While solving a boundary value problem (background provided in the Context section) I reach the following variable separated two equations ($F(x)$ and $G(y)$)

\begin{eqnarray} \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2 F &=& 0, \tag 1\\ V \lambda_c G''' - 2 V \lambda_c \beta_c G'' + \left( (\lambda_c \beta_c - 1) V \beta_c + \mu \right) G' + V \beta_c^2 G &=& 0,\tag 2 \end{eqnarray} with some separation constant $\mu \in \mathbb{R}$.

with boundary conditions as:

For G: $G'(0)=0, G(0)=0$ and $\frac{G''(1)}{G'(1)}=\beta_c$

For F: $F'(0)=0$ and $\frac{F''(1)}{F'(1)}=\beta_h$

The non-homogeneous condition in $F$ is: $\beta_h e^{-\beta_c y}G'(y)F(0)=1$

Since the boundary conditions for $(2)$ are all homogeneous, I have calculated the eigenvalues $\mu_i$ for $(2)$ numerically. For some realistic parameters like βc = 0.921, λc = 1.775*10^-4, V=1, these eigen values are

0.834041, 0.845661, 0.864286, 0.888675, 0.916951, 0.94696, 0.977271, 1.0079, 1.03972, 1.07361, 1.11015,...

Now since these eigenvalues are numeric in nature, I cannot figure out how to move forward with this problem.

I know from standard PDE problems that these eigen values should be utilized to build the $G$ solution and then should be employed in the $F$ non-homogeneous condition to determine the constants. But how am I supposed to even use orthogonality here ? I would really appreciate if someone could give a step-wise way forward for such problems where the EVs are numeric.


I had the following system of PDEs $$\frac{\partial \theta_h}{\partial x}+\beta_h (\theta_h-\theta_w) = 0 \tag A$$

$$\frac{\partial \theta_c}{\partial y} + \beta_c (\theta_c-\theta_w) = 0 \tag B$$

$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V\frac{\partial^2 \theta_w}{\partial y^2}-\frac{\partial \theta_h}{\partial x} - V\frac{\partial \theta_c}{\partial y} = 0 \tag C$$

with the boundary conditions ($\beta_h, \beta_c, V, \lambda_h, \lambda_c$ are constants)

$$\theta_w(0,y)=1, \theta_w(x,0)=0$$ $$\frac{\partial \theta_w(1,y)}{\partial x}=\frac{\partial \theta_w(x,1)}{\partial y}=0$$

$$\theta_h(0,y)=1, \theta_c(x,0)=0$$ Using the transformation $\theta_{h1}(x,y):=\theta_h (x,y)-1$ (so that $\theta_w(0,y)=0$ which is needed to get an additional homogeneous condition on $F$) where

\begin{eqnarray} \theta_{h1}(x,y) &=& \beta_h e^{-\beta_h x} \int e^{\beta_h x} (\theta_w(x,y)-1) \, \mathrm{d}x,\\ \theta_c(x,y) &=& \beta_c e^{-\beta_c y} \int e^{\beta_c y} \theta_w(x,y) \, \mathrm{d}y. \end{eqnarray}

which are substituted in $(C)$ to get:

$$\lambda_h \frac{\partial^2 \theta_w}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta_w}{\partial y^2} +( -\beta_h - V \beta_c )\theta_w +\beta_h^2 e^{-\beta_h x} \int e^{\beta_h x} \theta_w(x,y) \mathrm{d}x + \beta_c^2 e^{-\beta_c y}\int e^{\beta_c y} \theta_w(x,y)\mathrm{d}y = 0 \tag D$$

\begin{eqnarray} \rightarrow 0 &=& e^{-\beta_h x} \left( \lambda_h e^{\beta_h x} \frac{\partial^2 \theta_w}{\partial x^2} - \beta_h e^{\beta_h x} \theta_w + \beta_h^2 \int e^{\beta_h x} \theta_w \, \mathrm{d}x \right) +\\ && + V e^{-\beta_c y} \left( \lambda_c e^{\beta_c y} \frac{\partial^2 \theta_w}{\partial y^2} - \beta_c e^{\beta_c y} \theta_w + \beta_c^2 \int e^{\beta_c y} \theta_w \, \mathrm{d}y \right). \tag E \end{eqnarray}

