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From a system of PDEs where i used the following ansatz: $$\theta_w(x,y) = e^{-\beta_h x} f(x) e^{-\beta_c y} g(y)$$. $F(x) := \int f(x) \, \mathrm{d}x$ and $G(y) := \int g(y) \, \mathrm{d}y$ So, $$\theta_w(x,y) = e^{-\beta_h x} F'(x) e^{-\beta_c y} G'(y)$$

I have the following two third order linear ODEs which have been arrived at after applying separation of variables \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,\\ 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, \end{eqnarray}

$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are

For $F$: $$F(0)=0$$ $$\frac{F''(0)}{F'(0)}=\beta_h$$ $$\frac{F''(1)}{F'(1)}=\beta_h$$ $\lambda_h$, $\beta_h$ and $V$ are constants $>0$, while $\mu$ is the constant of separation. I decided to apply the Laplace transform to each ODE and find the functions $F$ and $G$ individually finally obtaining $\theta_w$.

Applying Laplace transform

($\bar F$ is $\mathcal{L}F(x)$) $$\lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2=0$$

Taking constants before each term as $a$, $b$, $c$ and $d$ makes $a=\lambda_h$, $b=-2\lambda_h\beta_h$, $c=(\lambda_h\beta_h-1)\beta_h-\mu$ and $d={\beta_h}^2$, we have

$$a[s^3\bar F(s)-s^2 F(0)-sF'(0)-F''(0)]+b[s^2\bar F(s)-sF(0)-F'(0)]+c[s\bar F(s)-F(0)]+d[\bar F(s)]=0$$ I know that the seperation constant $\mu$ is contained in $c$ and is not a known quantity and can attain values $<,=,>0$. Ultimately after applying the first 2 b.c. i arrive at:

$$\bar F(s)[as^3+bs^2+cs+d]=F'(0)[s+b+a\beta_h]$$

I cannot proceed from here, as $F'(0)$ is not known. Also i have doubts whether my approach here to solve two variable separated ODEs with a separation constant using Laplace transform is correct in the first place or not ?

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  • $\begingroup$ @Christoph any thoughts ? $\endgroup$ – Indrasis Mitra Jan 13 at 9:05
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I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.

In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(\lambda) = a \lambda^3 + b \lambda^2 + c \lambda + d$.

Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.

  1. If all zeros are real, then the solution takes one of the forms \begin{equation} y = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} + C_3 e^{\lambda_3 x}, \quad y = (C_1 + C_2 x) e^{\lambda_1 x} + C_3 e^{\lambda_3 x}, \quad y = (C_1 + C_2 x + C_3 x^2) e^{\lambda_1 x}, \end{equation} depending on the multiplicity of the zeros.

  2. If only one zero is real then the solution will be of the form \begin{equation} y = C_1 e^{\lambda_1 x} + e^{-b_1 x} \left( C_2 \cos(b_2 x) + C_3 \sin(b_2 x) \right). \end{equation}

You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!

Finally, this distinction should not be made according to the sign of the separation constant $\mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $\mu$.

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  • $\begingroup$ much appreciation for these points. So if i am understanding correctly, no general $F$ and $G$ expressions are possible and one needs to take individual $\lambda_h$ $\beta_h$ values, see the behaviour of the discriminant of $P$, arriving at a general expression for F, repeat the same for $G$ and ultimately arrive at a $\theta_w$ expression for those particular set of values $\endgroup$ – Indrasis Mitra Jan 13 at 17:00
  • $\begingroup$ Well, in principle I think that you can derive conditions on the parameters $\lambda_h, \beta_h$ in order to fall within a specific case, but I guess with third-order polynomials this would become rather complicated. It may be easier the way you described it. $\endgroup$ – Christoph Jan 13 at 17:38

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