# Wave equation of $2-$ dimensional

Consider the Wave equation in $$2-$$ dimensional $$u_{tt}=u_{xx}+u_{yy}, (x,y)\in D,t>0\\u(x,y,0)=f(x),u_t(x,y,0)=g(x),(x,y)\in D\\ u(x,y,t)=0 , (x,y)\in \partial D,t>0$$

then show that if the solution exists then it unique

my idea is::-

suppose $$w=u_1-u_2$$ where $$u_1$$ and $$u_2$$ are two solutions then

then

$$w_{tt}=w_{xx}+w_{yy}\\w(x,y,0)=0,w_t(x,y,0)=0\\ w(x,y,t)=0$$

consider $$E(t)=\int \int _D (w_t^2+u_x^2+w_y^2)dxdy$$

then how to we prove that remaining part that i,.e

$$E(0)=0$$ and $$E=constant$$ (how to prove)

So I think you might have some errors. So $$u(x,y,t) = 0$$ will not make sense unless both $$f(x) = g(x) = 0$$. Also, are you solving over the whole of $$\mathbb{R}^n$$? Are you just solving over some domain $$D$$ and the boundary data is specified just for that domain? I'll assume now that you are trying to solve the 2D wave equation $$u_{tt} = \Delta u$$ with $$u = f$$ on $$\partial D$$ and $$u_t = g$$ on $$D$$ at $$t=0$$.

With this, let's take you idea to consider two solutions $$u_1,u_2$$ and look at $$w = u_2 - u_1$$ and define

$$E(t) = \iint_D w_t^2 + |\nabla w|^2 dx dy$$

So differentiating under the integral sign we have

$$E'(t) = 2\iint_D w_tw_{tt} + \nabla w \cdot \nabla w_t dxdy$$

An important identity is Green's First Identity which tells

$$\int_{\Omega} \nabla \psi \cdot \nabla \phi dV = \int_{\partial \Omega} \psi \nabla \phi \cdot \nu dS - \int_{\Omega} \psi \Delta \phi dV$$

where $$\Omega \subset \mathbb{R}^n$$ and $$\nu$$ is the outward surface normal. The $$dS$$ integral is a surface integral. Thus from this we have

$$\iint_D \nabla w \cdot \nabla w_t dxdy = \iint_{\partial D} w_t \nabla w \cdot \nu dS - \iint_D w_t \Delta w dxdy$$

(so we have taken $$\psi = w_t$$ and $$\phi = w$$). The boundary term will disappear since $$w = 0$$ on $$\partial D$$ and so $$w_t = 0$$ on $$\partial D$$ so we have

$$E'(t) = 2\iint_D w_tw_{tt} - w_t \Delta w dx = 2 \iint_D w_t(w_{tt} - \Delta w) dxdy = 0$$

So we see that $$E(t)$$ is constant and

$$E(0) = \iint_D w_t^2(x,y,0) + w_{x}^2(x,y,0) + w_{y}^2(x,y,0) dxdy$$

We have $$w(x,y,0) = 0$$ thus, $$w_x(x,y,0) = w_y(x,y,0) = 0$$ and also we have $$w_t(x,y,0) = 0$$ thus, $$E(0) = 0$$. So we conclude that $$w_t = \nabla w = 0$$ and so we conclude $$w = 0$$ on $$D$$ and therefore $$u_1 = u_2$$.

• ...here what i mean is $u(x,y,t)=0$ on $\partial D$ Nov 28 '18 at 3:39
• Ah I see. I think what I've written is sufficient then. Nov 28 '18 at 4:35
• .....Any way thanks its really nice explanation Nov 28 '18 at 4:45