Using the ansatz $\theta_w(x,y) = e^{-\beta_h x} f(x) e^{-\beta_c y} g(y)$ on $(E)$ we obtain

$$\Bigg(\frac{\lambda_h (f^{''}-2\beta_h f^{'}+\beta_h^2 f)}{f} - \beta_h + \frac{\beta_h^2 \int f}{f}\Bigg) = \\ -V\Bigg(\frac{\lambda_c (g^{''}-2\beta_c g^{'}+\beta_c^2 g)}{g} - \beta_c + \frac{\beta_c^2 \int g}{g}\Bigg) = \mu \tag F $$

Each term on $(F)$ is a function of either only $x$ or only $y$ which is only possible if they are equal to a constant $\mu$

Thus taking:

$$ \Bigg(\frac{\lambda_h (f^{''}-2\beta_h f^{'}+\beta_h^2 f)}{f} - \beta_h + \frac{\beta_h^2 \int f}{f}\Bigg) = \mu \tag i $$

On solving $i$, and using $F(x) := \int f(x)$ to remove the integral we get:

$$ \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2 F = 0, \tag 1\\ $$

Similar manipulations for $G$ gives $(2)$

  • $\begingroup$ Just to make sure, we are solving a system of two linear higher order PDE's with a non-linear boundary condition, where $F$ and $G$ are functions, $x$ and $y$ are unknowns, $\mu$ are the eigenvalues and everything else is a known constant. Is that right? So, what do you want - an analytic or numeric solution? $\endgroup$ Jun 1, 2020 at 22:30
  • $\begingroup$ More questions. Your original problem looks to be a simple linear PDE. Isn't it simpler to just solve the original PDE numerically, or via separation of variables? Finally, the original PDE does seem to be an eigenvalue problem. At what point is $\mu$ introduced? $\endgroup$ Jun 1, 2020 at 22:35
  • $\begingroup$ @AleksejsFomins Thanks for your interest. I am seeking analytical solutions using the method of separation of variables. I have added more steps in the context section to explain how I reached $(1)$ and $(2)$. In other words the ODEs posted are a result of applying SOV. Now one of these ODEs, the $(2)$, has only homogeneous conditions for which I calculated the Eigen values using a numerical method. It is at this point where I get stuck. I must now concede that the solution will be semi-analytical since, EVs are being calculated numerically $\endgroup$
    – Avrana
    Jun 2, 2020 at 6:46

2 Answers 2


here's an attempt at solution with slightly different path than yours. I was taught to first solve all of the equations in a general form, and only care about boundary conditions at the very end when fixing constants. I would start directly with of separation of variables

$$ \begin{eqnarray} \theta_w (x,y) &=& X_w(x)Y_w(y) \\ \theta_h (x,y) &=& X_h(x)Y_h(y) \\ \theta_c (x,y) &=& X_c(x)Y_c(y) \end{eqnarray} $$

Substituting these into the original PDE we get

$$ \begin{eqnarray} \dot{X}_h Y_h + \beta_h (X_h Y_h - X_w Y_w) &=& 0 \tag{A} \\ X_c \dot Y_c + \beta_c (X_c Y_c - X_w Y_w) &=& 0 \\ \lambda_h \ddot{X}_w Y_w + \lambda_c V X_w \ddot Y_w - \dot{X}_h Y_h - V X_c \dot{Y}_c &=& 0 \end{eqnarray} $$

Rearranging the terms so that X's are on one side, and Y's are on the other, we get the following for the first two equations

$$ \begin{eqnarray} \frac{\dot{X}_h}{X_h} + \beta_h &=& \beta_h \frac{X_w}{X_h} \frac{Y_w}{Y_h} \\ \frac{\dot{Y}_c}{Y_c} + \beta_c &=& \beta_c \frac{X_w}{X_c} \frac{Y_w}{Y_c} \end{eqnarray} $$

