# Initial-boundary value problem for PDE

I need a little help with solving IBVP for hyperbolic and parabolic equations like these: $$hyperbolic: \left\{\begin{matrix} \frac{\partial^2u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}+1\\ u(0,t) =u(1, t) =0\\ u(x,0) = 0\\ u_t(x,0)=x \end{matrix}\right. \\parabolic: \left\{\begin{matrix} 3\frac{\partial u}{\partial t}=4\frac{\partial^2 u}{\partial x^2}\\ u(0,t) =u(5, t) =0\\ u(x,0) = x \end{matrix}\right.$$ Don't even know where to start.

I'm not actually asking for full solution, but would appreciate it if you could give me direction. Links to tutorials and examples are also highly welcomed.

Update:

Second equation looks like 1 dimensional heat conduction equation: $u_t = c^2u_{xx}$ with $c=\frac{2\sqrt{3}}{3}$.

Ok, for the second one method of separation of variables could be applied.

We assume that $u$ can be written as a product of single variable functions of each independent variable, $u(x, t) = X (x)T (t)$. Substituting this guess into the heat equation, we find that $XT′ =c^2X′′T$. Dividing both sides by $c^2$ and $u = XT$, we then get $\frac{1}{c^2}\frac{T'}{T} = \frac{X''}{X} = \lambda$. This leads to two equations: $$T′ =c^2\lambda T\\ X'' = \lambda X$$ giving $$T(t) = Ae^{c^2λt}\\ X(x) = c_1e^{\sqrt{\lambda}x} + c_2e^{\sqrt{-\lambda}x}$$ The aim is to force our product solutions to satisfy both the boundary conditions and initial conditions.

Is it possible to use this method for the first equation?

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And the first one is a nonhomogeneous wave equation. Both problems can be solved by eigenfunction expansion method (see, e.g., Farlow, Partial differential equations for scientists and engineers) –  Artem May 13 '12 at 16:34
Yes, the same method works. You first need to find the eigenfunctions of the corresponding homogeneous problem (let me denote them as $e_i(x)$, these are $X(x)$ in your solution, there is a countable number of them), then you represent your non-homogeneous term (in your case $f(x,t)=1$) as a series using the found eigenfunctions $f(x,t)=\sum f_i(t)e_i(x)$, and then look for the solution in the form $u(x,t)=\sum T_i(t)e_i(x)$ (just plug this expression in your equation). For the unknown functions $T_i(t)$ you'll get differential equations which take into account $f_i(t)$.