Question

By the method of separation of variables, solve the boundary value problem

∂u/∂x=4∂u/∂y, u(0,y)=8e^-3y

Answer 1

let the solution is of the form u(x,y)=X(x)Y(y).

put u(x,y) in the given pde.

$Y\frac{\partial X}{\partial x}=4X\frac{\partial Y}{\partial y}\\ \frac{1}{X}\frac{\partial X}{\partial x}=\frac{4}{Y}\frac{\partial Y}{\partial y}=\lambda(say)\\ \text{first solve}\\ \frac{1}{X}\frac{\partial X}{\partial x}=\lambda\\ \implies X=c_1e^{\lambda x}\\ \text{Second solve}\\ \frac{4}{Y}\frac{\partial Y}{\partial y}=\lambda\\ \implies Y=c_1e^{ \frac{\lambda y}{4}}\\ \text{Therefore, the solution is given by}\\ u(x,y)=c_1c_2e^{\lambda(x+\frac{ y}{4})} =ce^{\lambda(x+\frac{ y}{4})}\\ \text{Now, apply the given condition, we get}\\ c=8e^{-y(3+ \frac{\lambda}{4})} \text{Therefore, the solution is}\\ u(x,y)=8e^{-y(3+ \frac{\lambda}{4})}e^{\lambda(x+y)}\\ \text{ Since, it is a boundary value problem. Therefore, when the boundary conditions given then we can calculate the value of lambda and get the solution u(x,y).}$