Boundary Value Problem: Does it Have a Solution?

[evinda]

Hello! (Wave)

I want to check if the following boundary value problem has a solution

$\left\{\begin{matrix}
-u_{xx}-4u=\sin {2x}, x \in (0,\pi)\\
u(0)=u(\pi)=0
\end{matrix}\right.$

I have thought the following:

We consider the corresponding homogeneous equation $-u_{xx}-4u=0$.

The characteristic polynomial is $-\lambda^2-4=0 \Rightarrow \lambda^2=-4$, contradiction.

Thus the given boundary value problem does not have a solution.

Is is it right? (Thinking)

 

[I like Serena]

Hello! (Wave)

I want to check if the following boundary value problem has a solution

$\left\{\begin{matrix}
-u_{xx}-4u=\sin {2x}, x \in (0,\pi)\\
u(0)=u(\pi)=0
\end{matrix}\right.$

I have thought the following:

We consider the corresponding homogeneous equation $-u_{xx}-4u=0$.

The characteristic polynomial is $-\lambda^2-4=0 \Rightarrow \lambda^2=-4$, contradiction.

Thus the given boundary value problem does not have a solution.

Is is it right? (Thinking)

Hey evinda! ;)

Isn't $\sin 2x$ a solution of the homogeneous equation? (Wondering)

 

[evinda]

Hey evinda! ;)

Isn't $\sin 2x$ a solution of the homogeneous equation? (Wondering)

Yes, that's right. (Nod) So since the homogeneous equation has a non-trivial solution, we deduce that the given boundary value problem has no solution. Right?

 

[I like Serena]

Yes, that's right. (Nod) So since the homogeneous equation has a non-trivial solution, we deduce that the given boundary value problem has no solution. Right?

Not really... (Thinking)

Since the solutions of the characteristic equation are $\lambda=\pm 2i$, it follows that the full solution of the homogeneous equation is:
$$u_h = c_1 e^{2ix} + c_2 e^{-2ix}$$
Or alternatively:
$$u_h = A\sin 2x + B\cos 2x$$

Now we can use for instance the Method of undetermined coefficients or Variation of Parameters to find a particular solution... (Thinking)

... or we can throw the equation at Wolfram... (Emo)

 

[evinda]

Not really... (Thinking)

Since the solutions of the characteristic equation are $\lambda=\pm 2i$, it follows that the full solution of the homogeneous equation is:
$$u_h = c_1 e^{2ix} + c_2 e^{-2ix}$$
Or alternatively:
$$u_h = A\sin 2x + B\cos 2x$$

Now we can use for instance the Method of undetermined coefficients or Variation of Parameters to find a particular solution... (Thinking)

Oh yes, that's right... (Nod) Thanks! (Smirk)

... or we can throw the equation at Wolfram... (Emo)

(Blush)