Solving a PDE ( wave equation ) $$u_{tt}- \Delta u= e^{t}, 0<x<2\pi, 0<y<1, t>0$$
$$u_x(0,y,t)=u_x(2\pi,y,t)=0, 0<x<2\pi, t>0$$
$$u_y(x,0,t)=u_y(x,1,t)=0, 0<x<2\pi, t>0$$
$$u(x,y,0)=0, 0<x<2\pi, 0<y<1$$
$$u_t(x,y,0)= 1+\cos(x), 0<x<2\pi, 0<y<1$$
Can anyone please help me? I am learning PDE and I have no idea how to solve this problem.
Thanks.
 A: The kind of wave equation you are dealing with, is non-homogeneous. We first try to solve the homogeneous wave equation $u_{tt}=\Delta u={\partial^2u\over \partial^2 x}+{\partial^2u\over \partial^2 y}$ given initial condition using Separation of Variables Method. We first define $u(x,y,t)=f_1(x)f_2(y)g(t)$. Then by substituting we obtain$$u_{tt}=f_1(x)f_2(y)g''(t)\\\Delta u={\partial^2u\over \partial^2 x}+{\partial^2u\over \partial^2 y}=f_1''(x)f_2(y)g(t)+f_1(x)f_2''(y)g(t)\\\implies\\f_1(x)f_2(y)g''(t)=f_1''(x)f_2(y)g(t)+f_1(x)f_2''(y)g(t)\\\implies \\{g''(t)\over g(t)}={f_1''(x)\over f_1(x)}+{f_2''(y)\over f_2(y)}$$The above equation states that a summation of three functions of pure $x,y,t$ is zero. This is possible only if they are all constants, i.e.$${g''(t)\over g(t)}=k\\{f_1''(x)\over f_1(x)}=k_1\\{f_2''(y)\over f_2(y)}=k_2\\k=k_1+k_2$$Now, let's turn back to our own equation. As the non-homogeneous term on RHS (i.e. $e^t$) is a function of $t$, not $x,y$ we neglect it for now (there is a very simple way to add it later). The most important thing now is to determine the signs of constants $k,k_1,k_2$, which has a dramatic effect on out answer, for example if $k$ is positive, the answer of the first equation takes an exponential form and with $k<0$ we attain to a sinusoidal shape. The case $k=0$ has no matter of significance, since it leads to linear unbounded answer $at+b$ for constants $a,b$. Furthermore, since the constraints are bounded along $x$ and $y$, we argue that our final answer should have a sinusoidal form along $x$ and $y$ leading to $k_1,k_2<0$. Define $$k_1\triangleq-\lambda_1^2\\k_2\triangleq-\lambda_2^2$$therefore $$k=-\lambda_1^2-\lambda_2^2=-\lambda^2$$and our partial answers for $f_1,f_2,g$ become:$$f_1(x)=a_1\cos \lambda_1 x+a_2\sin \lambda_1 x\\f_2(y)=b_1\cos \lambda_2 y+b_2\sin \lambda_2 y\\g(t)=c_1\cos \lambda t+c_2\sin \lambda t$$therefore$$u_0(x,y,t;a_1,a_2,b_1,b_2,c_1,c_2)=\\(a_1\cos \lambda_1 x+a_2\sin \lambda_1 x)\cdot(b_1\cos \lambda_2 y+b_2\sin \lambda_2 y)\cdot(c_1\cos \lambda t+c_2\sin \lambda t)$$but we are not finished yet! Since the equation is linear, any combination of the partial answers is an answer itself (that's why we called the answers found as partial beforehand), finally$$\hat u(x,y,t)=\sum_{n=0}^{\infty}(a_{1n}\cos \lambda_{1n} x+a_{2n}\sin \lambda_{1n} x)\cdot(b_{1n}\cos \lambda_{1n} y+b_{2n}\sin \lambda_{2n} y)\cdot(c_{1n}\cos \lambda_n t+c_{2n}\sin \lambda_n t)$$But why changing $\lambda$ to $\lambda_n$? To include higher harmonics of the partial answers!

Final kicking off
We solved the equation when it was homogeneous. To include $e^t$ in
  our answer, note that if $u(x,y,t)$ is an answer to the homogeneous
  equation, then $u(x,y,t)+e^t+at+b$ is an answer to the main,
  non-homogeneous wave equation. Therefore
  $$u(x,y,t)=\\e^t+C_1t+C_2+\sum_{n=1}^{\infty}(a_{1n}\cos \lambda_{1n}
 x+a_{2n}\sin \lambda_{1n} x)\cdot(b_{1n}\cos \lambda_{2n} y+b_{2n}\sin
 \lambda_{2n} y)\cdot(c_{1n}\cos \lambda_n t+c_{2n}\sin \lambda_n t)$$

Applying initial condition
Note that$$u_x(x,y,t)=\sum_{n=1}^{\infty}(-a_{1n}\lambda_{1n}\sin \lambda_{1n}
 x+a_{2n}\lambda_{1n}\cos \lambda_{1n} x)\cdot(b_{1n}\cos \lambda_{1n} y+b_{2n}\sin
 \lambda_{2n} y)\cdot(c_{1n}\cos \lambda_n t+c_{2n}\sin \lambda_n t)$$ and $$u_y(x,y,t)=\sum_{n=1}^{\infty}(a_{1n}\cos \lambda_{1n}
 x+a_{2n}\sin \lambda_{1n} x)\cdot(-b_{1n}\lambda_{2n}\sin \lambda_{2n} y+b_{2n}\lambda_{2n}\cos
 \lambda_{2n} y)\cdot(c_{1n}\cos \lambda_n t+c_{2n}\sin \lambda_n t)$$Now by throwing $x=0$ and $x=2\pi$ in $u_x(x,y,t)$ we obtain$$x=0\implies a_{2n}=0\\x=2\pi\implies \lambda_{1n}={n\over 2}$$similarly from $u_y(x,y,t)$ we obtain$$y=0\implies b_{2n}=0\\y=1\implies \lambda_{2n}={n\pi}$$also $u(x,y,0)=0$ therefore$$c_{1n}=0\\C_2=-1$$and finally$$u(x,y,t)=e^t-1+C_2t+\sum_{n=1}^{\infty}d_n\cos{\lambda_{x,n}}x\cos{\lambda_{y,n}}y\sin{\lambda_{n}}t$$Now from our last condition we have$$u_t(x,y,0)=1+C_2+\sum_{n=1}^{\infty}{\lambda_{n}}d_n\cos{\lambda_{x,n}}x\cos{\lambda_{y,n}}y=1+\cos x$$which yields to $$C_2=0\\\lambda_n=\lambda_{x,n}=1\\d_n=1\\\lambda_{y,n}=0$$

Final answer$$u(x,y,t)=e^t-1+\cos x\sin t$$

