Difficult partial differential equation

Problem $$\Omega\frac{\partial}{\partial t}A(y,t) +6\Lambda\Omega\left(y^2-y\right) \sin(t) = \frac{\partial^2}{\partial y^2}A(y,t)$$ Boundary conditions $$\frac{\partial}{\partial y}A(t,0) = \frac{\partial}{\partial y}A(t,1) = 0$$ Solution $$A(y,t) =6\Lambda \cdot \operatorname{Im}\, \left\{ \left[\frac{i\sinh(\alpha y)}{\alpha}-\frac{i\big(1-\cosh(\alpha)\big)\cosh(\alpha y)}{\alpha\sinh(\alpha)}+i{y}^{2}-iy+2\Omega^{-1}\right]e^{it} \right\}$$ where $$\alpha=\frac{1}{2}(1+i)\sqrt{2\Omega}$$

but I don't know how to get the solution. Maple didn't show anything. please help. • What is the domain of your problem? – DaveNine Dec 23 '17 at 2:43
• @DaveNine I assume it's $0<y<1$, $t >0$ – Dylan Dec 23 '17 at 3:45

The equation is the imaginary part of $$\Omega\frac{\partial Z}{\partial t} + 6\lambda\Omega (y^2-y)e^{it} = \frac{\partial^2Z}{\partial y^2}$$

where $$Z(y,t) = Y(y)e^{it}$$

Then the problem reduces to the ODE

$$Y'' - i\Omega Y = 6\lambda\Omega(y^2-y)$$

This is linear and non-homogeneous, so we can apply the method of undetermined coefficients. Let $Y = Y_h + Y_p$, with homogeneous and particular solutions, respectively.

For $Y_h$, the characteristic polynomial has roots $r = \pm \sqrt{i\Omega} = \pm\sqrt{\frac\Omega2}(1+i) = \pm \alpha$, therefore

$$Y_h = E\cosh (\alpha y) + F\sinh(\alpha y)$$

For $Y_p$, we take the ansatz $Y_p = By^2 + Cy + D$. After substitution $$2B - i\Omega(By^2+Cy+D) = 6\lambda\Omega(y^2-y)$$ Equating coefficients gives $B = -C = 6\lambda i$, $D = 12\lambda\Omega^{-1}$. Therefore your general solution takes the form

$$Y(y) = 6\lambda \left[c_1\cosh(\alpha y) + c_2\sinh(\alpha y) + iy^2 - iy + \frac2\Omega \right]$$

Now you can use the BC $Y'(0) = Y'(1) = 0$ to determine the remaining constants, which are given in the solution as

$$c_1 = -\frac{i(1-\cosh\alpha)}{\alpha\sinh\alpha}, \quad c_2 = \frac{i}{\alpha}$$