For the PDE:

$\frac{\partial u}{\partial t} = \frac{\partial ^2u}{\partial x^2}-e^{rx}$ , $0<x<L, t>0$

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

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


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

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 $c_2$?


You can make a conclusion:

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.

  • $\begingroup$ and we find $c_2$ if we're given another initial condition like $$u(x,0)=f(x)_ ,0<x<L$$, correct? $\endgroup$ – MarissaB Feb 22 at 18:22

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