Since LHS only depends on one variable, and RHS depends on both, we conclude that

$$ \begin{eqnarray} Y_w &=& \eta_h Y_h \\ X_w &=& \eta_c X_c \end{eqnarray} $$

Where $\eta$'s are some proportionality constants. With this in mind, the first two equations become

$$ \begin{eqnarray} \dot{X}_h + \beta_h X_h &=& \beta_h \eta_h X_w \tag B\\ \dot{Y}_c + \beta_c Y_c &=& \beta_c \eta_c Y_w \end{eqnarray} $$

Rearranging the terms in the 3rd equation, and substituting the two identities we have just found, we arrive at

$$ \begin{eqnarray} \lambda_n \ddot{X}_w - \mu X_w &=& \dot{X}_h \tag C\\ \lambda_c \ddot{Y}_w + \frac{\mu}{V} Y_w &=& \dot{Y}_c \\ \end{eqnarray} $$

where $\mu$ is a separation constant (you called it eigenvalue) in your solution.

Now, we could proceed as you suggest, solving the first two equations using an integrating factor. I have tried, but it gets messy. Instead, why don't we try to eliminate w's from the equations. This way, we still have to solve 3rd order ODE's, but at least our constraints are still very simple. Substituting $X_w$ and $Y_w$ into the last equation, we get

$$ \begin{eqnarray} \lambda_n (\dddot{X}_h + \beta_h \ddot{X}_h) - \mu (\dot{X}_h + \beta_h X_h) &=& \beta_h \eta_h \dot{X}_h \\ \lambda_c (\dddot{Y}_c + \beta_c \ddot{Y}_c) + \frac{\mu}{V} (\dot{Y}_c + \beta_c Y_c) &=& \beta_c \eta_c \dot{Y}_c \\ \end{eqnarray} $$

From here on it gets messy, so I will only outline the further solution, not do it fully. We will solve each of the linear 1st order ODE's using ansatz $X_h = e^{a_h x}$ and $Y_c = e^{a_c y}$. This will give you a 3rd order characteristic polynomial, which you (maybe) will solve for 3 roots that satisfy it. Thus, your solutions will be

$$ \begin{eqnarray} X_h &=& A_h e^{a_h x} + B_h e^{b_h x} + C_h e^{c_h x} \\ Y_c &=& A_c e^{a_c y} + B_c e^{b_c y} + C_c e^{c_c y} \end{eqnarray} $$

where the lower-case constants are the roots of your polynomial, that is, functions of $\mu$ and $\beta$'s, $\lambda$'s and $\eta$'s, and the upper-case constants are still unknown. For each equation, you can eliminate one of the 3 capital constants by applying the corresponding boundary condition.

Once this has been done, you can substitute $X_h$ and $Y_c$ into (B), obtaining expressions for $X_w$ and $Y_w$. Then, you can apply the remaining 4 boundary conditions to $X_w$ and $Y_w$, eliminating the remaining 4 capital constants and potentially discretizing the values of $\mu$. Further, you can substitute all of the resulting equations into (A) in order to eliminate $\eta$'s

I expect the final solution to be in the form of series

$$ \begin{eqnarray} \theta_w(x,y) &=& \sum_\mu \alpha_\mu X_w(x, \mu) Y_w(y, \mu) \\ \theta_h(x,y) &=& \sum_\mu \alpha_\mu X_h(x, \mu) Y_w(y, \mu) \\ \theta_c(x,y) &=& \sum_\mu \alpha_\mu X_w(x, \mu) Y_c(y, \mu) \end{eqnarray} $$

I think that the series constants $\alpha_\mu$ will have to be the same for all 3 equations for this to work.

