Solve the steady state problem $\frac{\partial u}{\partial t} = \frac{\partial ^2u}{\partial x^2}-e^{rx}$

For the PDE:

$$\frac{\partial u}{\partial t} = \frac{\partial ^2u}{\partial x^2}-e^{rx}$$ , $$00$$

together with the boundary conditions ($$r$$ and $$\alpha$$ are constants)

$$\frac{\partial u}{\partial x}(0,t)=0$$ , $$\frac{\partial u}{\partial x}(L,t)=\alpha, t>0$$

I'm trying to find the relationship between $$r,\alpha$$, and $$L$$ that have to hold in order for a solution to exist.

Since this is a steady-state problem, and the PDE does not depend on time, then

$$\frac{\partial u}{\partial t}=0$$, which gives $$\frac{d ^2u}{d x^2}=e^{rx}$$

Integrating with respect to $$x$$ gives: $$\frac{du}{dx}=\frac{1}{r} e^{rx}+c_1$$

, and integrating with respect to $$x$$ again gives: $$u(x)=\frac{1}{r^2}e^{rx}+c_1x+c_2$$

What confuses me is how to apply the boundary conditions, which are essentially

$$\frac{du}{dx}(0)=0$$$$\frac{du}{dx}(L)=\alpha$$

So going back to the first derivative to apply the conditions, would it be correct to say that

$$\frac{du}{dx}(0)=\frac{1}{r}e^{r(0)}+c_1=0 \implies \frac{1}{r}+c_1=0 \implies c_1=- \frac{1}{r}$$

and

$$\frac{du}{dx}(L)=\frac{1}{r}e^{r(L)}+c_1=\alpha \implies c_1=\alpha -\frac{1}{r}e^{rL}$$

which implies that

$$- \frac{1}{r} = \alpha -\frac{1}{r}e^{rL}$$

must hold in order for a solution to exist?

Then what exactly is the constant $$c_1$$ and how do you find $$c_2$$?

If $$r\alpha+1=e^{rL}$$ steady-state solution is $$u(x)=\frac{1}{r^2}e^{rx}-\frac{x}{r}+c_2.$$ Here $$c_2$$ is any constant.
• and we find $c_2$ if we're given another initial condition like $$u(x,0)=f(x)_ ,0<x<L$$, correct? – MarissaB Feb 22 at 18:22