{xx}(x)g(t) - af(x)g(t) \\ \\ \rightarrow~~~& \frac{g_t(t)}{g(t)} + a = \frac{f_{xx}(x)}{f(x)} \end{split} \end{equation} Setting each side of the equation equal to some positive constant $\lambda$ yields two ODEs \begin{equation} \begin{split} g_t(t) &= \left(\lambda - a \right) g(t) \\ \\ f_{xx}(x) &= \lambda f(x) \end{split} \end{equation} The solutions to the above ODEs are \begin{equation} \begin{split} g(t) &= Be^{\left(\lambda - a \right)t} \\ \\ f(x) &= A_1e^{x\sqrt{\lambda}} + A_2e^{-x\sqrt{\lambda}} \end{split} \end{equation} where $A_1$, $A_2$ and $B$ are constants of integration. Combining the two solutions yields \begin{equation} \begin{split} y(x,t) &= \left( A_1e^{x\sqrt{\lambda}} + A_2e^{-x\sqrt{\lambda}} \right)Be^{\left(\lambda - a \right)t} \\ &= \left( C_1e^{x\sqrt{\lambda}} + C_2e^{-x\sqrt{\lambda}} \right)e^{\left(\lambda - a \right)t} \end{split} \end{equation} where $C_1$ and $C_2$ are arbitrary constants.

Unique solution

A unique solution will require expressions for $C_1$, $C_2$ and $\lambda$. The boundary condition $y_x(x = X, t) = 0$ at the outside boundary results in $$ 0 = \left( C_1e^{X\sqrt{\lambda}} - C_2e^{-X\sqrt{\lambda}} \right)e^{\left(\lambda - a \right)t}\sqrt{\lambda} $$ which simplifies to $$ C_2 = C_1e^{2X\sqrt{\lambda}} $$

$C_1$ and $\lambda$ can be solved for given known values of $y(0,0)$ and $y_t(0,0)$; however, this results in expressions \begin{equation} \begin{split} C_1 &= \frac{y_{00}}{1 + e^{2X\sqrt{\lambda}}} \\ \\ \lambda &= \frac{y'_{00}}{y_{00}} + a \end{split} \end{equation} yielding $$ y(x,t) = \frac{y_{00}}{1 + e^{2X\sqrt{\frac{y'_{00}}{y_{00}} + a}}} \left( e^{x\sqrt{\frac{y'_{00}}{y_{00}} + a}} + e^{\left(2X - x \right)\sqrt{\frac{y'_{00}}{y_{00}} + a}} \right)e^{\left(\frac{y'_{00}}{y_{00}} \right)t} $$ This solution does not intuitively seem correct to me because it does not depend on $y(0,t)$ when $t > 0$ (only depends on the initial value). $y(0,t)$ is continuous but not constant, and I would expect it to factor into the solution for all $t$.

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