  • $\begingroup$ Thanks for this. Really appreciate it. I will go through this once I sit in front of my primary computer (writing this from phone ) and work out the steps. Will revert back then. $\endgroup$
    – Avrana
    Jun 2, 2020 at 8:55
  • 1
    $\begingroup$ Sure, take your time, this is a biggon to swallow :D $\endgroup$ Jun 2, 2020 at 8:56
  • $\begingroup$ One of the first problem I run into is at the step where I need to determine the capital constants from the derived $X_h$ and $Y_c$ expressions using the respective bc(s). The condition on $\theta_c$ is $\theta_c(x,0)=0$ which easily gives $Y_c(0)=0$ and a capital constant from the $Y_c$ expression can be determined. In contrast the condition on $\theta_h$ is $\theta_h(0,y)=1$ which only translates to $X_h(0)Y_h(y)=1$ and stops me from evaluating a constant in the $X_h$ expression. Nevertheless, I will keep trying. $\endgroup$
    – Avrana
    Jun 2, 2020 at 12:52
  • $\begingroup$ A change of the form $\theta_{h1}(x,y)=\theta_h(x,y) - 1$ might make the bc homogeneous such that $\theta_{h1} (0,y)=0$ but I guess it upends the entire derivation you have performed since the $\theta_h$ equation itself changes. $\endgroup$
    – Avrana
    Jun 2, 2020 at 12:56
  • $\begingroup$ @IndrasisMitra To be perfectly honest, I am not convinced that it is possible to have a nicely looking analytic solution in this case. I have outlined how you can convert differential equations to regular equations. However, these regular equations contain exponents and roots of 3rd order polynomials - it is not guaranteed that a nice analytic solution is possible, even for a sum. Unless you have very strict reasons to seek an analytical series solution, I suspect you would have much easier time solving the whole thing numerically from the start $\endgroup$ Jun 2, 2020 at 13:14

I've been thinking a bit about your problem, and I think I know of another approach that has high likelihood of working. It will be a bit of a pain, but I don't see any impossible steps in it.

  1. Represent each of the three functions using 2D Fourier series
  2. Observe that all equations are linear
    • Thus there is no frequency coupling
    • Thus for every pair of frequencies $\omega_x$, $\omega_y$ there will be a solution from a linear combination of only those terms
  3. Apply boundary conditions directly to each of the three series
    • Note that by orthogonality the boundary condition has to apply to each term of the fourier series
    • Perhaps here your substitution for $\theta_h' = \theta_h-1$ will be good, because
  4. Plug in Fourier series into PDE and solve coefficient matching (see here for example in 1D). Make sure you treat separately the cases when one or both of the frequencies are zero.
  5. If you consider all equations for a given frequency pair, you can arrange them into an equation $M\alpha = 0$, where $\alpha$ are fourier coefficients for that those frequencies, and $M$ is a small sparse matrix (sth like 12x12) that will only depend on the constants.
  6. For each frequency, the allowed solutions will be in the Null space of that matrix. In case you are not able to solve for the null space analytically, it is not a big deal - calculating null space numerically is easy, especially for small matrices.
  • $\begingroup$ Really appreciate your added inputs here. I have been trying to unpack this answer from the last few hours, but honestly speaking it has been quite difficult for me to grasp the steps in its entirety. Although the video link has been really helpful. Conceding the fact that this forum is not meant for full answers and I am supposed to work it out myself, I would nevertheless request you to work out the steps for me at a time you feel convenient. I apologize if this goes against the MSE ethos. Meanwhile I will try to learn more about the 2D Fourier series. $\endgroup$
    – Avrana
    Jun 4, 2020 at 4:11
  • $\begingroup$ @IndrasisMitra I would really like to do it. It sounds cool. However, I am very busy for the next month. I might come back afterwards to have a look at it $\endgroup$ Jun 4, 2020 at 12:37
  • $\begingroup$ I am grateful. Thanks. Meanwhile I will keep on trying.This problem and your answers have been a great learning experience. $\endgroup$
    – Avrana
    Jun 4, 2020 at 12:44
  • $\begingroup$ Hi Aleksejs, This is a gentle inquiry if you could look into this problem if time permits you. Have a nice day. $\endgroup$
    – Avrana
    Jul 14, 2020 at 16:00
  • 1
    $\begingroup$ Oh oh, no problems ! Completely understandable, I too just started my fourth year and can relate how swiftly time passes. Thanks and best wishes to you for your work ahead. $\endgroup$
    – Avrana
    Jul 14, 2020 at 18:07

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