# Solving the canonical form of an elliptic PDE [HEAT EQUATION]

I have the following PDE

$$\lambda_h \frac{\partial^2 \theta}{\partial x^2} + \lambda_c V \frac{\partial^2 \theta}{\partial y^2} + \theta (k_1) + k_2 = 0$$

This turns out to be an Elliptic second order linear PDE. On converting to the canonical form it takes the following shape

$$\frac{\partial^2 \theta}{\partial \alpha^2}+\frac{\partial^2 \theta}{\partial \beta^2}=q_1\theta+q_2$$ where $$q_1=-k_1/(\lambda_cV),q_2=-k_2/(\lambda_cV), \beta=-px, \alpha=y$$ and $$p^2=\frac{\lambda_cV}{\lambda_h}$$

What should be the approach now to move forward with this canonical form ? Should I introduce a Laplace transform or is there a standard method for such a PDE ?

[The Laplace equation would have a zero on the RHS with no $$\theta$$ term and the Poisson equation a function of $$x$$ and $$y$$ on RHS with no $$\theta$$ term. So this canonical form doesn't correspond to any of them ?]

The PDE needs to be solved on a rectangular region where $$x$$ varies between $$0$$ to $$1$$ and $$y$$ varies between $$0$$ to $$1$$. The boundary conditions are of Neumann type.

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

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

Attempt

As suggested i tried as following using seperation of variables: $$\theta_{\alpha\alpha}+\theta_{\beta\beta}-q_1\theta=0$$

let $$\theta=X(\alpha)Y(\beta)$$

This gives: $$X^{''}Y+XY^{''}-q_1XY=0$$ $$\frac{X^{''}}{X}+\frac{Y^{''}}{Y}-q_1=0$$

This is not turning towards a usual variable separation problem due to the variable $$q_1$$. Any further suggestions ?

• What is the domain you're trying to solve your PDE on? Do you have boundary conditions? – Dmoreno Jan 6 at 11:22
• @Dmoreno I have edited the question to add the boundary conditions. – Indrasis Mitra Jan 6 at 12:19
• Have you tried using separation of variables? – Dmoreno Jan 6 at 15:34
• @Dmoreno No , i encountered this problem while analysing a heat exchanger system. I did go through some material for equations of similar type from standard PDE textbooks but encountered homogenous type and with no $u$ term. They all do use seperation of variables. Should i try the standard u= XY seperation route ? – Indrasis Mitra Jan 6 at 17:28
• Yes, but first notice that your PDE is non-homogenous because of the $q_2$ term. I would first solve for the eigenmodes satisfying the same equation and BCs but for $q_2 = 0$ and then expand. – Dmoreno Jan 6 at 17:47