# How to use the Ritz method with the weak form to approximate solution of differential equation

I am having trouble understanding how to approximate the solution to this problem using the Ritz method and the weak form:

$$\frac{d^2u}{dx^2} - u=0; \ \ x \in [0,1]$$ $$u(x=0)=0; \ \ \frac{du}{dx} \bigg|_{x=1} = 20$$

We multiply the strong form of the equation by a weight function $$w$$ and integrate over the domain by parts to get

$$\int_0^1w\left[\frac{d^2u}{dx^2} - u \right] dx = \int_0^1 w \frac{d^2 u}{dx^2} dx - \int_0^1wu \ dx$$

$$\int_0^1 \frac{du}{dx} \frac{dw}{dx} dx - \int_0^1 wu \ dx - 20 w(1) = 0$$

which is the weak form. Now, I am confused because when I try to change this into a matrix equation, I practically try to solve

$$\int_0^1 \frac{du}{dx} \frac{dw}{dx} dx = \int_0^1 wu \ dx + 20 w(1)$$

If I try to use, say $$u\approx\hat{u}=a_1 \phi_1 + a_2 \phi_2$$

I should be able to form a linear problem $$\mathbf{K} \vec{a} = \vec{b}$$

where $$\mathbf{K}$$ is a $$2 \times 2$$ matrix obtained from the left hand side of the equation where $$w \to \phi_i$$ and $$u \to \phi_j$$ to get the entries $$K_{ij}$$, but this substitution would not work to obtain the components of $$\vec{b}$$ since I this substitution gives me terms of $$\phi_i$$ and $$\phi_j$$ and this makes no sense since $$\vec{b}$$ is indexed only by one subindex. I know I am not understanding something, but this was my professor's explanation and I am confused.

Thank you so much for your help!

Everything in your OP looks correct.

I think you have just got into a muddle with the notation, and what is known and what is unknown.

$$\phi_1$$ and $$\phi_2$$ are known functions, i.e. you choose two sensible functions for the problem you want to solve, to give a "good" approximation to the solution you expect.

You can differentiate them with respect to $$x$$. Let $$\dfrac{d\phi_i}{dx} = \phi'_i$$.

If you let $$w = \phi_1$$, your integral equation becomes $$\int_0^1 (a_1 \phi'_1 + a_2 \phi'_2)\phi'_1\,dx - \int_0^1(a_1\phi_1 + a_2\phi_2)\phi_1\,dx = 20 \phi_1(1).$$ You can evaluate the integrals $$\int\phi_1^2$$, $$\int \phi_2\phi_1$$, $$\int {\phi'_1}^2$$, and $$\int \phi'_2\phi'_1$$ (either analytically or numerically) since the functions $$\phi_i$$ and $$\phi'_i$$ are known.

You then get an equation of the form $$K_{11} a_1 + K_{12} a_2 = b_1$$ where you know the $$K$$ and $$b$$ terms.

Similarly letting $$w = \phi_2$$ gives you the equation $$K_{21} a_1 + K_{22} a_2 = b_2.$$

These are the two rows of the matrix equation $$\mathbf{K}\vec a = \vec b$$.

• Oh I see how it works now! I was very confused with my professor's notes. Thank you so much! – The Bosco Feb 12 '20 at 22:08